What is $\left[\mathrm{H}^{+}\right]$in $\mathrm{mol} / \mathrm{L}$ of a solution that is $0.20 \mathrm{M}$…
What is $\left[\mathrm{H}^{+}\right]$in $\mathrm{mol} / \mathrm{L}$ of a solution that is $0.20 \mathrm{M}$ in $\mathrm{CH}_3 \mathrm{COONa}$ and $0.10 \mathrm{M}$ in $\mathrm{CH}_3 \mathrm{COOH}$ ? $\left(\mathrm{K}_{\mathrm{a}}\right.$ for $\mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5}$ )
$3.5 \times 10^{-4}$
$1.1 \times 10^{-5}$
$1.8 \times 10^{-5}$
$9.0 \times 10^{-6}$
Solution
Key Idea $\mathrm{CH}_3 \mathrm{COOH}$ (weak acid) and $\mathrm{CH}_3 \mathrm{COONa}$ (conjugated salt) form acidic buffer and for acidic buffer,
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{[\text { salt }]}{[\text { acid }]}$
and
$\begin{aligned}
{\left[\mathrm{H}^{+}\right] } & =-\operatorname{antilog} \mathrm{pH} \\
\mathrm{pH} & =-\log [\mathrm{K}_{\mathrm{a}}]+\log \frac{[\text { salt }]}{[\text { acid }]} \\
& \left.\quad \because \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}}\right] \\
& =-\log \left(1.8 \times 10^{-5}\right)+\log \frac{(0.20)}{(0.10)} \\
& =4.74+\log 2 \\
& =4.74+0.3010=5.041
\end{aligned}$
Now,
$\begin{gathered}
{\left[\mathrm{H}^{+}\right]=\operatorname{antilog}(-5.045)} \\
=9.0 \times 10^{-6} \mathrm{~mol} / \mathrm{L}
\end{gathered}$