0.2 M acetic acid is diluted to 0.02 M. The ratio of % of ionization of acetic acid in its 0.02 M solution to that in its 0.2 M solution is
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Let x M is the concentration of acetic acid that is ionised.
CH
3
−COOH⇌CH
3
−COO
−
+H
+
Out of 0.2 M acetic acid, x M acetic acid ionises to give x M acetate ions and x M protons. (0.2−x) M acetic acid will remain.
K
a
=
[CH
3
−COOH]
[CH
3
−COO
−
][H
+
]
1.8×10
−5
=
0.2−x
x×x
Since K
a
is very small, 0.2−x can be approximated to 0.2.
1.8×10
−5
=
0.2
x×x
x=0.00189 M
The percent ionisation
=
0.2
x
×100
=
0.2
0.00189
×100
=0.95
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