Question

Calculate$\left[{\mathbf{H}}_3{\mathbf{O}}^{+}\right]$in a solution that contains$0.100\text{mol}$ of${\mathbf{NH}}_4\mathbf{CN}$per liter.

$\begin{array}{c}{\mathbf{NH}}_4^{+}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\rightleftharpoons {\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{N}{\mathbf{H}}_3\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ {\mathbf{K}}_a\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{-10}\\ \mathbf{HCN}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\rightleftharpoons {\mathbf{H}}_3{\mathbf{O}}^{+}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{N}}^{-}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ {\mathbf{K}}_a\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{17}\mathbf{\times }{\mathbf{10}}^{-10}\end{array}$

Short answer

The required value of$\left[{\text{H}}_3{\text{O}}^{+}\right]$ is$5.36\times {10}^{-11}\text{M}$

Step-by-step solution

Given Reactions

Consider the following are the reactions:

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

The$K_{a_1}=5.6\times {10}^{-10}$

$\text{HCN}+{\text{H}}_2\text{O}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}+{\text{CN}}^{-}$

The$K_{a_2}=6.17\times {10}^{-10}$

Subtract the second reaction from the first to arrive at:

${{\text{NH}}_3}^{+}+{\text{CN}}^{-}\to {\text{NH}}_3+\text{HCN}$

Finding the value of K

To get the value of

$\begin{array}{c}K=\frac{\left[{\text{NH}}_3\right]\left[\text{HCN}\right]}{\left[{\text{CN}}^{-}\right]\left[{{\text{NH}}_4}^{+}\right]}\\ =\frac{K_{a_1}}{K_{a_2}}K\\ =\frac{5.6\times {10}^{-10}}{6.17\times {10}^{-10}}K\\ =0.90762\end{array}$

Substitute the value to get the equilibrium concentrations

The following are the equilibrium concentrations:

$\begin{array}{c}\left[{\text{NH}}_3\right]=\left[\text{HCN}\right]\\ =x\\ \left[{\text{CN}}^{-}\right]=\left[{{\text{NH}}_4}^{+}\right]\\ =0.1-x\end{array}$

Letinsert numbers into the equation, we get:

$\begin{array}{c}0.90762=\frac{x^2}{{(0.1-x)}^2}\\ 0.95269=\frac{x^2}{0.1-x}\end{array}$

$x^2+0.95269x-0.095269=0$

$x=0.091258$

The$\left[{\text{NH}}_3\right]$ concentrationis as follows:

$\begin{array}{c}\left[{\text{NH}}_3\right]=x\\ =0.091258\text{M}\end{array}$

The$\left[{{\text{NH}}_4}^{+}\right]$ concentrationis as follows:

$\begin{array}{c}\left[{{\text{NH}}_4}^{+}\right]=0.1-x\\ =0.1-0.091258\\ =8.742\times {10}^{-3}\text{M}\end{array}$

Finding the value of H3O+

Consider that$K_{a_1}$is:

$K_{a_1}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{NH}}_3\right]}{\left[{{\text{NH}}_4}^{+}\right]}$

The required concentration of$\left[{\text{H}}_3{\text{O}}^{+}\right]$ is:

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{K_{a_1}\cdot \left[{{\text{NH}}_4}^{+}\right]}{\left[{\text{NH}}_3\right]}\left[{\text{H}}_3{\text{O}}^{+}\right]\\ =\frac{5.6\times {10}^{-10}\times 8.742\times {10}^{-3}}{0.091258}\left[{\text{H}}_3{\text{O}}^{+}\right]\\ =5.36\times {10}^{-11}\text{M}\end{array}$

Hence, the$\left[{\text{H}}_3{\text{O}}^{+}\right]$ concentration is $5.36\times {10}^{-11}\text{M}$.