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A monoprotic acid in 1.00 M solution is 0.01% ionised. The dissociation constant of this acid is

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By Expert Faculty of Sri Chaitanya
a
1 × 10-8
b
1 × 10-4
c
1 × 10-6
d
10-5
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detailed solution

Correct option is A

given   α = 0.01100 = 10-4 ; C=1  ∴   K = α2C = 10-42×1 =(10-4)2              =1×10-8

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