$(K_w = 10^{-14})$
$K_{w} \text { of } H_{2} O=10^{-14}=\left[H^{+}\right]\left[O H^{-}\right]$
or $\left[H^{+}\right]=10^{-7} M$ (due to its neutral behaviour)
So, in aqueous solution of $10^{-8} \mathrm{MHCl}$
$\left[H^{+}\right] =\left[H^{+}\right] \text {of } \mathrm{HCl}+\left[H^{+}\right] \text {of water }$
$=10^{-8}+10^{-7}$
$=11 \times 10^{-8} M \approx 1.10 \times 10^{-7} M$
[અહીં $K_b\,(NH_4OH) = 10^{-5}$ અને $log\,2 = 0.301$ ]