Chapter 8, Problem 87E

(a)

Interpretation Introduction

Interpretation: The pH of the solution before addition of KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

Before KOH is added, the pH of the solution depends on concentration of acid.

The first dissociation of ${\text{H}}_{\text{3}}\text{X}$ is represented as follows:

  ${\text{H}}_{\text{3}}\text{X}\rightleftharpoons H^{+}+H_2{X}^{-}$

The given concentration of ${\text{H}}_{\text{3}}\text{X}$ is 0.05 M thus, the ICE table can be represented as follows:

  $\begin{array}{l}{\text{ H}}_{\text{3}}\text{X}\rightleftharpoons H^{+}+H_2{X}^{-}\\ I\text{ 0}\text{.05 0 0}\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.05-x x x}\end{array}$

The first acid dissociation constant can be represented as follows:

  $\begin{array}{l}{K}_{a1}=\frac{\left[H^{+}\right]\left[H_2{X}^{-}\right]}{\left[H_{3}X\right]}\\ 1.0\times {10}^{-3}=\frac{x^2}{0.05-x}\end{array}$

Or,

  $x^2+{10}^{-3}x-5\times {10}^{-5}=0$

On solving,

  $x=6.6\times {10}^{-3}$

This is the concentration of hydrogen ion in the solution. Thus, pH can be calculated as follows:

  $pH=-\log \left[{\text{H}}^{+}\right]=-\log \left(6.6\times {10}^{-3}\right)=2.18$

(b)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 10 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The number of moles of ${\text{H}}_{\text{3}}\text{X}$ and KOH in the solution can be calculated as follows:

  $\begin{array}{l}{n}_{{\text{H}}_{\text{3}}\text{X}}=\left(0.05\text{ M}\right)\left(100\text{ mL}\right)=5\times {10}^{-3}\text{ mol}\\ n_{O{H}^{-}}=\left(0.1\text{ M}\right)\left(100\text{ mL}\right)=1\times {10}^{-3}\text{ mol}\end{array}$

Now, hydroxide ion will react with ${\text{H}}_{\text{3}}\text{X}$ and ${\text{H}}_2{\text{X}}^{-}$ is produced which is equal to the hydroxide ion concentration used.

Thus,

  $M=\frac{n}{V}=\frac{1\times {10}^{-3}\text{ mol}}{110\times {10}^{-3}\text{ L}}=0.00909\text{ M}$

Thus, the amount of ${\text{H}}_{\text{3}}\text{X}$ left in the solution will be:

  $n_{H_{3}X}=5\times {\text{10}}^{-3}\text{ mol}-1\times {\text{10}}^{-3}\text{ mol}=4\times {\text{10}}^{-3}\text{ mol}$

The total volume will be 110 mL thus, concentration can be calculated as follows:

  $M=\frac{n}{V}=\frac{4\times {10}^{-3}\text{ mol}}{110\times {10}^{-3}\text{ L}}=0.0364\text{ M}$

The ICE table can be prepared as follows:

  $\begin{array}{l}{H}_{3}X\rightleftharpoons H^{+}+H_2{X}^{-}\\ I\text{ 0}\text{.0364 - -}\\ C\text{ -x x x}\\ E\text{ 0}\text{.0364-x x 0}\text{.00909+x }\end{array}$

The equilibrium expression can be represented as follows:

  $K_{a_1}=1\times {10}^{-3}=\frac{x\left(0.00909+x\right)}{\left(0.0364-x\right)}$

Or,

  $x^2+\left(1.01\times {10}^{-2}\right)x-3.6\times {10}^{-5}=0$

On solving,

  $x=2.8\times {10}^{-3}$

The pH of the solution will be:

  $pH=-\log x=-\log \left(2.8\times {10}^{-3}\right)=2.55$

(c)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 25 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The molar amount of hydroxide ion initially present can be calculated as follows:

  $n=M\times V=0.1\text{ M}\times \text{25 mL=2}\text{.5 mmol}$

The reaction of hydroxide ion and H₃X to produce $H_2{X}^{-}$ is equal to amount of hydroxide ion concentration.

Thus, the amount of H₃X left can be calculated as follows:

  $5-2.5=2.5\text{ mmol}$

The concentration of H₃X is equal to concentration of $H_2{X}^{-}$ at first equivalent Point. At this point, the pH is equal to $p{K}_a$ .

  $pH=p{K}_a=-\log \left(1.0\times {10}^{-3}\right)=3$

(d)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 50 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The number of moles of hydroxide ion can be calculated as follows:

  $n=\left(0.1\text{ M}\right)\left(50\text{ mL}\right)=5\text{ mmol}$

The hydrogen ion reacts with H₃X and the amount is equal to that of hydroxide ion.

The remaining amount of H₃X will be:

  $5-5=0$

Now, all the acid exists as ${\text{H}}_{\text{2}}{\text{X}}^{-}$ and this is the first stoichiometric point.

The pH can be calculated as follows:

  $\begin{array}{l}pH=\frac{p{K}_{a1}+p{K}_{a2}}{2}=\frac{3-\log \left(1.0\times {10}^{-7}\right)}{2}\\ =5\end{array}$

(e)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 60 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The number of moles of hydroxide ion can be calculated as follows:

  $n=\left(0.1\text{ M}\right)\left(60\text{ mL}\right)=6\text{ mmol}$

Initially 5 mmol of hydroxide ion react with H₃X and the remaining will react to with $H_2{X}^{-}$ to form $H{X}^{2-}$ .

The amount of $H_2{X}^{-}$ left will be:

  $\left(5-1\right)\text{ mmol=4 mmol }{H}_2{X}^{-}$

The molarity can be calculated as follows:

  $M=\frac{n}{V}=\frac{4\text{ mmol}}{160\text{ mL}}=0.025$

Also, the molarity of $H{X}^{2-}$ will be:

  $M=\frac{n}{V}=\frac{\text{1 mmol}}{160\text{ mL}}=0.00625$

The ICE table can be prepared as follows:

  $\begin{array}{l}{H}_2{X}^{-}\rightleftharpoons H^{+}+H{X}^{2-}\\ I\text{ 0}\text{.025 0 0}\text{.00625}\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.025-x x 0}\text{.00625+x}\end{array}$

The expression for $K{a}_2$ will be:

  $1.0\times {10}^{-7}=\frac{x\left(0.00625+x\right)}{0.025-x}$

The $K{a}_2$ value is very small thus,

  $\begin{array}{l}1.0\times {10}^{-7}=\frac{x\left(0.00625\right)}{0.025}\\ x=4.0\times {10}^{-7}\end{array}$

This is the concentration of hydrogen ion.

The pH value can be calculated as follows:

  $\begin{array}{l}pH=-\log \left[H^{+}\right]=-\log \left(4\times {10}^{-7}\right)\\ =6.4\end{array}$

(f)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 75 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The number of moles of hydroxide ion can be calculated as follows:

  $n=M\times V=\left(0.1\text{ M}\right)\left(75\text{ mL}\right)=7.5\text{ mmol}$

Initially 5 mmol of hydroxide ion react with H₃X and the remaining will react to with $H_2{X}^{-}$ to form $H{X}^{2-}$ .

Thus, the amount of $H_2{X}^{-}$ left will be:

  $\left(5-2.5\right)\text{ mmol=2}\text{.5mmol }{H}_2{X}^{-}$

Here, the concentration of $H_2{X}^{-}$ is equal to $H{X}^{2-}$ which is the second equivalence point. Thus, pH value will be equal to $p{K}_{a2}$ .

Thus, pH value can be calculated as follows:

  $pH=-\log K_{a2}=-\log \left(1.0\times {10}^{-7}\right)=7$

(g)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 100 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The initial number of moles of hydroxide ion can be calculated as follows:

  $n=M\times V=\left(0.1\text{ M}\right)\left(100\text{ mL}\right)=10\text{ mmol}$

Initially 5 mmol of hydroxide ion react with H₃X and the remaining will react to with $H_2{X}^{-}$ to form $H{X}^{2-}$ .

The amount of $H_2{X}^{-}$ left will be:

  $\left(5-5\right)\text{ mmol=0 mmol }{H}_2{X}^{-}$

Now, only $H{X}^{2-}$ exists thus, it is the second equivalence point.

The pH can be calculated as follows:

  $pH=\frac{p{K}_{a2}+p{K}_{a3}}{2}$

Putting the values,

  $\begin{array}{l}pH=\frac{7-\log \left(1\times {10}^{-12}\right)}{2}\\ =9.5\end{array}$

(h)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 125 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The initial number of moles of hydroxide ion can be calculated as follows:

  $n=M\times V=\left(0.1\text{ M}\right)\left(125\text{ mL}\right)=12.5\text{ mmol}$

Initially 5 mmol of hydroxide ion react with H₃X and the remaining will react to with $H_2{X}^{-}$ to form $H{X}^{2-}$ .

The remaining hydroxide ion will react to form $X^{3-}$ which is equal to the amount of hydroxide ion formed.

The amount of $H{X}^{2-}$ left can be calculated as follows:

  $\left(5-2.5\right)\text{ mmol=2}\text{.5 mmol }{H}_2{X}^{-}$

The molarity can be calculated as follows:

  $M=\frac{n}{V}=\frac{2.5\text{ mmol}}{225\text{ mL}}=0.011\text{ M}$

Similarly, for ${\text{HX}}^{2-}$ ,

  $M=\frac{n}{V}=\frac{2.5\text{ mmol}}{225\text{ mL}}=0.011\text{ M}$

This is third equivalence point and the ICE table can be represented as follows:

  $\begin{array}{l}{X}^{3-}\rightleftharpoons O{H}^{-}+H{X}^{2-}\\ I\text{ 0}\text{.011 - 0}\text{.011 }\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.011-x x 0}\text{.011+x}\end{array}$

The base dissociation constant can be represented as follows:

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[H{X}^{2-}\right]}{\left[X^{3-}\right]}\\ \frac{{10}^{-14}}{{10}^{-12}}=\frac{x\left(0.011+x\right)}{\left(0.011-x\right)}\end{array}$

The equation will be:

  $x^2+0.022x-1.1\times {10}^{-4}=0$

On solving, the value of x will be:

  $x=4.3\times {10}^{-3}$

This is concentration of hydroxide ion, the pH value can be calculated as follows:

  $\begin{array}{l}pH=14-pOH\\ pH=14+\log \left[O{H}^{-}\right]\end{array}$

Putting the values,

  $pH=14+\log \left[O{H}^{-}\right]=14+\log \left(4.3\times {10}^{-3}\right)=11.63$

(i)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 150 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The hydrogen ion concentration in the beginning can be calculated as follows:

  ${\text{OH}}^{-}=0.1\text{ M}\times 150\text{ mL=15 mmol}$

Initially 5 mmol of hydroxide ion react with H₃X and the remaining will react with $H_2{X}^{-}$ to form $H{X}^{2-}$ .

The remaining hydroxide ion will react to form $X^{3-}$ which is equal to the amount of hydroxide ion formed.

The amount of $H{X}^{2-}$ left can be calculated as follows:

  $\left(5-5\right)\text{ mmol=0 mmol }{H}_2{X}^{-}$

The molarity for $X^{3-}$ can be calculated as follows:

  $M=\frac{n}{V}=\frac{5\text{ mmol}}{250\text{ mL}}=0.02\text{ M}$

The ICE tale will be:

  $\begin{array}{l}{X}^{3-}\rightleftharpoons O{H}^{-}+H{X}^{2-}\\ I\text{ 0}\text{.02 - - }\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.02-x x x}\end{array}$

The base dissociation constant can be represented as follows:

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[H{X}^{2-}\right]}{\left[X^{3-}\right]}\\ \frac{{10}^{-14}}{{10}^{-12}}=\frac{x^2}{\left(0.02-x\right)}\end{array}$

The equation will be:

  $x^2+0.01x-2\times {10}^{-4}=0$

On solving, the value of x will be:

  $x=0.01$

This is concentration of hydroxide ion, the pH value can be calculated as follows:

  $\begin{array}{l}pH=14-pOH\\ pH=14+\log \left[O{H}^{-}\right]\end{array}$

Putting the values,

  $pH=14+\log \left[O{H}^{-}\right]=14+\log \left(0.010\right)=12$

(j)

Interpretation Introduction

Interpretation: The pH of the solution after addition of 200 mL of 0.1 M KOH needs to be determined.

Concept Introduction: The acid dissociation reaction of ${\text{H}}_{\text{3}}\text{X}$ can be represented as follows:

  $\begin{array}{l}{\text{H}}_{\text{3}}\text{X}\rightleftharpoons {\text{H}}^{+}+{\text{H}}_{\text{2}}{\text{X}}^{-}\\ {\text{H}}_{\text{2}}{\text{X}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{HX}}^{2-}\\ {\text{HX}}^{2-}\rightleftharpoons {\text{H}}^{+}+{\text{X}}^{3-}\end{array}$

The acid dissociation constant for the above three reactions will be $K_{a1},K_{a2}\text{ and }{K}_{a3}$ respectively.

Explanation of Solution

The hydrogen ion concentration in the beginning can be calculated as follows:

  ${\text{OH}}^{-}=0.1\text{ M}\times 200\text{ mL=20 mmol}$

Now, to convert ${\text{H}}_{\text{3}}\text{X}$ to $X^{3-}$ , 15 mmol of hydroxide ion is used.

The remaining number of moles of hydroxide ion will be 5 mmol.

The molarity of $X^{3-}$ and hydroxide ion can be calculated as follows:

  $\begin{array}{l}{M}_{X^{3-}}=\frac{n}{V}=\frac{5\text{ mmol}}{300\text{ mL}}=0.017\text{ M}\\ M_{O{H}^{-}}=\frac{n}{V}=\frac{5\text{ mmol}}{300\text{ mL}}=0.017\text{ M}\end{array}$

The reaction of hydroxide ion with $X^{3-}$ can be represented as follows:

  $\begin{array}{l}{X}^{3-}\rightleftharpoons O{H}^{-}+H{X}^{2-}\\ I\text{ 0}\text{.017 0}\text{.017 - }\\ C\text{ -x +x +x}\\ E\text{ 0}\text{.017 -x 0}\text{.017 x}\end{array}$

The base dissociation constant can be represented as follows:

  $\begin{array}{l}{K}_b=\frac{\left[O{H}^{-}\right]\left[H{X}^{2-}\right]}{\left[X^{3-}\right]}\\ \frac{{10}^{-14}}{{10}^{-12}}=\frac{x\left(0.017+x\right)}{\left(0.017-x\right)}\end{array}$

The equation will be:

  $x^2+0.022x-1.7\times {10}^{-4}=0$

On solving, the value of x will be:

  $x=0.0053$

The hydroxide ion concentration will be:

  $\left[{\text{OH}}^{-}\right]=0.017+0.053=0.022\text{ M}$

This is concentration of hydroxide ion, the pH value can be calculated as follows:

  $\begin{array}{l}pH=14-pOH\\ pH=14+\log \left(O{H}^{-}\right)\end{array}$

Putting the values,

  $pH=14+\log \left[O{H}^{-}\right]=14+\log \left(0.022\right)=12.34$