Question

$20\mathrm{ml}$ of $0.1\mathrm{M}{\mathrm{H}}_2{\mathrm{SO}}_4$ is added to $30\mathrm{ml}$ of $0.2\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$ then calculate $\mathrm{pH}$ of resultant solution. (Given that ${\mathrm{pK}}_{\mathrm{b}}\mathrm{of}{\mathrm{NH}}_4\mathrm{OH}\mathrm{is}4.7$)

• A. $9$
• B. $9.4$
• C. $5.2$
• D. $5$

Answer 1

The correct option is A

$9$

Explanation for the correct option:

Step 1: Given data:

Volume of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is ${\mathrm{V}}_1=20\mathrm{mL}$

Volume of ${\mathrm{NH}}_4\mathrm{OH}$ is ${\mathrm{V}}_2=30\mathrm{mL}$

Concentration of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is $M_1=0.1\mathrm{M}$

Concentration of ${\mathrm{NH}}_4\mathrm{OH}$ is $M_2=0.2\mathrm{M}$

Base dissociation constant ${\left({\mathrm{pK}}_{\mathrm{b}}\right)}_{\left({\mathrm{NH}}_4\mathrm{OH}\right)}=4.7$

Step 2: Find the number of moles of the reactants and products:

1. $$\begin{gathered} {\mathrm{n}}_{\left(H_{2}S{O}_4\right)}=\mathrm{Concentration}\mathrm{of}{H}_{2}S{O}_4\mathit{}\mathit{\times }\mathit{}Volume\mathit{}of\mathit{}{H}_{\mathit{2}}S{O}_{\mathit{4}} \\ \mathit{\Rightarrow }{\mathrm{n}}_{\left(H_{\mathrm{2}}S{O}_{\mathrm{4}}\right)}=0.2\mathrm{M}\times 20\mathrm{mL}\left[\because 1H_{\mathit{2}}S{O}_{\mathit{4}}\mathit{}\mathit{\to }\mathit{}\mathit{2}\mathit{}{H}^{\mathit{+}}\right] \\ \Rightarrow {\mathrm{n}}_{\left(H_{\mathrm{2}}S{O}_{\mathrm{4}}\right)}=4\mathrm{m}\mathrm{mol} \end{gathered}$$
2. $$\begin{gathered} {\mathrm{n}}_{\left(N{H}_{4}OH\right)}=\mathrm{Concentration}\mathrm{of}N{H}_{\mathit{4}}OH\mathit{}\mathit{\times }\mathit{}Volume\mathit{}of\mathit{}N{H}_{\mathit{4}}OH \\ \mathit{\Rightarrow }{\mathrm{n}}_{\left(N{H}_{\mathrm{4}}OH\right)}=0.2\mathrm{M}\times 30\mathrm{mL} \\ \Rightarrow {\mathrm{n}}_{\left(N{H}_{\mathrm{4}}OH\right)}=6\mathrm{m}\mathrm{mol} \end{gathered}$$
3. The balanced equation is

$\underset{Final}{\underset{Initial}{}}\underset{0\mathrm{m}\mathrm{mol}}{\underset{4\mathrm{m}\mathrm{mol}}{{\mathrm{H}}_2{\mathrm{SO}}_4}}+\underset{2\mathrm{m}\mathrm{mo}}{\underset{6\mathrm{m}\mathrm{mol}}{2{\mathrm{NH}}_4\mathrm{OH}}}\to \underset{4\mathrm{m}\mathrm{mol}}{\underset{0\mathrm{m}\mathrm{mol}}{{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4}}+\hspace{0.17em}2{\mathrm{H}}_2\mathrm{O}$

4. The number of moles of ${\left({\mathrm{NH}}_4\right)}_{2}S{O}_4$ is ${\mathrm{n}}_{\mathit{\left(}{\left({\mathrm{NH}}_{\mathrm{4}}\right)}_{\mathit{2}}\mathit{S}{\mathit{O}}_{\mathit{4}}\mathit{\right)}}=4\mathrm{m}\mathrm{mol}$

Here, the number of moles of ${\mathrm{NH}}_4\mathrm{OH}$ is more than the number of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$.

Therefore, the obtained product is a Basic Buffer solution.

Step 3: Find the $\mathbf{pOH}$ of the solution obtained:

$$\begin{gathered} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]} \\ \Rightarrow \mathrm{pOH}=4.7+\log \frac{4}{2} \\ \Rightarrow \mathrm{pOH}=4.7+0.3 \\ \Rightarrow \mathrm{pOH}=5 \end{gathered}$$

Step 4 Find the $\mathbf{pH}$ of resultant solution:

$$\begin{gathered} \mathrm{pH}+\mathrm{pOH}=14 \\ \Rightarrow \mathrm{pH}=14-5 \\ \Rightarrow \mathrm{pH}=9 \end{gathered}$$

Hence, option A is correct. The $\mathbf{pH}$ of resultant solution is $\mathbf{9}$.