$(K_w = 10^{-14})$
\(K_{w} \text { of } H_{2} O=10^{-14}=\left[H^{+}\right]\left[O H^{-}\right]\)
or \(\left[H^{+}\right]=10^{-7} M\) (due to its neutral behaviour)
So, in aqueous solution of \(10^{-8} \mathrm{MHCl}\)
\(\left[H^{+}\right] =\left[H^{+}\right] \text {of } \mathrm{HCl}+\left[H^{+}\right] \text {of water }\)
\(=10^{-8}+10^{-7}\)
\(=11 \times 10^{-8} M \approx 1.10 \times 10^{-7} M\)
$(1) 0.005\, M\, H_2SO_4\,(2) 0.1\, M\, Na_2SO_4\,(3) 10^{-2}\, M\, NaOH\,(4) 0.01 \,M\, HCl$