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Q.

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

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a

1 × 10-8 

b

1 × 10-6 

c

10-5 

d

1 × 10-4 

answer is A.

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Detailed Solution

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

                              

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