If the percentage dissociation of an acid in 0.1 M solution is $0.01 %$ , what it is percentage of dissociation in 0.001 M solution ?

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Solution

The correct option is B

$K_{a} = C α^{2}; From the equation (K_{a} is independent of concentration) α_{1} = 0.01 % = \frac{0.01}{100} = 10^{- 4} C_{1} α_{1}^{2} = C_{2} α_{2}^{2}; 0.1 \times (10^{- 4})^{2} = 0.001 \times α_{2}^{2} α_{2} = 10^{- 3} = 10^{- 3} \times 100 = 0.1 %$

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If the percentage dissociation of an acid in 0.1 M solution is 0.01%, what it is percentage of dissociation in 0.001 M solution ?

The correct option is B Ka=Cα2;From the equation(Kais independent of concentration)α1=0.01%=0.01100=10−4C1α21=C2α22;0.1×(10−4)2=0.001×α22α2=10−3=10−3×100=0.1%

If the percentage dissociation of an acid in 0.1 M solution is $0.01 %$ , what it is percentage of dissociation in 0.001 M solution ?

The correct option is B

$K_{a} = C α^{2}; From the equation (K_{a} is independent of concentration) α_{1} = 0.01 % = \frac{0.01}{100} = 10^{- 4} C_{1} α_{1}^{2} = C_{2} α_{2}^{2}; 0.1 \times (10^{- 4})^{2} = 0.001 \times α_{2}^{2} α_{2} = 10^{- 3} = 10^{- 3} \times 100 = 0.1 %$