Chapter 7, Problem 125AE

(a)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{NH}}_3{\text{ + H}}_3{\text{O}}^{+}\rightleftharpoons {\text{ NH}}_4^{\text{+}}{\text{ + H}}_2\text{O}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=1.8\times {10}^9$

Explanation of Solution

  ${\text{NH}}_3{\text{ + H}}_3{\text{O}}^{+}\rightleftharpoons {\text{ NH}}_4^{\text{+}}{\text{ + H}}_2\text{O}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{\left[{\text{NH}}_4^{\text{+}}\right]}{\left[{\text{NH}}_3\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}\\ {\text{K}}_{eq}=\frac{1}{K_a{\text{ for NH}}_4^{\text{+}}}=\frac{K_b}{K_w}=\frac{1.8\times {10}^{-5}}{1.0\times {10}^{-14}}\\ {\text{K}}_{eq}=1.8\times {10}^9\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{NO}}_2{}^{-}{\text{ + H}}_3{\text{O}}^{+}\rightleftharpoons {\text{ HNO}}_2{\text{ + H}}_2\text{O}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=2.5\times {10}^3$

Explanation of Solution

  ${\text{NO}}_2{}^{-}{\text{ + H}}_3{\text{O}}^{+}\rightleftharpoons {\text{ HNO}}_2{\text{ + H}}_2\text{O}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{\left[{\text{HNO}}_2\right]}{\left[{\text{NO}}_2{}^{-}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}\\ {\text{K}}_{eq}=\frac{1}{K_a{\text{ for HNO}}_2}=\frac{1}{4.0\times {10}^{-4}}\\ {\text{K}}_{eq}=2.5\times {10}^3\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{NH}}_4{}^{+}{\text{ + CH}}_3{\text{CO}}_2{}^{-}\rightleftharpoons {\text{ NH}}_3{\text{ + CH}}_3\text{COOH}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=3.1\times {10}^{-5}$

Explanation of Solution

  ${\text{NH}}_4{}^{+}{\text{ + CH}}_3{\text{CO}}_2{}^{-}\rightleftharpoons {\text{ NH}}_3{\text{ + CH}}_3\text{COOH}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{\left[{\text{NH}}_3\right]\left[{\text{CH}}_3\text{COOH}\right]}{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{CH}}_3{\text{CO}}_2{}^{-}\right]}\\ {\text{K}}_{eq}=\frac{\left[{\text{NH}}_3\right]\left[{\text{CH}}_3\text{COOH}\right]}{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{CH}}_3{\text{CO}}_2{}^{-}\right]}\times \frac{\left[{\text{H}}^{+}\right]}{\left[{\text{H}}^{+}\right]}\\ {\text{K}}_{eq}=\frac{K_a{\text{ for NH}}_4^{+}}{K_a{\text{ for CH}}_3\text{COOH}}=\frac{K_w}{\left(K_b{\text{ for NH}}_3\right)\left(K_a{\text{ for CH}}_3\text{COOH}\right)}\\ {\text{K}}_{eq}=\frac{1.0\times {10}^{-14}}{\left(1.8\times {10}^{-5}\right)\left(1.8\times {10}^{-5}\right)}\\ {\text{K}}_{eq}=3.1\times {10}^{-5}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{H}}_3{\text{O}}^{+}{\text{ + OH}}^{-}\rightleftharpoons {\text{ 2 H}}_2\text{O}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=1.0\times {10}^{14}$

Explanation of Solution

  ${\text{H}}_3{\text{O}}^{+}{\text{ + OH}}^{-}\rightleftharpoons {\text{ 2 H}}_2\text{O}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{1}{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}\\ {\text{K}}_{eq}=\frac{1}{K_w}=\frac{1}{1.0\times {10}^{-14}}\\ {\text{K}}_{eq}=1.0\times {10}^{14}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{NH}}_4{}^{+}{\text{ + OH}}^{-}\rightleftharpoons {\text{ NH}}_3{\text{ + H}}_2\text{O}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=5.6\times {10}^4$

Explanation of Solution

  ${\text{NH}}_4{}^{+}{\text{ + OH}}^{-}\rightleftharpoons {\text{ NH}}_3{\text{ + H}}_2\text{O}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{\left[{\text{NH}}_3\right]}{\left[{\text{NH}}_4{}^{+}\right]\left[{\text{OH}}^{-}\right]}\\ {\text{K}}_{eq}=\frac{1}{K_b{\text{ for NH}}_3}=\frac{1}{1.8\times {10}^{-5}}\\ {\text{K}}_{eq}=5.6\times {10}^4\end{array}$

(f)

Interpretation Introduction

Interpretation:

The value for the equilibrium constant for the reaction ${\text{HNO}}_2{\text{ + OH}}^{-}\rightleftharpoons {\text{ NO}}_2{}^{-}{\text{ + H}}_2\text{O}$ should be calculated.

Concept Introduction :

The equilibrium constant expression is the ratio of the concentration of products and reactants of a reaction at equilibrium. The concentrations are raised to the power equal to the stoichiometric coefficient in the balanced equation.

Answer to Problem 125AE

  ${\text{K}}_{eq}=4.0\times {10}^{10}$

Explanation of Solution

  ${\text{HNO}}_2{\text{ + OH}}^{-}\rightleftharpoons {\text{ NO}}_2{}^{-}{\text{ + H}}_2\text{O}$

  $\begin{array}{l}{\text{K}}_{eq}=\frac{\left[{\text{NO}}_2{}^{-}\right]}{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}\\ {\text{K}}_{eq}=\frac{\left[{\text{NO}}_2{}^{-}\right]}{\left[{\text{HNO}}_2\right]\left[{\text{OH}}^{-}\right]}\times \frac{\left[{\text{H}}^{+}\right]}{\left[{\text{H}}^{+}\right]}\\ {\text{K}}_{eq}=\frac{K_a{\text{ for HNO}}_2}{K_w}=\frac{4.0\times {10}^{-4}}{1.0\times {10}^{-14}}\\ {\text{K}}_{eq}=4.0\times {10}^{10}\end{array}$