    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)=2×10−4 Ka2(CH3COOH)=2×10−5 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 2×10−3 MDenoting 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)+NO−2 (aq)C1(1−α1)C1α1+C2α2C1α1 [NO−2]=C1α1=0.01×10−1=1×10−3 CH3COOH⇌H++CH3COO−C2(1−α2)C1α1+C2α2C2α2 Dissociation constant for HNO2 i.e. Ka1 : Ka1=(C1α1+C2α2)(C1α1)C1(1−α1) Dissociation constant for CH3COOH i.e. Ka2 : Ka2=(C1α1+C2α2)(C2α2)C2(1−α2) Since, both are weak acids so α1 and α2 are very small in comparison to unity for weak monoprotic acids. So 1−α1≈1 and 1−α2≈1. C1Ka1+C2Ka2 = (C1α1+C2α2)2 [H+] = C1α1+C2α2 = √C1Ka1+C2Ka2 [H+]=√Ka1C1+Ka2C2=√2×10−4×0.01+2×10−5×0.1[H+]=√2×10−6+2×10−6[H+]=2×10−3 M  Suggest Corrections  1      Similar questions  Explore more