Consider a diprotic acid H2A. The dissociation constants for H2A and HA− are 10−6and10−13 respectively. Then, the pH of 0.25M aqueous solution of H2A will be :
A
3.3
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B
4.8
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C
5.6
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D
7.4
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Solution
The correct option is A3.3 Given, Ka1(H2A)=10−6Ka2(HA−)=10−13[H2A]=0.25M
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.