Chapter 16, Problem 129SCQ

Consider a salt of a weak base and a weak acid such as ammonium cyanide. Both the NH₄⁺ and CN⁻ ions interact with water in aqueous solution, but the net reaction can be considered as a proton transfer from NH₄⁺ to CN⁻.

NH₄⁺(aq) + CN⁻(aq) ⇌ NH₃(aq) + HCN(aq)

1. (a) Show that the equilibrium constant for this reaction, K_act, is
2. 1. $K_{\text{net}}=\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}$

   where K_a is the ionization constant for the weak acid NH₄⁺ and K_b, is the constant for the weak base CN⁻.

3. (b) Calculate K_net values fur each of the following, NH₄CN, NH₄CH₃CO₂, and NH₄K Which salt has the largest value of K_act, and why?
4. (c) Predict whether a solution of each of the compounds in (b) is acidic or basic. Explain how you made this prediction (Yon do not need a calculation to make this prediction.)

Interpretation Introduction

Interpretation:

The equilibrium constant for the reaction has to be determined.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

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

${\text{K}}_{\text{b}}$ is base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

The value of $K_{\text{net}}$ is derived from three equations. The three equations represent acid ionization constant for a weak acid, base ionization constant for a weak base and autoionization of water.

Answer to Problem 129SCQ

The value of the equilibrium constant is derived to be,

$K_{\text{net}}\text{=}\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}$

Explanation of Solution

For ammonium ion;

${\text{NH}}_{\text{4}}{}^{+}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftarrows {\text{NH}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)$                                                         (1)

From equation $1$,

$K_1=K_{\text{a}}$

For cyanide ion;

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

From equation 2,

$K_2=K_{\text{b}}$

For autoionization of water;

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

From equation 3,

$K_3=1/K_{\text{w}}$

Add equation 1,2 and 3,

${\text{NH}}_{\text{4}}{}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftarrows {\text{NH}}_{\text{3}}\left(aq\right)+\text{HCN}\left(aq\right)$                                                       (4)

Therefore,

$\begin{array}{c}{K}_{\text{net}}=K_1{K}_2{K}_3\\ =\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}\end{array}$

Interpretation Introduction

Interpretation:

The $K_{\text{net}}$ values has to be calculated for three salts and salt with the largest value of the equilibrium constant is to be identified.

Concept Introduction:

The value of $K_{\text{net}}$ is derived from three equations. The three equations represent acid ionization constant for a weak acid, base ionization constant for a weak base and autoionization of water.

Value of $K_{\text{net}}$ is calculated for the three salts (of a weak acid and weak base) by using the equation given below:

$K_{\text{net}}\text{=}\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}$

From a comparison of values of the acid dissociation constant and base dissociation constant, nature of the solution is predicted.

If the value of acid dissociation constant is more than the value of base dissociation constant, then the products will have more acidic species and therefore the nature of solution will be acidic.

On the same concept, if the value of base dissociation constant is more than acid dissociation constant then the solution will be basic.

If the value of acid and base dissociation constant is equal then the solution becomes neutral.

Answer to Problem 129SCQ

The value of $K_{\text{net}}$ for ${\text{NH}}_{\text{4}}\text{CN}$ is $1.4$.

The value of $K_{\text{net}}$ for ${\text{NH}}_{\text{4}}{\text{CH}}_3{\text{CO}}_2$ is $3.1\times {10}^{-5}$

The value of $K_{\text{net}}$ for ${\text{NH}}_{\text{4}}\text{F}$ is $7.8\times {10}^{-7}$

Explanation of Solution

For values of acid or base ionization constant refer to Table $16.2$.

The value for $K_{\text{w}}$ is ${10}^{-14}$.

It is derived that;

$K_{\text{net}}=\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}$

For ${\text{NH}}_{\text{4}}\text{CN}$,

${\text{NH}}_{\text{4}}{}^{+}\left(aq\right)+{\text{CN}}^{-}\left(aq\right)\rightleftarrows {\text{NH}}_{\text{3}}\left(aq\right)+\text{HCN}\left(aq\right)$

$\begin{array}{c}{K}_{\text{net}}=\frac{\left(5.6\times {10}^{-10}\right)\left(2.5\times {10}^{-5}\right)}{1\times {10}^{-14}}\\ =1.4\end{array}$

For ${\text{NH}}_{\text{4}}{\text{CH}}_3{\text{CO}}_2$,

${\text{NH}}_{\text{4}}{}^{+}\left(aq\right)+{\text{CH}}_3{\text{COO}}^{-}\left(aq\right)\rightleftarrows {\text{NH}}_{\text{3}}\left(aq\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)$

$\begin{array}{c}{K}_{\text{net}}=\frac{\left(5.6\times {10}^{-10}\right)\left(5.6\times {10}^{-10}\right)}{{10}^{-14}}\\ =3.1\times {10}^{-5}\end{array}$

For ${\text{NH}}_{\text{4}}\text{F}$,

${\text{NH}}_{\text{4}}{}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)\rightleftarrows {\text{NH}}_{\text{3}}\left(aq\right)+\text{HF}\left(aq\right)$

$\begin{array}{c}{K}_{\text{net}}=\frac{\left(5.6\times {10}^{-10}\right)\left(1.4\times {10}^{-11}\right)}{{10}^{-14}}\\ =7.8\times {10}^{-7}\end{array}$

Value of $K_{\text{net}}$ is largest for ${\text{NH}}_{\text{4}}\text{CN}$ among the three-given species.

It is inferred from the value of $K_{\text{net}}$ that only cyanide ion can abstract a proton from ammonium ion strongly enough to give a significant amount of products

Interpretation Introduction

Interpretation:

Nature of solution of different salts is predicted according to their acid or base ionization constant.

Concept Introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

${\text{K}}_{\text{a}}$ is an acid constant for equilibrium reactions.

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

${\text{K}}_{\text{b}}$ is a base constant for equilibrium reaction.

$\begin{array}{l}{\text{BOH + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{B}}^{\text{+}}{\text{+ OH}}^{\text{-}}\\ {\text{K}}_{\text{a}}\text{= }\frac{{\text{[B}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}}{\text{[BOH]}}\end{array}$

Ion product constant for water

$\begin{array}{l}{\text{ K}}_{\text{w}}{\text{= [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ \text{ =1}{\text{.00×10}}^{\text{-14}}\\ {\text{ pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ {\text{pOH= -log[OH}}^{\text{-}}\text{]}\end{array}$

Relation between $pHandpOH$

$\text{ pH + pOH =14}$

The value of $K_{\text{net}}$ is derived from three equations. The three equations represent acid ionization constant for a weak acid, base ionization constant for a weak base and autoionization of water.

Value of $K_{\text{net}}$ is calculated for the three salts (of a weak acid and weak base) by using the equation given below:

$K_{\text{net}}\text{=}\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}$

From a comparison of values of the acid dissociation constant and base dissociation constant, nature of the solution is predicted.

If the value of acid dissociation constant is more than the value of base dissociation constant, then the products will have more acidic species and therefore the nature of solution will be acidic.

On the same concept, if the value of base dissociation constant is more than acid dissociation constant then the solution will be basic.

If the value of acid and base dissociation constant is equal then the solution becomes neutral.

Answer to Problem 129SCQ

${\text{NH}}_{\text{4}}\text{CN}$ is a basic solution.

${\text{NH}}_{\text{4}}{\text{CH}}_3{\text{CO}}_2$ is the neutral solution.

${\text{NH}}_{\text{4}}\text{F}$ is an acidic solution.

Explanation of Solution

If the value of acid dissociation constant is more than the value of base dissociation constant, then the products will have more acidic species and therefore the nature of solution will be acidic.

On the same concept, if the value of base dissociation constant is more than acid dissociation constant then the solution will be basic.

If the value of acid and base dissociation constant is equal then the solution becomes neutral.

${\text{NH}}_{\text{4}}\text{CN}$ is basic in nature because

$\begin{array}{l}{K}_{\text{b}}\left({\text{CN}}^{-}\right)>K_{\text{a}}\left({\text{NH}}_{\text{4}}{}^{+}\right)\\ \left(2.5\times {10}^{-5}\right)>\left(5.6\times {10}^{-10}\right)\end{array}$

${\text{NH}}_{\text{4}}{\text{CH}}_3{\text{CO}}_2$ is a neutral solution.

$\begin{array}{l}{K}_{\text{b}}\left({\text{CH}}_3{\text{COO}}^{-}\right)=K_{\text{a}}\left({\text{NH}}_{\text{4}}{}^{+}\right)\\ \left(5.6\times {10}^{-10}\right)=\left(5.6\times {10}^{-10}\right)\end{array}$

${\text{NH}}_{\text{4}}\text{F}$ is an acidic solution

$\begin{array}{l}{K}_{\text{b}}\left({\text{F}}^{-}\right)<K_{\text{a}}\left({\text{NH}}_{\text{4}}{}^{+}\right)\\ \left(1.4\times {10}^{-11}\right)<\left(5.6\times {10}^{-10}\right)\end{array}$