b
\((a)\,\mathop {\mathop {{H_2}S{O_4}}\limits_{400 \times 1 = 40} }\limits_{20}  + \mathop {\mathop {2NaOH}\limits_{400 \times 1 = 40} }\limits_0  \to N{a_2}S{O_4} + 2{H_2}O\)

\(\therefore [{H^ + }] = \frac{{20 \times 2}}{{800}} = \frac{1}{{20}} \Rightarrow pH =  - log\left( {\frac{1}{{20}}} \right)\)

\(\therefore \,pH = 1.3\,\) so \((a)\) is correct

\((b)\,\log \,\left( {\frac{{{K_{{w_2}}}}}{{{K_{{w_1}}}}}} \right) = \frac{{\Delta H}}{{2.303\,R}}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]\)

so ionic product of water is temp. dependent hence \((b)\) is correct.

\((c)\,{K_a} = {10^{ - 5}},\,pH = 5 \Rightarrow [{H^ + }] = {10^{ - 5}}\)

\({K_a} = \frac{{c{\alpha ^2}}}{{(1 - \alpha )}} \Rightarrow {K_a} = \frac{{{{[H]}^ + }.\alpha }}{{(1 - \alpha )}}\)

\(\therefore \,{10^{ - 5}} = \frac{{{{10}^{ - 5}}.\alpha }}{{(1 - \alpha )}} \Rightarrow 1 - \alpha  = \alpha  = \frac{1}{2} = 50\% \)

so \((c)\) is correct

\((d)\) Le-chatelier's principle is applicable to cpmmon-Ion effect so option \((d)\) is wrong