Consider a diprotic acid H2A. The dissociation constants for H2A and HA− are 10−6 and 10−13 respectively. Then, the pH of 0.25 M aqueous solution of H2A will be :
Given, Ka1(H2A)=10−6Ka2(HA−)=10−13 [H2A]=0.25 M
Since, the second acid dissociation constant is very small in comparison to the first acid dissociation constant of H2A i.e. Ka2<<Ka1. Thus H+ ions contribution can be taken mainly from the first step i.e. [H+]total=[H+] from H2A dissociation.
H2A (aq)⇌HA− (aq)+H+ (aq)Initially: C 0 0Equilibrium:C(1−α) Cα CαKa1=[H+][HA−][H2A][H+]eq=[HA−]eq=Cα1−α≈1[H+]2=Ka1×C∴[H+]=√Ka1×C[H+]=√10−6×0.25[H+]=√25×10−8=5×10−4 MpH=−log[H+]=−log(5×10−4)=(4−log5)pH=(4−0.7)=3.3