Question

Consider a diprotic acid $H_{2}A$. The dissociation constants for $H_{2}A$ and $H{A}^{-}$ are ${10}^{-6}and{10}^{-13}$ respectively. Then, the pH of $0.25M$ aqueous solution of $H_{2}A$ will be :

Answer 1

Given, $K_{a1}\left(H_{2}A\right)={10}^{-6}{K}_{a2}\left(H{A}^{-}\right)={10}^{-13}$ $\left[H_{2}A\right]=0.25M$

Since, the second acid dissociation constant is very small in comparison to the first acid dissociation constant of $H_{2}A$ i.e. $K_{a_2}<<K_{a_1}$. Thus $H^{+}$ ions contribution can be taken mainly from the first step i.e. $[H^{+}{]}_{total}=[H^{+}]$ from $H_{2}A$ dissociation.

$H_{2}A\left(aq\right)\rightleftharpoons H{A}^{-}\left(aq\right)+H^{+}\left(aq\right)Initially:C00Equilibrium:C(1-\alpha )C\alpha C\alpha K_{a1}=\frac{\left[H^{+}\right]\left[H{A}^{-}\right]}{\left[H_{2}A\right]}[H^{+}{]}_{eq}=[H{A}^{-}{]}_{eq}=C\alpha 1-\alpha \approx 1[H^{+}{]}^2=K_{a1}\times C\therefore [H^{+}]=\sqrt{K_{a1}\times C}[H^{+}]=\sqrt{{10}^{-6}\times 0.25}[H^{+}]=\sqrt{25\times {10}^{-8}}=5\times {10}^{-4}MpH=-\log [H^{+}]=-\log (5\times {10}^{-4})=(4-log5)pH=(4-0.7)=3.3$