Chapter 3, Problem 3.19P

Interpretation Introduction

(a)

Interpretation:

The equation for the equilibria is to be predicted. The $\text{p}{K}_{\text{a}}$ value associated with the acidic species in the equilibrium is to be predicted.

Concept introduction:

Dissociation constant $K_{\text{a}}$ is used to measure the strength of a Brønsted acid. There is an inverse relationship between the acidity of a Brønsted acid and its $\text{p}{K}_{\text{a}}$ value. The relation between $K_{\text{eq}}$ with $\text{p}{K}_{\text{a}}$ is given by an expression shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{\text{BH}}/\text{p}{K}_{\text{AH}}}$

Where,

• $K_{\text{AH}}$ is dissociation constant for an acid, $\text{AH}$.

• $K_{\text{BH}}$ is dissociation constant for conjugate acid, $\text{BH}$.

Answer to Problem 3.19P

The equation for the equilibria is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$. The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_4{}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ is $9.25$ and $15.7$ respectively.

Explanation of Solution

The relation between $K_{\text{eq}}$ with $\text{p}{K}_{\text{a}}$ is given by an expression shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{\text{BH}}/\text{p}{K}_{\text{AH}}}$…(1)

Where,

• $K_{\text{AH}}$ is dissociation constant for an acid, $\text{AH}$.

• $K_{\text{BH}}$ is dissociation constant for conjugate acid, $\text{BH}$.

The base and acid is given as ${\text{NH}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$. Ammonia accepts a proton from ${\text{H}}_{\text{2}}\text{O}$ to form conjugate acid, ${\text{NH}}_4{}^{+}$. The ${\text{H}}_{\text{2}}\text{O}$ donates its proton to form conjugate base, ${}^{-}\text{OH}$.

The corresponding reaction is shown below.

${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$

The equation (1) for the reaction is written as shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{{\text{NH}}_4{}^{+}}/\text{p}{K}_{{\text{H}}_{\text{2}}\text{O}}}$…(2)

Where,

• $K_{{\text{NH}}_4{}^{+}}$ is the dissociation constant for conjugate acid.

• $K_{{\text{H}}_{\text{2}}\text{O}}$ is the dissociation constant for an acid.

The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_4{}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ from the Table 3.1 is $9.25$ and $15.7$ respectively.

Substitute the value of $\text{p}{K}_{\text{a}}$ for ${\text{NH}}_4{}^{+}$ and $\text{p}{K}_{\text{a}}$ for ${\text{H}}_{\text{2}}\text{O}$ in equation (2).

$\begin{array}{rll}{K}_{\text{eq}}& =& {10}^{9.25/15.7}\\ & =& {10}^{0.589}\\ & =& 3.88\end{array}$

Therefore, the equilibrium constant for the reaction is $3.88$.

Conclusion

The equation for the equilibria is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$. The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_4{}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ is $9.25$ and $15.7$ respectively.

Interpretation Introduction

(b)

Interpretation:

The equation for the equilibria is to be predicted. The $\text{p}{K}_{\text{a}}$ value associated with the acidic species in each equilibrium is to be predicted using Table 3.1. The reaction that has the larger $K_{\text{eq}}$ and is more important in an aqueous solution of ammonia is to be stated.

Concept introduction:

Dissociation constant $K_{\text{a}}$ is used to measure the strength of a Brønsted acid. There is an inverse relationship between the acidity of a Brønsted acid and its $\text{p}{K}_{\text{a}}$ value. The relation between $K_{\text{eq}}$ with $\text{p}{K}_{\text{a}}$ is given by an expression shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{\text{BH}}/\text{p}{K}_{\text{AH}}}$

Where,

• $K_{\text{AH}}$ is dissociation constant for an acid, $\text{AH}$.

• $K_{\text{BH}}$ is dissociation constant for conjugate acid, $\text{BH}$.

Answer to Problem 3.19P

The equation for the equilibria is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{H}}_3{\text{O}}^{+}+{\text{NH}}_2{}^{-}$. The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_{\text{3}}$ and ${\text{H}}_3{\text{O}}^{+}$ is $35$ and $-1.7$ respectively.

The equilibrium constant for the reaction is $0.89$. The reaction in part (a) has the larger value of $K_{\text{eq}}$, $3.88$. The reaction that is more important in an aqueous solution of ammonia is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$.

Explanation of Solution

The relation between $K_{\text{eq}}$ with $\text{p}{K}_{\text{a}}$ is given by an expression shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{\text{BH}}/\text{p}{K}_{\text{AH}}}$…(1)

Where,

• $K_{\text{AH}}$ is dissociation constant for an acid, $\text{AH}$.

• $K_{\text{BH}}$ is dissociation constant for conjugate acid, $\text{BH}$.

The acid and base is given as ${\text{NH}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$. Water accepts a proton from ${\text{NH}}_{\text{3}}$ to form conjugate acid, ${\text{H}}_3{\text{O}}^{+}$. The ${\text{NH}}_{\text{3}}$ donates its proton to form conjugate base, ${\text{NH}}_2{}^{-}$.

The corresponding reaction is shown below.

${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{H}}_3{\text{O}}^{+}+{\text{NH}}_2{}^{-}$

The equation (1) for the reaction is written as shown below.

$K_{\text{eq}}={10}^{\text{p}{K}_{{\text{H}}_3{\text{O}}^{+}}/\text{p}{K}_{{\text{NH}}_{\text{3}}}}$…(2)

Where,

• $K_{{\text{H}}_3{\text{O}}^{+}}$ is the dissociation constant for conjugate acid.

• $K_{{\text{NH}}_{\text{3}}}$ is the dissociation constant for an acid.

The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_{\text{3}}$ and ${\text{H}}_3{\text{O}}^{+}$ from Table 3.1 is $35$ and $-1.7$ respectively.

Substitute the value of $\text{p}{K}_{\text{a}}$ for ${\text{NH}}_4{}^{+}$ and $\text{p}{K}_{\text{a}}$ for ${\text{H}}_{\text{2}}\text{O}$ in equation (2).

$\begin{array}{rll}{K}_{\text{eq}}& =& {10}^{-1.7/35}\\ & =& {10}^{-0.049}\\ & =& 0.89\end{array}$

Therefore, the equilibrium constant for the reaction is $0.89$.

The reaction in part (a) has the larger value of $K_{\text{eq}}$, $3.88$. The reaction that is more important in an aqueous solution of ammonia is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$.

Conclusion

The equation for the equilibria is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{H}}_3{\text{O}}^{+}+{\text{NH}}_2{}^{-}$. The $\text{p}{K}_{\text{a}}$ of ${\text{NH}}_{\text{3}}$ and ${\text{H}}_3{\text{O}}^{+}$ is $35$ and $-1.7$ respectively.

The reaction in part (a) has the larger value of $K_{\text{eq}}$, $3.88$. The reaction that is more important in an aqueous solution of ammonia is ${\text{NH}}_{\text{3}}+{\text{H}}_{\text{2}}\text{O}\rightleftarrows {\text{NH}}_4{}^{+}+{}^{-}\text{OH}$.