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Question

Find the percentage ionization of 0.2M acetic acid solution, whose disassociation constant is 1.8×105

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Solution

Let x M is the concentration of acetic acid that is ionised.
CH3COOHCH3COO+H+
Out of 0.2 M acetic acid, x M acetic acid ionises to give x M acetate ions and x M protons. (0.2x) M acetic acid will remain.
Ka=[CH3COO][H+][CH3COOH]
1.8×105=x×x0.2x
Since Ka is very small, 0.2x can be approximated to 0.2.
1.8×105=x×x0.2
x=0.00189 M
The percent ionisation
=x0.2×100
=0.001890.2×100
=0.95

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