Explanation

(b)
Given, for $NaOH,V = 10\ mL,\ N = 0.1\ N$
For $H_{2}SO_{4},\ V = 10\ mL,\ N = 0.05\ N$
Miliequivalents of $NaOH = 10 \times 0.1 = 1$
Miliequivalents of $H_{2}SO_{4} = 10 \times 0.05 = 0.5$
$H_{2}SO_{4}\ \ \ \ + \ \ \ \ 2NaOH\ \ \ \ \rightarrow \ \ \ \ Na_{2}SO_{4}\ \ \ \ \ + \ \ \ \ \ \ H_{2}O$
1 equivalent 2 equivalent
$\because 0.5\ equivalent\ of\ H_{2}SO_{4}\ will\ react\ with\ 1\$
$equivalent\ of\ NaOH$
$\therefore The\ pH\ of\ solution = 7\ (neutral)$