    Question

# The [H+] of the resulting solution that is 0.01 M in acetic acid (Ka=1.8×10−5 M) and 0.01 M in benzoic acid. (Ka=6.3×10−5 M) will be:

A
4×104 M
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B
16×104 M
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C
81×104 M
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D
9×104 M
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Solution

## The correct option is D 9×10−4 MLet component 1 be acetic acid and component 2 be benzoic acid. C1=C2=0.01 MKa1=1.8×10−5 MKa2=6.3×10−5 M For two weak acid solutions [H+]=(Ka1C1+Ka2C2)0.5[H+]=[(0.01)(1.8×10−5+6.3×10−5)]0.5[H+]=(8.1×10−7)0.5[H+]=9×10−4 M 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 C1−C1α1 (C1α1+C2α2) C1α1 For HB : HB(aq)⇌H+(aq)+B−(aq) at t=0 C2 0 0 at t=teq C2−C2α2 (C1α1+C2α2) C2α2 Dissociation constant for HA Ka1 : Ka1=(C1α1+C2α2)(C1α1)C1(1−α1) Dissociation constant for HB Ka2 : Ka2=(C1α1+C2α2)(C2α2)C2(1−α2) Since, 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  Suggest Corrections  4      Similar questions  Related Videos   Degree of Dissociation
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