Chapter 7, Problem 153CP

(a)

Interpretation Introduction

Interpretation: The value of the equilibrium constant for the given reaction needs to be determined.

  ${\text{HCO}}_3^{-}{}_{\text{(aq)}}{\text{ + HCO}}_3^{-}{}_{\text{(aq)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{+ CO}}_3^{2-}{}_{\text{(aq)}}$

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

Answer to Problem 153CP

K_eq = $1.3\times {\text{10}}^{\text{-4}}$

Explanation of Solution

The equilibrium constant for the given reaction can be written as:

  $\begin{array}{l}{\text{HCO}}_3^{-}{}_{\text{(aq)}}{\text{ + HCO}}_3^{-}{}_{\text{(aq)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{+ CO}}_3^{2-}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{] [ CO}}_3^{2-}{}_{\text{(aq)}}\text{]}}{{\text{[HCO}}_3^{-}{}_{\text{(aq)}}{\text{][HCO}}_3^{-}{}_{\text{(aq)}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{] [ CO}}_3^{2-}{}_{\text{(aq)}}\text{]}}{{\text{[HCO}}_3^{-}{}_{\text{(aq)}}{\text{][HCO}}_3^{-}{}_{\text{(aq)}}\text{]}}\times \frac{{\text{[H}}^{\text{+}}{}_{\text{(aq)}}\text{]}}{{\text{[H}}^{\text{+}}{}_{\text{(aq)}}\text{]}}=\frac{{\text{K}}_{\text{a2}}}{{\text{K}}_{\text{a1}}}=\frac{\text{5}\text{.6}\times {\text{10}}^{\text{-11}}}{\text{4}\text{.3}\times {\text{10}}^{\text{-7}}}=1.3\times {\text{10}}^{\text{-4}}\end{array}$

(b)

Interpretation Introduction

Interpretation: The relation between ${\text{H}}_{\text{2}}{\text{CO}}_3^{}$ and ${\text{CO}}_3^{2-}$ concentration for the given reaction needs to be determined.

  ${\text{HCO}}_3^{-}{}_{\text{(aq)}}{\text{ + HCO}}_3^{-}{}_{\text{(aq)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{+ CO}}_3^{2-}{}_{\text{(aq)}}$

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

Answer to Problem 153CP

At equilibrium, [ ${\text{H}}_{\text{2}}{\text{CO}}_3^{}$ ]= [ ${\text{CO}}_3^{2-}$ ]

Explanation of Solution

The equilibrium constant for the given reaction can be written as:

  $\begin{array}{l}{\text{HCO}}_3^{-}{}_{\text{(aq)}}{\text{ + HCO}}_3^{-}{}_{\text{(aq)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{+ CO}}_3^{2-}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{] [ CO}}_3^{2-}{}_{\text{(aq)}}\text{]}}{{\text{[HCO}}_3^{-}{}_{\text{(aq)}}{\text{][HCO}}_3^{-}{}_{\text{(aq)}}\text{]}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}{\text{] [ CO}}_3^{2-}{}_{\text{(aq)}}\text{]}}{{\text{[HCO}}_3^{-}{}_{\text{(aq)}}{\text{][HCO}}_3^{-}{}_{\text{(aq)}}\text{]}}\end{array}$

Hence at equilibrium, [ ${\text{H}}_{\text{2}}{\text{CO}}_3^{}$ ]= [ ${\text{CO}}_3^{2-}$ ]

(c)

Interpretation Introduction

Interpretation: The expression for the pH of the solution in terms of Ka₁and Ka₂ for the given reaction needs to be determined.

  ${\text{H}}_{\text{2}}{\text{CO}}_3^{}{}_{\text{(aq)}}\rightleftharpoons {\text{2 H}}_{}^{+}{}_{\text{(aq)}}{\text{+ CO}}_3^{2-}{}_{\text{(aq)}}$

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

Answer to Problem 153CP

  $\text{pH = }\frac{{\text{pK}}_{\text{a1}}{\text{+ pK}}_{\text{a2}}}{\text{2}}$

Explanation of Solution

The equilibrium constant for the given reaction can be written as:

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}^{}{}_{\text{(aq)}}\rightleftharpoons {\text{2 H}}_{}^{\text{+}}{}_{\text{(aq)}}{\text{+ CO}}_{\text{3}}^{\text{2-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{{\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{]}}^{\text{2}}{\text{ [ CO}}_{\text{3}}^{\text{2-}}{}_{\text{(aq)}}\text{]}}{{\text{[H}}_{\text{2}}{\text{CO}}_{\text{3}}^{}{}_{\text{(aq)}}\text{]}}{\text{= K}}_{\text{a1}}{\text{× K}}_{\text{a2}}\\ {\text{Since[H}}_{\text{2}}{\text{CO}}_{\text{3}}^{}{}_{\text{(aq)}}{\text{]=[ CO}}_{\text{3}}^{\text{2-}}{}_{\text{(aq)}}\text{]}\\ {\text{Hence [H}}_{}^{\text{+}}{}_{\text{(aq)}}{\text{]}}^{\text{2}}{\text{=K}}_{\text{a1}}{\text{× K}}_{\text{a2}}\\ {\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] = }\sqrt{{\text{K}}_{\text{a1}}{\text{× K}}_{\text{a2}}}\\ {\text{-log[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] =-log }\sqrt{{\text{K}}_{\text{a1}}{\text{× K}}_{\text{a2}}}\\ \text{pH = }\frac{{\text{pK}}_{\text{a1}}{\text{+ pK}}_{\text{a2}}}{\text{2}}\end{array}$

(d)

Interpretation Introduction

Interpretation: The value of pH of ${\text{NaHCO}}_{\text{3}}$ solution needs to be determined.

Concept Introduction: An acid is the substance that gives H+ or ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in its aqueous solution. A weak acid ionized partially and reaches to equilibrium.

  ${\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The acid dissociation constant is the ratio of concentration of product and reactant for the equilibrium reaction of weak acid in solution. For the given weak acid HA, it can be written as:

  $\begin{array}{l}{\text{HA}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l) }}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}\\ {\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{] [A}}^{\text{-}}\text{]}}{\text{[HA]}}\end{array}$

Answer to Problem 153CP

  $pH=8.3$

Explanation of Solution

The hydrogen ion concentration can be calculated as follows:

  ${\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] = }\sqrt{{\text{K}}_{\text{a1}}{\text{× K}}_{\text{a2}}}$

Substitute the values of Ka1 and Ka2:

  $\begin{array}{l}{\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] = }\sqrt{{\text{K}}_{\text{a1}}{\text{× K}}_{\text{a2}}}\\ {\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] = }\sqrt{\text{4}\text{.3}\times {\text{10}}^{\text{-7}}\text{ × 5}\text{.6}\times {\text{10}}^{\text{-11}}}\\ {\text{[H}}_{}^{\text{+}}{}_{\text{(aq)}}\text{] = 4}\text{.9}\times {\text{10}}^{\text{-9}}\end{array}$

Calculate pH:

  $\begin{array}{l}{\text{pH = - log[H}}^{\text{+}}\text{]}\\ \text{pH = - log[4}{\text{.9×10}}^{\text{-9}}\text{]}\\ \text{pH=8}\text{.3}\end{array}$