What is $\left[H^{+}\right]$ in $mol/L$ of a solution that is $0.20M$ in $C{H}_{3}COONa$ and $0.1M$ in $C{H}_{3}COOH$? $K_a$ for $C{H}_{3}COOH$ is $1.8\times {10}^{-5}$?

Key Idea - $C{H}_{3}COOH$ (weak acid) and $C{H}_{3}COONa$ (conjugated salt) form acidic buffer and for acidic buffer,
$pH=p{K}_a+\log .\frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$
and $\left[H^{+}\right]=-$ antilog pH
$pH=-{\log K}_a+\frac{\log \left(\left[\text{salt}\right]\right)}{\left[\text{acid}\right]}$
$\left[\because p{K}_a=-{\log K}_a\right]$
$=-\log \left(1.8\times {10}^{-5}\right)+\log .\frac{\left(0.20\right)}{\left(0.10\right)}$
$=4.74+\log 2$
$=4.74+0.3010=5.41$
Now, $\left[H^{+}\right]=$ antilog (-5.045)
$9.0\times {10}^{-6}mol/L$