pH of acid solution $[H^{+}]$ = $\sqrt{K_a \times C}$ $K_a$ = 1.35 × $10^{-3}$ , C = 0.1 M $[H^{+}]$ = $\sqrt{1.35 \times 10^{-3} \times 0.1}$ = $\sqrt{1.35 \times 10^{-4}}$ = 1.16 × $10^{-2}$ pH = – log $[H^{+}]$ = – log(1.16 × $10^{-2}$) = 2 – 0.064 = 1.936 pH of 0.1 M sodium salt solution Sodium salt of chloroacetic acid is salt of weak acid and strong base. Hence, pH = $\frac{1}{2}[pK_w+pK_a+\log C]$ $K_a$ = 1.35 × $10^{-3}$ $pK_a$ = - log $K_a$ = - log (1.35 × $10^{-3}$) = - (0.1303 - 3) = 2.8697 $pK_w$ = - log $10^{-14}$ = 14 C = 0.1 M , log C = log (0.1) = - 1 ⇒ pH = $\frac{1}{2}$ [14 + 2.8697 - 1] = 7.935