Explanation

(b)
$NaOH = \left\lbrack OH^{-} \right\rbrack = 10^{- 3}$
$\left\lbrack H^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack = 10^{- 14}$
$\left\lbrack H^{+} \right\rbrack = 10^{- 11}$
$pH = - \log{\lbrack H^{+}\rbrack}$
$= - \log{\left\lbrack 10^{- 11} \right\rbrack = 11}$
$HCl(aq) = \left\lbrack H^{+} \right\rbrack = 10^{- 3}$
$pH = - \log{\left\lbrack 10^{- 3} \right\rbrack = 3}$
$NaCl(aq) =$Neutral; $\left\lbrack H^{+} \right\rbrack = \left\lbrack OH^{-} \right\rbrack = 10^{- 7}$
$ie,\ pH = 7$