\(p H=p K_{a}+\log \frac{| \text {salt}|}{\text { lacid }}\) 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 \((-5.045)\)
\(=9.0 \times 10^{-6}\, \mathrm{mol} / L\)
$(I)\,H_2O$ $(II)\,\,NH_2$ $(III)\,\,H_2PO_4^-$ $(IV)\,\,HCO_3^-$