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Question

Calculate the percentage ionization of 0.01M acetic acid in 0.1M 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 C 0.018%
HClH++Cl
initial 0.1 0 0
0 0.1 0.1 after it's complete dissociation

H2O is also added in case of weak acid
CH3COOH+H2O(l)CH3COO+H+
initial 0.01 0 0 0.1 (from HCl dissociation)
0.01x 0 x x+0.1 (after dissociation)
Ka=[CH3COO][H+][CH3COOH] given, Ka of acetic acid =1.8×105
1.8×105=x(0.1+x)0.01
1.8×107=x(0.1)+x2
negligible
x=1.8×1070.1
x=1.8×106M
So, % ionisation of acid =x0.01×100=1.8×102%
=0.018%
Percentage ionisation is 0.018%

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