A weak monobasic acid is 1% ionised in 0.1 M | vedclass

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(B) For a weak monobasic acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.In the first case,$C_1 = 0.1 \ M$ and $\

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(B) For a weak monobasic acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.In the first case,$C_1 = 0.1 \ M$ and $\alpha_1 = 0.01$ (since $1\%$ ionised).So,$K_a = C_1 \alpha_1^2 = 0.1 	imes (0.01)^2 = 10^{-5}$.In the second case,$C_2 = 0.025 \ M$.The new degree of dissociation $\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{10^{-5}}{0.025}} = \sqrt{\frac{10^{-5}}{2.5 	imes 10^{-2}}} = \sqrt{4 	imes 10^{-4}} = 0.02$.Therefore,the percentage ionisation is $0.02 	imes 100 = 2\%$.

$A$ weak monobasic acid is $1\%$ ionised in $0.1 \ M$ solution at $25 \ ^{\circ}C$. The $\%$ ionisation in its $0.025 \ M$ solution is

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