$p H=p K_{a}+\log \frac{| \text {salt}|}{\text { lacid }}$ and $\left[H^{+}\right]=-$ antilogp $H$

$p H=-\log K_{a}+\log \frac{| \text {eg } t |}{| \text { accid }}$

$\left[\therefore p K_{a}=-\log K_{a}\right]$

$=-\log \left(1.8 \times 10^{-5}\right)+\log \frac{0.20}{0.10}$

$=4.74+\log 2$

$=4.74+0.3010=5.041$

Now, $\left[H^{+}\right]=$ antilog $(-5.045)$

$=9.0 \times 10^{-6}\, \mathrm{mol} / L$