Chapter 15, Problem 15.121QP

Calculate the pH of a 2.00 M NH₄CN solution.

Interpretation Introduction

Interpretation:

The pH of a 2.00 M ${\text{NH}}_{\text{4}}\text{CN}$ solution has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}\stackrel{}{\to }{\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between ${\text{K}}_{\text{a}}$and $K_b$

${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

To Calculate: The pH of a 2.00 M ${\text{NH}}_{\text{4}}\text{CN}$ solution

Answer to Problem 15.121QP

The pH of 2.00 M ${\text{NH}}_{\text{4}}\text{CN}$ solution is 11.80

Explanation of Solution

We examine the hydrolysis of the cation and anion separately

${\text{NH}}_{\text{4}}{\text{CN}}_{\text{(aq)}}\to {\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+CN}}^{\text{-}}{}_{\text{(aq)}}$

Hydrolysis of cation: ${\text{NH}}_{\text{4}}{{}^{+}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{NH}}_{\text{3}}{}_{(aq)}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{(aq)}$

	${\text{NH}}_{\text{4}}{{}^{+}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{\rightleftarrows }{\to }{\text{NH}}_{\text{3}}{}_{(aq)}+{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{(aq)}$
Initial $(M)$	2.00 -x 2.00-x	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

Concentration of hydronium ion:

The ${\text{K}}_{\text{b}}$ value for ${\text{NH}}_{\text{3}}$ is $1.8\times {10}^{-5}$

We know that ${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

${\text{K}}_{\text{a}}$ for ${\text{NH}}_{\text{4}}{}^{\text{+}}$ can be calculated as follows,

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{{\text{K}}_w}{{\text{K}}_b}\\ \text{ =}\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ \text{ =5}\text{.6}\times {\text{10}}^{\text{-10}}\end{array}$

Using the obtained ${\text{K}}_{\text{a}}$ for ${\text{NH}}_{\text{4}}{}^{\text{+}}$, hydronium ion concentration is calculated as follows,

 $\begin{array}{l}{\text{K}}_{\text{a}}=\frac{{\text{[NH}}_{\text{3}}{\text{][H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}}\\ 5.6\times {10}^{-10}=\frac{x^2}{2.00-x}\\ \approx \frac{x^2}{2.00}\\ x=3.35\times {10}^{-5}\text{ M}\\ {\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=}3.35\times {10}^{-5}\text{ M}\end{array}$

Therefore, the hydronium ion concentration ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]=}3.35\times {10}^{-5}\text{ M}$

Hydrolysis of anion: ${\text{CN}}^{-}{}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HCN}}_{(aq)}+{\text{OH}}^{-}{}_{(aq)}$

	${\text{CN}}^{-}{}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HCN}}_{(aq)}+{\text{OH}}^{-}{}_{(aq)}$
Initial $(M)$	2.00 -y 2.00-y	$0.00$	$0.00$
Change $(M)$		$+y$	$+y$
Equilibrium $(M)$		$y$	$y$

Concentration of hydroxide ion:

The ${\text{K}}_{\text{a}}$ value for $\text{HCN}$ is $4.9\times {10}^{-10}$

We know that ${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}{\text{=K}}_{\text{w}}$

${\text{K}}_b$ for ${\text{CN}}^{\text{-}}$ can be calculated as follows,

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \text{ =}\frac{1.0\times {10}^{-14}}{4.9\times {10}^{-10}}\\ \text{ =2}\text{.0}\times {\text{10}}^{\text{-5}}\end{array}$

Using the obtained ${\text{K}}_b$ for ${\text{CN}}^{\text{-}}$, hydroxide ion concentration is calculated as follows,

$\begin{array}{l}{\text{K}}_b=\frac{{\text{[HCN][OH}}^{\text{-}}\text{]}}{{\text{[CN}}^{\text{-}}\text{]}}\\ 2.0\times {10}^{-5}=\frac{y^2}{2.00-y}\\ \approx \frac{y^2}{2.00}\\ x=6.32\times {10}^{-3}\text{ M}\\ {\text{[OH}}^{-}\text{]=6}\text{.32}\times {10}^{-3}\text{ M}\end{array}$

Therefore, concentration of hydroxide ion is ${\text{[OH}}^{-}\text{]=6}\text{.32}\times {10}^{-3}\text{ M}$

pH calculation:

${\text{CN}}^{\text{-}}$ is stronger as a base than ${\text{NH}}_{\text{4}}{}^{\text{+}}$ is as an acid.

Some ${\text{OH}}^{\text{-}}$ produced from the hydrolysis of ${\text{CN}}^{\text{-}}$ will be neutralized by ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ produced from the hydrolysis of ${\text{NH}}_{\text{4}}{}^{\text{+}}$

	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{(aq)}+{\text{OH-}}_{(aq)}\to 2{\text{H}}_{\text{2}}{\text{O}}_{(l)}$
Initial $(M)$	$3.35\times {10}^{-5}$ $-3.35\times {10}^{-5}$ 0	$6.32\times {10}^{-3}$
Change $(M)$		$-3.35\times {10}^{-5}$
Equilibrium $(M)$		$6.29\times {10}^{-3}$

$\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]=6}\text{.29}\times {\text{10}}^{\text{-3}}M\\ \text{pOH =-log(6}\text{.29}\times {\text{10}}^{\text{-3}})\\ \text{ =2}\text{.20}\end{array}$

The pH can be calculated as follows,

$\begin{array}{l}\text{pH+pOH=14}\\ \text{pH=14-pOH}\\ \text{ =14-2}\text{.20}\\ \text{ =11}\text{.80}\end{array}$

Therefore, the pH of the given ${\text{NH}}_{\text{4}}\text{CN}$ solution is 11.80

Conclusion

The pH of a 2.00 M ${\text{NH}}_{\text{4}}\text{CN}$ solution was calculated