Explanation

(b)
$H_{2}SO_{4} \rightleftharpoons 2H^{+} + SO_{4}^{2 -}$
1 mol 2 mol
0.005 M $H_{2}SO_{4} = 2 \times 0.005\ N\ H_{2}SO_{4}$
$= 0.01\ N\ H_{2}SO_{4}$
$\left\lbrack H^{+} \right\rbrack = 0.01$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 10^{- 2}$
$pH = - \log{\lbrack H^{+}\rbrack}$
$= - \log\left\lbrack 1 \times 10^{- 2} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ pH = 2$