A monoprotic acid in $1.00 M$ solution is $0.001$ % ionised. The dissociation constant of acid is:

1 \times 10^{- 3}

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1 \times 10^{- 6}

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1 \times 10^{- 8}

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1 \times 10^{- 10}

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Solution

The correct option is D $1 \times 10^{- 10}$
According to Ostwald's dilution law
$K = \frac{α^{2} C}{(1 - α)}$
As degree of ionisation is very less than 1 so above equation reduces to
$K = C α^{2}$
$K = 1 (10^{- 5})^{2}$
$K = 10^{- 10}$

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A monoprotic acid in 1.00M solution is 0.001% ionised. The dissociation constant of acid is:

The correct option is D 1×10−10According to Ostwald's dilution lawK=α2C(1−α)As degree of ionisation is very less than 1 so above equation reduces toK=Cα2K=1(10−5)2K=10−10

A monoprotic acid in $1.00 M$ solution is $0.001$ % ionised. The dissociation constant of acid is:

The correct option is D $1 \times 10^{- 10}$
According to Ostwald's dilution law
$K = \frac{α^{2} C}{(1 - α)}$
As degree of ionisation is very less than 1 so above equation reduces to
$K = C α^{2}$
$K = 1 (10^{- 5})^{2}$
$K = 10^{- 10}$