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Consider a diprotic acid H2A. The dissociation constants for H2A and HA are 106 and 1013 respectively. Then, the pH of 0.25 M aqueous solution of H2A will be :

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Solution

Given, Ka1(H2A)=106Ka2(HA)=1013 [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(1α) Cα CαKa1=[H+][HA][H2A][H+]eq=[HA]eq=Cα1α1[H+]2=Ka1×C[H+]=Ka1×C[H+]=106×0.25[H+]=25×108=5×104 MpH=log[H+]=log(5×104)=(4log5)pH=(40.7)=3.3

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