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Question

Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. Ka of acetic acid is 1.8×105:


A
0.18%
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B
0.018%
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C
1.8%
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D
18%
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Solution

The correct option is B 0.018%
[H+]=101
CH3COOHCH3COO+H+

initially: 0.01M
After dissociation: 0.01xxx+101

Ka=[CH3COO][H+][CH3COOH]=x(x+101)0.01x=1.8×105

Approximation: The acid is weak, hence 0.01x0.01 and x+0.10.1

Ka=x×0.10.01=1.8×105

x=1.8×106

% ionization =x×1000.01=1.8×1060.01×100=0.018%

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