Explanation

(c)
$m = \frac{\mathrm{\Delta}T}{k_{b}} = \frac{- 0.060}{- 1.86} = 3.2 \times 10^{- 2} = 0.032$
$ie$, 0.032 = total particle
$\therefore$ The number of $H^{+} = (0.032 - 0.025)m = 0.007\ MH^{+}$
$HA \rightleftharpoons H^{+} + A^{-}$
$\left\lbrack H^{+} \right\rbrack = \left\lbrack A^{-} \right\rbrack = 0.007\ M$
$HA = 0.018$
$\therefore K_{a} = \frac{(0.007)^{2}}{0.018} = 3 \times 10^{- 3}$
$pK_{a} = 2.5$