Chapter 16, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The equilibrium constant for the reaction,

${\text{NH}}_{\text{2}}\text{OH}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{NH}}_{\text{3}}{\text{OH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$, is to be stated.

Concept introduction:

In a chemical reaction, when the rates of both forward and reverse reaction are equal then the chemical reaction is said to be at equilibrium. They are reversible in nature. Equilibrium constant is denoted as $K_{eq}$. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients,

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Answer to Problem 26E

The expression for equilibrium constant for the dissociation of ${\text{NH}}_{\text{2}}\text{OH}$ is shown below.

$K_{\text{i}}=\frac{\left[{\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NH}}_2\text{OH}\right]}$

Explanation of Solution

The given reaction is shown below.

${\text{NH}}_{\text{2}}\text{OH}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{NH}}_{\text{3}}{\text{OH}}^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

For weak bases, the equilibrium constant $K_{eq}$ is represented as the ionization equilibrium constant, $K_{\text{i}}$. The expression for equilibrium constant for the dissociation of ${\text{NH}}_{\text{2}}\text{OH}$ is given as shown below.

$K_{\text{i}}=\frac{\left[{\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[{\text{NH}}_2\text{OH}\right]}$

As the concentration of liquid and solid is not considered while writing the expression, therefore water is not included in the equilibrium expression.

Conclusion

The expression for equilibrium constant for the dissociation of ${\text{NH}}_{\text{2}}\text{OH}$ is stated above.

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant for the reaction,

${\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$, is to be stated.

Concept introduction:

In a chemical reaction, when the rates of both forward and reverse reaction are equal then the chemical reaction is said to be at equilibrium. They are reversible in nature. Equilibrium constant is denoted as $K_{eq}$. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients,

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Answer to Problem 26E

The expression for equilibrium constant for the dissociation of ${\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}$ is shown below.

$K_{\text{i}}=\frac{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\right]}$

Explanation of Solution

The given reaction is shown below.

${\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

For weak bases, the equilibrium constant $K_{eq}$ is represented as the ionization equilibrium constant, $K_{\text{i}}$. The expression for equilibrium constant for the dissociation of ${\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}$ is given as shown below.

$K_{\text{i}}=\frac{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_3^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}\right]}$

As the concentration of liquid and solid is not considered while writing the expression, therefore water is not included in the equilibrium expression.

Conclusion

The expression for equilibrium constant for the dissociation of ${\text{C}}_6{\text{H}}_5{\text{NH}}_{\text{2}}$ is stated above.

Interpretation Introduction

(c)

Interpretation:

The equilibrium constant for the reaction,

${\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_2{\text{NH}}_2^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$, is to be stated.

Concept introduction:

In a chemical reaction, when the rates of both forward and reverse reaction are equal then the chemical reaction is said to be at equilibrium. They are reversible in nature. Equilibrium constant is denoted as $K_{eq}$. Mathematically, it is represented as the ratio of product concentration to reactant concentration raised to their stoichiometric coefficients,

$K_{eq}=\frac{{\left[\text{Product}\text{concentration}\right]}^{\text{a}}}{{\left[\text{Reactant}\text{concentration}\right]}^{\text{b}}}$

Answer to Problem 26E

The expression for equilibrium constant for the dissociation of ${\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}$ is shown below.

$K_{\text{i}}=\frac{\left[{\left({\text{CH}}_{\text{3}}\right)}_2{\text{NH}}_2^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}\right]}$

Explanation of Solution

The given reaction is shown below.

${\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\left({\text{CH}}_{\text{3}}\right)}_2{\text{NH}}_2^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

For weak bases, the equilibrium constant $K_{eq}$ is represented as the ionization equilibrium constant, $K_{\text{i}}$. The expression for equilibrium constant for the dissociation of ${\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}$ is given as shown below.

$K_{\text{i}}=\frac{\left[{\left({\text{CH}}_{\text{3}}\right)}_2{\text{NH}}_2^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}\right]}$

As the concentration of liquid and solid is not considered while writing the expression, therefore water is not included in the equilibrium expression.

Conclusion

The expression for equilibrium constant for the dissociation of ${\left({\text{CH}}_{\text{3}}\right)}_2\text{NH}$ is stated above.