Chapter 8, Problem 167AE

(a)

Interpretation Introduction

Interpretation:

The initial pH value of the weak base before adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

The pH of the initial weak base solution is 9.67.

Explanation of Solution

Initial pH of the analyte solution; $HON{H}_2$ is a weak base that forms an equilibrium when dissolved in water. The equilibrium is as follows.

  $HON{H}_2+H_{2}O\rightleftharpoons HON{H}_3{}^{+}+O{H}^{-}$

The amount of weak base at the beginning $=0.2mol/L\times 100.0\times {10}^{-3}L=20\times {10}^{-3}mol$ . By constructing an ICE table, the amount of lactate ion in the solution after the acid dissociation can be determined.

Reaction	Weak base	Conjugate acid	OH-
Initial	0.2	0	0
Change	-x	+x	+x
Equilibrium	(0.2-x)	x	x

  $\begin{array}{l}{K}_b=\frac{[HON{H}_3{}^{+}][O{H}^{-}]}{[HON{H}_2]}\\ 1.1\times {10}^{-8}=\frac{[x][x]}{[0.2-x]}=\frac{x^2}{[0.2-x]}\end{array}$

For small base dissociation constant value, the value of x can be neglected from the denominator.

Thus,

  $\begin{array}{l}1.1\times {10}^{-8}=\frac{x^2}{0.2}\\ x=\left[O{H}^{-}\right]=4.7\times {10}^{-5}\end{array}$

Then the pH of the initial solution can be determined.

  $\begin{array}{l}pOH=-\log [O{H}^{-}]=-log[4.7\times {10}^{-5}]=4.33\\ pH=14.00-4.33=9.67\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.88.

Explanation of Solution

Addition of $25.0mL$ofacid:

Total amount of base to be neutralized $=0.2mol/L\times 100.0\times {10}^{-3}L=20\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 25.0\times {10}^{-3}L=25\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.02	0	0	0
Add	0	0.0025
Change	-0.0025	-0.0025	0.0025	0.0025
Equilibrium	0.0175	0	0.0025	0.0025

The base dissociation constant value is $1.1\times {10}^{-8}$

Thus,

  $p{K}_b=-\log K_b=-\log \left(1.1\times {10}^{-8}\right)=7.96$

Since, volume is same thus, number of moles can be used instead of concentrations.

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pOH=p{K}_b+\log \frac{[Conjugateacid]}{[weak\text{ base}]}\\ pOH=7.96+\log \frac{[2.50]}{[17.5]}=7.12\end{array}$

Now, pH can be calculated as follows:

  $pH=14-pOH=14-7.12=6.88$

Thus, the pH of the solution is 6.88.

(c)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (70 mL HCl) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

The pH of the solution after adding the acid is 6.31.

Explanation of Solution

Addition of $70.0mL$ofacid:

Total amount of base to be neutralized $=0.2mol/L\times 100.0\times {10}^{-3}L=20\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 70.0\times {10}^{-3}L=70\times {10}^{-4}mol$

Then the ICE table after the addition of base is created in order to determine the pH of the solution using Henderson-Hasselbalch equation.

Reaction	Weak base	H+	Conjugate acid	OH-
Initial	0.0200	0	0	0
Add	0	0.0070
Change	-0.0070	-0.0070	0.0070	0.0070
Equilibrium	0.0130	0	0.0070	0.0070

Applying the Henderson-Hasselbalch equation,

  $\begin{array}{l}pOH=pKb+\log \frac{[Conjugateacid]}{[weak\text{ base}]}\\ pOH=7.96+\log \frac{[0.007]}{[0.013]}=7.69\\ pH=14-7.69=6.31\end{array}$

Thus, the pH of the solution is 6.31.

(d)

Interpretation Introduction

Interpretation:

The pH value of the weak base at equivalent point needs to be determined.

Concept introduction:

At equivalent point the concentration of acid and base (in which one of the species is weak) is equal in the solution. Thus, the pH will depend on the acidic or basic salt formed.

Answer to Problem 167AE

The pH of the solution at the equivalence point is 3.60.

Explanation of Solution

At the equivalence point, The amount of acid added $=20\times {10}^{-3}mol$

The volume of acid added $=\frac{20\times {10}^{-3}mol}{0.1mol/L}=0.2L=200.0mL$

At this point, there is no excess acid or base. Therefore, the only possible reaction here is the dissociation of the conjugate acid of the ammonia (that is ammonium ion).

  $HON{H}_3{}^{+}+H_{2}O\rightleftharpoons HON{H}_2+H_3{O}^{+}$

Thereafter, using the Ka value for this weak base, the amount of hydrogen ions in the solution can be determined to get the pH value at this point.

Now, total volume of the solution is 300 mL thus, molarity of HONH₃+ solution will be:

  $M=\frac{n}{V}=\frac{0.02\text{ mol}}{300\text{ mL}}=0.0667\text{ M}$

Reaction	HONH3+	HONH2	H+
Initial	0.0667	0	0
Change	-x	x	x
Equilibrium	(0.0667-x)	x	x

Then the pH can be calculated as follows:

  $\begin{array}{l}Ka=\frac{[HON{H}_2][H^{+}]}{[HON{H}_3{}^{+}]}\\ 9.09\times {10}^{-7}=\frac{[x][x]}{0.0667}=\frac{x^2}{0.0067}\\ x=2.5\times {10}^{-4}\text{ M}\end{array}$

This value is equal to the amount of hydrogen ions in the solution.

  $pH=-\log [H^{+}]=-log[2.5\times {10}^{-4}]=3.60$

Thus, the pH value at equivalence point is 3.60.

(e)

Interpretation Introduction

Interpretation:

The pH value of the weak base after adding the acid (300 mL) is to be determined.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

The pH of the solution after adding acid is $1.60$ .

Explanation of Solution

Addition of $300.0mL$acid, The amount of acid added $=0.1mol/L\times 300.0\times {10}^{-3}L=0.03mol$

At this point, there is excess acid in the solution, thus, the pH can be calculated as follows:

Excess amount of acid in the solution $=0.03-0.02=0.01mol$

The amount of hydrogen ions $=0.01mol$

The concentration of hydrogen ions $=\frac{0.01mol}{(100.0+300.0)\times {10}^{-3}L}=0.025mol/L$

  $pH=-\log [H^{+}]=-log[0.025]=1.60$

(f)

Interpretation Introduction

Interpretation:

The volume of HCl needs to be determined that is added to the solution if pH is 6.04.

Concept introduction:

A strong acid is the substance which completely dissociates into hydrogen ion and conjugate base in water. The weak base partially dissociate into its respective ions.

Answer to Problem 167AE

100 mL

Explanation of Solution

The $p{K}_b$ value for NH₂OH is 7.96. From the given pH value, pOH of the solution can be calculated as follows:

  $pOH=14-pH=14-6.04=7.96$

The Henderson Hesselbalch equation can be represented as follows:

  $pOH=p{K}_b+\log \frac{\left[salt\right]}{\left[base\right]}$

Since, the volume is same thus, concentration can be replaced by number of moles.

  $7.96=7.96+\log \frac{\left[salt\right]}{\left[base\right]}$

Thus, the number of moles of salt and base will be same that is $N{H}_{3}O{H}^{+}$ and NH₂OH respectively.

This is the condition which is halfway to the equivalence point. For the equivalence point, volume of HCl required was 200 mL thus, at halfway equivalence point the volume will be half to it. Therefore, volume of HCl is 100 mL.