Question

A buffered solution is made by adding$\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{g}\mathbf{ }\mathit{N}{\mathit{H}}_{\mathbf{4}}\mathit{C}\mathit{l}\mathbf{ }\mathit{t}\mathit{o}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathit{L}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{a}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{75}\mathbf{ }\mathit{M}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{N}{\mathit{H}}_{\mathbf{3}}$. Calculate the pH of the final solution. (Assume no volume change)

Short answer

The pH of the solution is prepared by adding $\text{50} \text{g} \text{of} {\text{NH}}_{\text{4}}\text{Cl} \text{to} \text{1.0} \text{L} \text{of} \text{0.75} \text{M} {\text{NH}}_{\text{3}}$is$9.16$ .

Step-by-step solution

Definition of pH

The pH is a quantitative measureof the acidity or basicity of aqueous or other liquid solutions.

Calculation of pH

Mass of${\text{NH}}_{\text{4}}\text{Cl}=50.0\text{g}$${\text{NH}}_{\text{4}}\text{Cl}=50.0\text{g}$

Molarmass of${\text{NH}}_{\text{4}}\text{Cl}=53.49\text{g/mol}$

Molarity$=0.935 \text{M}$

Calculation of pH

$$\begin{gathered} \text{pOH}={\text{pK}}_{\text{b}}{\text{(NH}}_{\text{3}}\text{)}+\log \frac{\left[{\text{NH}}_{\text{4}}\text{Cl}\right]}{\left[{\text{NH}}_{\text{3}}\right]} \\ {\text{K}}_{\text{b}}{\text{(NH}}_{\text{3}}\text{)}=1.8\times {10}^{-5} \\ {\text{pK}}_{\text{b}}=-\log \left({\text{K}}_{\text{b}}\right) \end{gathered}$$

$$\begin{gathered} \text{pOH}=-\log \left({\text{K}}_{\text{b}}\right)+\log \frac{\left[{\text{NH}}_{\text{4}}\text{Cl}\right]}{\left[{\text{NH}}_{\text{3}}\right]} \\ \text{pOH}=-\log (1.8\times {10}^{-5})+\log \frac{\left[{\text{NH}}_{\text{4}}\text{Cl}\right]}{\left[{\text{NH}}_{\text{3}}\right]} \\ \text{pOH}=4.745+0.0957 \\ \text{pOH}=4.84 \end{gathered}$$

$$\begin{gathered} \text{pH + pOH}=14 \\ \text{pH}=14-\text{pOH} \\ \text{pH}=14-4.84 \\ \text{pH}=9.16 \end{gathered}$$

The $\text{pH}$of the solution is $9.16$.