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Question

Calculate [H+] from a CH3COOH in a solution which is 101 M in HCl and 103 M in CH3COOH; [Ka=2×105].

A
2×104
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B
4×107
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C
2×107
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D
None of the above
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Solution

The correct option is C 2×107
CH3COOHCH3COO+H+At equ. C(1α) Cα (103+Cα)
H+ ion can be considered completely from HCl, due to less dissociation of CH3COOH (because of common ion effect by H+ of HCl,α<<1) and its low concentration, So, [H+]=101M pH=log [H+]=
pH=log 101=1
From above equilibrium,
K=[CH3COO][H+][CH3COOH]2×105=Cα×101C
α=2×104
[H+]from CH3COOH=Cα=103×2×104
[H+]=2×107M

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