Solution of $0.1 \mathrm{~N} \mathrm{NH}_{4} \mathrm{OH}$ and $0.1 \mathrm{~N} \mathrm{NH}_{4} \math

Question

Solution of $0.1 \mathrm{~N} \mathrm{NH}_{4} \mathrm{OH}$ and $0.1 \mathrm{~N} \mathrm{NH}_{4} \mathrm{Cl}$ has pH 9.25. Then find out $\mathrm{p} K_{b}$ of $\mathrm{NH}_{4} \mathrm{OH}$.\

Professor PC Thomas Classes · Answer

(b) : Solution of $0.1 \mathrm{~N} \mathrm{NH}_{4} \mathrm{OH}$ and $0.1 \mathrm{~N} \mathrm{NH}_{4} \mathrm{Cl}$ is a buffer solution.\ According to Henderson equation, the pH of a basic buffer,\ $\mathrm{pH}=14-\mathrm{p} K_{b}-\log \frac{[	ext { Salt }]}{[	ext { Base }]} \Rightarrow \mathrm{p} K_{b}=14-\mathrm{pH}-\log \frac{[	ext { Salt }]}{[	ext { Base }]}$\ $\Rightarrow \mathrm{p} K_{b}=14-9.25-\log \frac{0.1}{0.1}$\ $\Rightarrow \mathrm{p} K_{b}=14-9.25=4.75 \quad 	herefore \mathrm{p} K_{b}$ of $\mathrm{NH}_{4} \mathrm{OH}=4.75$