A buffer solution is prepared in which the concentration of $\mathrm{NH}_{3}$ is $0.30 \mathrm{M}$ and the concentration of $\mathrm{NH}_{4}^{+}$is $0.20 \mathrm{M}$. If the equilibrium constant, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ equals $1.8 \mathrm{x}$ $10^{-5}$, what is the pH of this solution? ( $\log 2.7=0.43$ )

Question

A buffer solution is prepared in which the concentration of $\mathrm{NH}_{3}$ is $0.30 \mathrm{M}$ and the concentration of $\mathrm{NH}_{4}^{+}$is $0.20 \mathrm{M}$. If the equilibrium constant, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{NH}_{3}$ equals $1.8 \mathrm{x}$ $10^{-5}$, what is the pH of this solution? ( $\log 2.7=0.43$ )

3 Answers

kritika · Answer 1 · 2021-12-20T08:27:05+0000

The correct option is: (b) $9.43$
Explanation:
$$
\begin{aligned}
&{\left[\mathrm{NH}_{3}\right]=0.30 \mathrm{M}, K_{b}=1.8 \times 10^{-5}} \\
&{\left[\mathrm{NH}_{4}^{+}\right]=0.20 \mathrm{M}} \\
&\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\text { salt }]}{[\text { base }]} \\
&=4.74+\log \frac{0.2}{0.3}=4.56 \\
&\mathrm{pH}=(14-4.56)=9.44
\end{aligned}
$$

Your answer

3 Answers

Your comment on this answer:

Your comment on this answer:

Your comment on this answer:

Related questions

Categories