Question

What concentration of ${\mathbf{\text{NH}}}_{\mathbf{\text{4}}}\mathbf{\text{Cl}}$ is necessary to buffer a $\mathbf{0.52}\mathbf{\text{-M}}$ ${\mathbf{\text{NH}}}_{\mathbf{3}}$solution at $\mathbf{\text{pH}}\mathbf{=}\mathbf{9.00}$?

(${\mathbf{\text{K}}}_{\mathbf{\text{b}}}{\mathbf{\text{\hspace{0.17em}for\hspace{0.17em}NH}}}_{\mathbf{3}}\mathbf{=}\mathbf{1.8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$.)

Short answer

The concentration of ${\text{NH}}_4\text{Cl}$ which requires a preparation of${\text{0.52\hspace{0.17em}M\hspace{0.17em}NH}}_3$ solution at$\text{pH\hspace{0.17em}=\hspace{0.17em}9}$ is$\text{0.94M}$ .

Step-by-step solution

Given information

The concentration of ammonia is 0.52 M. The buffer and${\mathrm{K}}_{\mathrm{b}}\left({\mathrm{NH}}_3\right)$ are 9 and $1.8\times {10}^{-5}.$Here, the pOH of the buffer is,

$\begin{array}{c}\mathrm{pH}+\mathrm{pOH}=14\\ \mathrm{pOH}=14-\mathrm{pH}\\ \mathrm{pOH}=14-9\\ \mathrm{pOH}=5\end{array}$

The concentration of NH4Cl.

As per the Henderson-Hasselbalch equation,

$\begin{array}{c}\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}\left({\mathrm{NH}}_3\right)+\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{\left[{\mathrm{NH}}_3\right]}\\ {\mathrm{K}}_{\mathrm{b}}\left({\mathrm{NH}}_3\right)=1.8\times {10}^{-5}\\ {\mathrm{pK}}_{\mathrm{b}}=-\log \left({\mathrm{K}}_{\mathrm{b}}\right)\end{array}$

Hence,

$\begin{array}{c}\mathrm{pOH}=-\log \left({\mathrm{K}}_{\mathrm{b}}\right)+\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{\left[{\mathrm{NH}}_3\right]}\\ 5=-\log (1.8\times {10}^{-5})+\log \frac{\left[{\mathrm{NH}}_4\mathrm{Cl}\right]}{[0.52\mathrm{M}]}\end{array}$

And the concentration is,

$\begin{array}{l}\left[{\mathrm{NH}}_4\mathrm{Cl}\right]=0.52\left({10}^{0.255}\right)\\ \left[{\mathrm{NH}}_4\mathrm{Cl}\right]=0.94\mathrm{M}\end{array}$