The pH during the titration 20.00 m L of 0.1000 M trimethylamine, ( K b =5 .2 × 10 − 4 ) with 0 .1000 M HCl solution after the addition of 0 mL titrant should be calculated. Concept introduction: pH : pH is the logarithm of the reciprocal of the concentration of H 3 O + in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions. pH = pK a + log [conjugate base] [acid] is Henderson-Hasselbalch equation. Henderson-Hasselbalch equation pKa = pH + log [ HA ] [ A - ] Henderson-Hasselbalch equation explains the relationship between p H of solution and p K a of acid. For a dissociation of acid ( H A ) in aqueous solution, H A + H 2 O ⇌ H 3 O + + A − p K a = p H + log [ H A ] [ A − ] During a dissociation of acid in aqueous solution, If p H = p K a , the concentration of compound in its acidic and basic form is equal. If p H < p K a , the compound exist in its acidic form. If p H > p K a , the compound exist in its basic form.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

presented by Morallen Lig Intermine the hand product for the given mution by adding atoms, bonds, nonhonding diarion panda скуль Step 3: Comp the draw the product Step 2: Agama workup Compithe 429 ملولة

Reaction A 0,0

presented by Morillon Leaning Predict the organic product for the min кусур HSC Adithane carved arnown to come than that to the condon slchroruis in acid in in aquishri with ною

Answer 2

Question

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Answer 6

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

Answer 7

Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $0 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 8

(a)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $0.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $11.86$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

$0 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

When a weak base reacts with a strong acid, it dissociates completely and behaves like a strong base.

Therefore, the amount of base reacts with equal amount of $H+$

Added amount of hydrogen ions $H+$ is as follows,

Initial number of moles of trimethylamine $(CH3CH2)3N$ ,

$= mole(M) × Volume (V) =(0.1000 mol (CH3CH2)3N/L)(10−3 L/1 mL)(20.00 mL)= 2.00×10-3 mol (CH3CH2)3N$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Ka = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = [x][x][0.1000−x]=x2[0.1000]Solve for (x),[OH−] = x= 7.2111×10-3 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(7.2111×10-3 )pOH= 2.141998$

Therefore, the given solution $pOH$ value is $2.14$

Calculation for $pH$

$pH =14.00−pOH= 14.00−2.141998=11.8580=11.86$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $10 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 15

(b)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.72$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Determine $10 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(10.00 mL)= 1.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.000×10−3 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.000×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03000 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.000×10−3 mol (CH3CH2)3N)/(0.03000 L)= 0.03333 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.000×10−3 mol (CH3CH2)3NH+)/(0.03000 L)= 0.03333 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0333+x)(0.0333−x)=x(0.03333)0.03333Solve for (x),[OH−] = x= 5.2×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(5.2×10-4 )pOH= 3.283997$

Therefore, the given solution $pOH$ value is $3.28$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.283997=10.7160=10.72$

Answer 20

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

Answer 21

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $15 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 22

(c)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $15.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $10.24.$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Determine $15 mL of HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(15.00 mL)= 1.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.500×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 10.00) mL] (10−3 L/1 mL)= 0.03500 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×10−4 mol (CH3CH2)3N)/(0.03500 L)= 0.0142857 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.500×10−3 mol (CH3CH2)3NH+)/(0.03500 L)= 0.0428571 M$

Since no acid has been added, only the weak base $(Kb)$ is important,

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0428571+x)(0.0142857−x)=x(0.0428571)0.0142857Solve for (x),[OH−] = x= 1.7333×10-4 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.7333×10-4 )pOH= 3.761126$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$pH =14.00−pOH= 14.00−3.761126=10.23887=10.24$

Answer 27

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

Answer 28

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 29

(d)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $9.44$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.00 mL)= 1.900×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $1.00×10−4 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.900×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.00) mL] (10−3 L/1 mL)= 0.03900 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(1.00×10−4 mol (CH3CH2)3N)/(0.03900 L)= 0.002564103 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.900×10−3 mol (CH3CH2)3NH+)/(0.03900 L)= 0.0487179 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0487179+x)(0.002564103−x)=x(0.0487179)0.002564103Solve for (x),[OH−] = x= 2.73684×10-5 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(2.73684×10-5 )pOH= 4.56275$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$pH =14.00−pOH= 14.00−4.56275=9.43725=9.44$

Answer 34

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

Answer 35

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $19.95 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 36

(e)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $8.1$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Determine for $19.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(19.95 mL)= 1.995×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $5.00×10−46 mol (CH3CH2)3N$ will remain, an equal number of moles of $1.995×10−3 mol (CH3CH2)3NH+$ will form.

The volume of the solution at this point is,

Volume

$= [(20.00 + 19.95) mL] (10−3 L/1 mL)= 0.03995 L$

Molarity of the excess $(CH3CH2)3N$ is,

$=(5.00×106 mol (CH3CH2)3N)/(0.03995 L)= 0.000125156 M$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(1.995×10−3 mol (CH3CH2)3NH+)/(0.03995 L)= 0.0499374 M$

Determination the base ion concentration from the $Kb$ and then determine the $pOH$ from the $pH$ .

$Kb = 5.2×10−4 = [(CH3CH2)3NH+][OH−][(CH3CH2)3N] = x(0.0499374+x)(0.000125156−x)=x(0.0499374)0.000125156Solve for (x),[OH−] = x= 1.303254×10-6 M$

Calculation for $pOH$

$pOH =−log[OH−]=−log(1.303254×10-6 )pOH= 5.88497$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$pH =14.00−pOH= 14.00−5.88497=8.11503=8.1$

Answer 41

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

Answer 42

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 43

(f)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $6.01$

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Determine for $20.00 mL HCl$ is added:

Trimethylamine is a weak base, when $HCl$ is added the $H+$ concentration is calculated using the following formula.

$pH = pKa + log[conjugate base][acid]$

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.00 mL)= 2.000×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $2.000×10−3 mol of (CH3CH2)3NH+$ will form. This is the equivalent point.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.00) mL] (10−3 L/1 mL)= 0.04000 L$

Molarity of the excess $(CH3CH2)3NH+$ formed is,

$=(2.000×10−3 mol (CH3CH2)3NH+)/(0.04000 L)= 0.05000 M$

Calculate $Ka$ for $(CH3CH2)3NH+$

$Kb = KW/Ka= (1.0×10−14)(5.2×10−4) = 1.9231×10−11$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

$Kb = 1.9231×10−11 = [H3O+][(CH3CH2)3N][(CH3CH2)3NH+] = [x][x](0.05000−x)=[x][x][0.05000]Solve for (x),[H3O+] = x= 9.80587×10-7 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(9.80587×10-7 )pH= 6.0085$

The given solution $pH$ value is $6.01$

Answer 48

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

Answer 49

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $20.05 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 50

(g)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $20.05 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $3.90.$

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Determine for $20.05 mL HCl$ is added:

After the equivalence point, the excess strong acid is the primary factor influencing the $pH$ .

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(20.05 mL)= 2.005×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.0×10−6 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 20.05) mL] (10−3 L/1 mL)= 0.04005 L$

Molarity of the excess $H3O+$ is,

$=(5.0×10−6 mol H3O+)/(0.04005 L)= 1.248×10−4 M$

Determination the base ion concentration from the $Ka$ and then determine the $pH$ from the $pOH$ .

Calculation for $pH$

$pH =−log[H3O+]=−log(1.248×10−4 )pH= 3.9036$

The given solution $pH$ value is $3.90$

Answer 55

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$

Explanation of Solution

Determine for $25.00 mL HCl$ is added:

Determine the moles of $HCl$ is added,

$= mole(M) × Volume (V) =(0.1000 mol HCl/L)(10−3 L/1 mL)(25.00 mL)= 2.500×10-3 mol HCl$

The $HCl$ will react with an equal amount of the base, and $0 mol (CH3CH2)3N$ will remain, and $5.00×10−4 mol of (CH3CH2)3NH+$ will be in excess.

The volume of the solution at this point is,

Volume

$= [(20.00 + 25.00) mL] (10−3 L/1 mL)= 0.04500 L$

Molarity of the excess $H3O+$ is,

$=(5.00×10−4 mol H3O+)/(0.04500 L)= 1.1111×10−2 M$

Calculation for $pH$

$pH =−log[H3O+]=−log(1.1111×10−2 )pH= 1.9542$

The given solution $pH$ value is $1.95.$

Answer 56

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00 mL$ of 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ solution after the addition of $25.00 mL$ titrant should be calculated.

Concept introduction:

$pH$ : $pH$ is the logarithm of the reciprocal of the concentration of $H3O+$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $pH$ due to more hydrogen ions and higher $pH$ due to less concentration of hydrogen ions.

$pH = pKa + log[conjugate base][acid]$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$pKa = pH + log [HA][A-]$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $pKa$ of acid. For a dissociation of acid $(HA)$ in aqueous solution,

$HA + H2O ⇌ H3O+ + A−$

$pKa = pH + log[HA][A−]$

During a dissociation of acid in aqueous solution,

If $pH = pKa$ , the concentration of compound in its acidic and basic form is equal.

If $pH < pKa$ , the compound exist in its acidic form.

If $pH > pKa$ , the compound exist in its basic form.

Answer 57

(h)

Expert Solution

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $10.00 mL$ titrant and addition of $25.00 mL$ and 0.1000 M trimethylamine, ( $Kb=5.2×10−4$ ) with $0.1000 M HCl$ is $1.95$