One litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_{3}$ and $\mathrm{NH}

Question

One litre buffer solution was prepared by adding 0.10 mol each of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4} \mathrm{Cl}$ in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is $\_\_\_\_$ $	imes 10^{-2}$. (Nearest integer)\ Given: $\mathrm{p} K_{b}$ of $\mathrm{NH}_{3}=4.745$ and $\log _{10} 3=0.477$\

Professor PC Thomas Classes · Answer

(48) : In resultant solution,\ $n_{\mathrm{NH}_{3}}=0.1-0.05=0.05 \mathrm{~mol}$\ $n_{\mathrm{NH}_{4} \mathrm{Cl}}=0.1+0.05=0.15 \mathrm{~mol}$ $$ \begin{aligned} \mathrm{pOH} & =\mathrm{p} K_{b}+\log \frac{\left[\mathrm{NH}_{4}^{+}ight]}{\left[\mathrm{NH}_{3}ight]} \ & =4.745+\log \frac{0.15}{0.05} \ & =4.745+\log 3 \ & =4.745+0.477=5.222 \end{aligned} $$ $\mathrm{pH}=14-5.222=8.778$\ Initialy $n_{\mathrm{NH}_{3}}=n_{\mathrm{NH}_{4}^{+}}=0.1$\ $	herefore \quad \mathrm{pOH}=\mathrm{p} K_{b}=4.745$\ $\mathrm{pH}=14-4.745=9.255$\ $\Delta \mathrm{pH}=9.255-8.778=0.477$\ or $\quad 47.7 	imes 10^{-2}$\