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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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$1 \times 10^{- 8}$

$1 \times 10^{- 6}$

$10^{- 5}$

$1 \times 10^{- 4}$

answer is A.

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Detailed Solution

$given α = \frac{0.01}{100} = 10^{- 4}; C = 1 ∴ K = α^{2} C = {[10^{- 4}]}^{2} \times 1 = {(10^{- 4})}^{2} = 1 \times 10^{- 8}$

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