Chapter 19, Problem 19.52P

(a)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{0 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{0}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $11.86$

Explanation of Solution

$0mLHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

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 ${\left(C{H}_{3}C{H}_2\right)}_{3}N$,

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000mol{\left(C{H}_{3}C{H}_2\right)}_{3}N/L\right)\left({10}^{-3}L/1mL\right)\left(20.00mL\right)\\ =2.00\times 10^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N\end{array}$ 

Since no acid has been added, only the weak base $\left(K_b\right)$ is important,

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

$\begin{array}{l}{K}_a=5.2\times {10}^{-4}=\frac{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}=\frac{\left[x\right]\left[x\right]}{\left[0.1000-x\right]}=\frac{x^2}{\left[0.1000\right]}\\ Solvefor\left(x\right),\\ \left[O{H}^{-}\right]=x=7.2111\times 10^{-3}M\end{array}$

Calculation for $pOH$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(\text{7}{\text{.2111×10}}^{\text{-3}}\right)\\ pOH=2.141998\end{array}$

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

Calculation for $pH$

$\begin{array}{c}pH=14.00-pOH\\ =14.00-2.141998\\ =11.8580\\ =11.86\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{10}\text{mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $10.72$

Explanation of Solution

Determine $10mLofHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(10.00mL\right)\\ =1.000\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $1.000\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, an equal number of moles of $1.000\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will form.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+10.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.03000L\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N$is,

$\begin{array}{l}=\left(1.000\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N\right)/\left(0.03000L\right)\\ =0.03333M\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$formed is,

$\begin{array}{l}=\left(1.000\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right)/\left(0.03000L\right)\\ =0.03333M\end{array}$

Since no acid has been added, only the weak base $\left(K_b\right)$ is important,

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

$\begin{array}{l}{K}_b=5.2\times {10}^{-4}=\frac{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}=\frac{x\left(0.0333+x\right)}{\left(0.0333-x\right)}=\frac{x\left(0.03333\right)}{0.03333}\\ Solvefor\left(x\right),\\ \left[O{H}^{-}\right]=x=5.2\times 10^{-4}M\end{array}$

Calculation for $pOH$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(\text{5}{\text{.2×10}}^{\text{-4}}\right)\\ pOH=3.283997\end{array}$

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

Calculation for $pH$

$\begin{array}{c}pH=14.00-pOH\\ =14.00-3.283997\\ =10.7160\\ =10.72\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{15 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $15.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $10.24.$

Explanation of Solution

Determine $15mLofHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ \\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(15.00mL\right)\\ \\ =1.500\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $5.00\times {10}^{-4}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, an equal number of moles of $1.500\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will form.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+10.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.03500L\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N$is,

$\begin{array}{l}=\left(5.00\times {10}^{-4}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N\right)/\left(0.03500L\right)\\ =0.0142857M\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$formed is,

$\begin{array}{l}=\left(1.500\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right)/\left(0.03500L\right)\\ =0.0428571M\end{array}$

Since no acid has been added, only the weak base $\left(K_b\right)$ is important,

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

$\begin{array}{l}{K}_b=5.2\times {10}^{-4}=\frac{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}=\frac{x\left(0.0428571+x\right)}{\left(0.0142857-x\right)}=\frac{x\left(0.0428571\right)}{0.0142857}\\ Solvefor\left(x\right),\\ \left[O{H}^{-}\right]=x=1.7333\times 10^{-4}M\end{array}$

Calculation for $pOH$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(\text{1}{\text{.7333×10}}^{\text{-4}}\right)\\ pOH=3.761126\end{array}$

The given solution $pOH$ value is $3.76$

Calculation for $pH$

$\begin{array}{c}pH=14.00-pOH\\ =14.00-3.761126\\ =10.23887\\ =10.24\end{array}$

(d)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{19 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $19.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $9.44$

Explanation of Solution

Determine for $19.00mLHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(19.00mL\right)\\ =1.900\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $1.00\times {10}^{-4}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, an equal number of moles of $1.900\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will form.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+19.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.03900L\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N$is,

$\begin{array}{l}=\left(1.00\times {10}^{-4}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N\right)/\left(0.03900L\right)\\ =0.002564103M\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$formed is,

$\begin{array}{l}=\left(1.900\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right)/\left(0.03900L\right)\\ =0.0487179M\end{array}$

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

$\begin{array}{l}{K}_b=5.2\times {10}^{-4}=\frac{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}=\frac{x\left(0.0487179+x\right)}{\left(0.002564103-x\right)}=\frac{x\left(0.0487179\right)}{0.002564103}\\ Solvefor\left(x\right),\\ \left[O{H}^{-}\right]=x=2.73684\times 10^{-5}M\end{array}$

Calculation for $pOH$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(\text{2}{\text{.73684×10}}^{\text{-5}}\right)\\ pOH=4.56275\end{array}$

The given solution $pOH$ value is $4.57$

Calculation for $pH$

$\begin{array}{c}pH=14.00-pOH\\ =14.00-4.56275\\ =9.43725\\ =9.44\end{array}$

(e)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{19}\text{.95 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $19.95mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $8.1$

Explanation of Solution

Determine for $19.00mLHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(19.95mL\right)\\ =1.995\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $5.00\times {10}^{-46}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, an equal number of moles of $1.995\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will form.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+19.95\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.03995L\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N$is,

$\begin{array}{l}=\left(5.00\times 106mol{\left(C{H}_{3}C{H}_2\right)}_{3}N\right)/\left(0.03995L\right)\\ =0.000125156M\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$formed is,

$\begin{array}{l}=\left(1.995\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right)/\left(0.03995L\right)\\ =0.0499374M\end{array}$

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

$\begin{array}{l}{K}_b=5.2\times {10}^{-4}=\frac{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}=\frac{x\left(0.0499374+x\right)}{\left(0.000125156-x\right)}=\frac{x\left(0.0499374\right)}{0.000125156}\\ Solvefor\left(x\right),\\ \left[O{H}^{-}\right]=x=1.303254\times 10^{-6}M\end{array}$

Calculation for $pOH$

$\begin{array}{c}pOH=-\log \left[O{H}^{-}\right]\\ =-\log \left(\text{1}{\text{.303254×10}}^{\text{-6}}\right)\\ pOH=5.88497\end{array}$

The given solution $pOH$ value is $5.90$

Calculation for $pH$

$\begin{array}{c}pH=14.00-pOH\\ =14.00-5.88497\\ =8.11503\\ =8.1\end{array}$

(f)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{20}\text{.00 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $20.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $6.01$

Explanation of Solution

Determine for $20.00mLHCl$is added:

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

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ \\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(20.00mL\right)\\ \\ =2.000\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $0mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, and $2.000\times {10}^{-3}molof{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will form. This is the equivalent point. 

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+20.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.04000L\end{array}$

Molarity of the excess ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$formed is,

$\begin{array}{l}=\left(2.000\times {10}^{-3}mol{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right)/\left(0.04000L\right)\\ =0.05000M\end{array}$

Calculate$K_a$for ${\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$

$K_b=K_W/K_a=\frac{\left(1.0\times {10}^{-14}\right)}{\left(5.2\times {10}^{-4}\right)}=1.9231\times {10}^{-11}$

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

$\begin{array}{l}{K}_b=1.9231\times {10}^{-11}=\frac{\left[H_3{O}^{+}\right]\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N\right]}{\left[{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}\right]}=\frac{\left[x\right]\left[x\right]}{\left(0.05000-x\right)}=\frac{\left[x\right]\left[x\right]}{\left[0.05000\right]}\\ Solvefor\left(x\right),\\ \left[H_3{O}^{+}\right]=x=9.80587\times 10^{-7}M\end{array}$

Calculation for $pH$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(\text{9}{\text{.80587×10}}^{\text{-7}}\right)\\ pH=6.0085\end{array}$

The given solution $pH$ value is $6.01$

(g)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{20}\text{.05 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $20.05mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $3.90.$

Explanation of Solution

Determine for $20.05mLHCl$is added:

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

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(20.05mL\right)\\ =2.005\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $0mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, and $5.0\times {10}^{-6}molof{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will be in excess.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+20.05\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.04005L\end{array}$

Molarity of the excess $H_3{O}^{+}$ is,

$\begin{array}{l}=\left(5.0\times {10}^{-6}mol{H}_3{O}^{+}\right)/\left(0.04005L\right)\\ =1.248\times {10}^{-4}M\end{array}$

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

Calculation for $pH$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(1.248\times {10}^{-4}\right)\\ pH=3.9036\end{array}$

The given solution $pH$ value is $3.90$

(h)

Interpretation Introduction

Interpretation:

The pH during the titration $20.00mL$ of 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ solution after the addition of $\text{25}\text{.00 mL}$ titrant should be calculated.

Concept introduction:

$\text{pH}$: $\text{pH}$ is the logarithm of the reciprocal of the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower $\text{pH}$ due to more hydrogen ions and higher $\text{pH}$ due to less concentration of hydrogen ions.

$\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\text{[conjugate base]}}{\text{[acid]}}$

It is a Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

$\text{pKa}\text{=}\text{pH}\text{+}\text{log}\frac{\left[\text{HA}\right]}{\left[{\text{A}}^{\text{-}}\right]}$

Henderson-Hasselbalch equation explains the relationship between $pH$ of solution and $p{K}_a$ of acid. For a dissociation of acid $\left(HA\right)$ in aqueous solution,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$p{K}_a=pH+\log \frac{\left[HA\right]}{\left[A^{-}\right]}$

During a dissociation of acid in aqueous solution,

• If $pH=p{K}_a$, the concentration of compound in its acidic and basic form is equal.
• If $pH<p{K}_a$, the compound exist in its acidic form.
• If $pH>p{K}_a$, the compound exist in its basic form.

Answer to Problem 19.52P

The pH of a given buffer solution after the addition $\text{10}\text{.00 mL}$ titrant and addition of $25.00mL$ and 0.1000 M trimethylamine, ($K_b\text{=5}\text{.2}\times {10}^{-4}$ ) with $\text{0}\text{.1000 M HCl}$ is $1.95$

Explanation of Solution

Determine for $25.00mLHCl$is added:

Determine the moles of $HCl$ is added, 

$\begin{array}{l}=mole\left(M\right)\times Volume\left(V\right)\\ =\left(0.1000molHCl/L\right)\left({10}^{-3}L/1mL\right)\left(25.00mL\right)\\ =2.500\times 10^{-3}molHCl\end{array}$ 

The $HCl$ will react with an equal amount of the base, and $0mol{\left(C{H}_{3}C{H}_2\right)}_{3}N$ will remain, and $5.00\times {10}^{-4}molof{\left(C{H}_{3}C{H}_2\right)}_{3}N{H}^{+}$ will be in excess.

The volume of the solution at this point is,

Volume

 $\begin{array}{l}=\left[\left(20.00+25.00\right)mL\right]\left({10}^{-3}L/1mL\right)\\ =0.04500L\end{array}$

Molarity of the excess $H_3{O}^{+}$ is,

$\begin{array}{l}=\left(5.00\times {10}^{-4}mol{H}_3{O}^{+}\right)/\left(0.04500L\right)\\ =1.1111\times {10}^{-2}M\end{array}$

Calculation for $pH$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(1.1111\times {10}^{-2}\right)\\ pH=1.9542\end{array}$

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