Milliequivalents of $20 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}=\mathrm{N}_{1} \mathrm{~V}_{1}$
$$
=20 \times 0.1=2
$$
(a) $\mathrm{pH}$ of distilled water $=7$
$$
\begin{aligned}
\therefore\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}=& 10^{-7} \mathrm{M} \\
\therefore \text { milliequivalents } &=\mathrm{NV} \quad \text { [for } \mathrm{HCl}, \mathrm{N}=\mathrm{M} \text { ] } \\
&=10^{-7} \times 20=2.0 \times 10^{-6}
\end{aligned}
$$
(b) Milliequivalents of $\mathrm{NaOH}=0.1 \times 1=0.1$
(c) $\mathrm{pH}$ of $\mathrm{HCl}=1$
$\therefore \quad\left[\mathrm{H}^{+}\right]=10^{-1}=0.1 \mathrm{M}$
$\therefore$ Milliequivalents $=0.1 \times 500=50$
(d) Milliequivalents of $\mathrm{HCl}=1 \times 1=1$
$\because$ Milliequivalents of $500 \mathrm{~mL}$ of $\mathrm{HCl}$ having $\mathrm{pH}=1$ is more than that of $20 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{HCl}$, therefore adding this, $\mathrm{pH}$ of $0.1 \mathrm{~N} \mathrm{HCl}$ solution, does not alter.