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Question

Two weak monobasic acids HA and HB having equal concentrations of 0.5 M are present in solution at 25oC having dissociation constants as 3.5×1010 and 4.5×1010 respectively. Calculate the pH of the solution at given temperature.

A
4.7
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B
5.7
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C
6.7
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D
2.5
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Solution

The correct option is A 4.7
C1=C2=0.5 MKa1=3.5×1010 MKa2=4.5×1010 M
For two weak acids :

[H+] = C1Ka1+C2Ka2
[H+]=0.5(3.5×1010+4.5×1010)
[H+]=2×105 MpH=log[H+]pH=log(2×105)=(5log2)pH=4.7
The pH of HA and HB in solution is 4.7

Theory :
Weak acid 1 (WA1) + Weak acid 2 (WA2) :

Mixture of two weak monoprotic acids HA and HB with concentration C1 and C2 and degree of dissociation α1 and α2

For HA :

HA(aq)H+(aq)+A(aq)

at t=0 C1 0 0

at t=teq C1C1α1 (C1α1+C2α2) C1α1

For HB :

HB(aq)H+(aq)+B(aq)

at t=0 C2 0 0

at t=teq C2C2α2 (C1α1+C2α2) C2α2

Dissociation constant for HA Ka1 :

Ka1=(C1α1+C2α2)(C1α1)C1(1α1)

Dissociation constant for HB Ka2 :

Ka2=(C1α1+C2α2)(C2α2)C2(1α2)

Since α1 and α2 are very small in comparison to unity for weak monoprotic acids.So 1α11 and 1α21.

C1Ka1+C2Ka2 = (C1α1+C2α2)2

[H+] = C1α1+C2α2 = C1Ka1+C2Ka2


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