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Question

Calculate [H+] of solution obtained by mixing equal volume of 0.02 M HNO2 ( a weak acid) and 0.2 M CH3COOH solution.
Given that:
Ka1(HNO2)=2×104
Ka2(CH3COOH)=2×105
Where , Ka1and Ka2 are the dissociation constants of HNO2 and CH3COOH respectively.

A
2×103 M
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B
4×104 M
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C
6×103 M
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D
8×104 M
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Solution

The correct option is A 2×103 M
Denoting a for HNO2 and b for CH3COOH.
Let the equal volumes of solution a and b as V , i.e. Va=Vb=V
So, C1=CaVaVa+Vb=0.02×VV+VC1=0.01 M
Similarly, C2=Cb×VbVa+Vb=0.2×VV+VC2=0.1 M
Now,
HNO2 (aq)H+ (aq)+NO2 (aq)C1(1α1)C1α1+C2α2C1α1
[NO2]=C1α1=0.01×101=1×103
CH3COOHH++CH3COOC2(1α2)C1α1+C2α2C2α2

Dissociation constant for HNO2 i.e. Ka1 :
Ka1=(C1α1+C2α2)(C1α1)C1(1α1)

Dissociation constant for CH3COOH i.e. Ka2 :
Ka2=(C1α1+C2α2)(C2α2)C2(1α2)
Since, both are weak acids so 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

[H+]=Ka1C1+Ka2C2=2×104×0.01+2×105×0.1[H+]=2×106+2×106[H+]=2×103 M

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