The weak acid, HA has a Ka of 1.00×10−5. If 0.1 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closet to:
1%
99.9%
0.1%
99%
A
99.9%
B
1%
C
0.1%
D
99%
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Solution
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0.1 mole of acid is dissolved in 1 litre of water means [HA]=0.1M
Let 'α' be degree of dissocition
HA⇌H++A−
0.1
0.1(1−α)0.1α0.1α Ka=[H+][A−][HA]=0.12α20.1(1−α)
Let α<<1 so 1−α=1
Ka=0.1α2=10−5
α=10−2
% of acid dissociated=1%
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