At what concentration of 0.2 M CH_3COOH will | vedclass

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(A) For a weak acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Let the initial concentration be $C_1 = 0.2 \ M$ a

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$0.05$

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(A) For a weak acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Let the initial concentration be $C_1 = 0.2 \ M$ and the initial degree of dissociation be $\alpha_1$.Then $\alpha_1 = \sqrt{\frac{K_a}{C_1}}$.We want the new degree of dissociation $\alpha_2 = 2\alpha_1$.So,$\alpha_2 = \sqrt{\frac{K_a}{C_2}} = 2 \sqrt{\frac{K_a}{C_1}}$.Squaring both sides,we get $\frac{K_a}{C_2} = 4 \frac{K_a}{C_1}$.This simplifies to $C_2 = \frac{C_1}{4}$.Substituting $C_1 = 0.2 \ M$,we get $C_2 = \frac{0.2}{4} = 0.05 \ M$.

At what concentration of $0.2 \ M$ $CH_3COOH$ will its degree of dissociation become double (in $M$)? (Given: $K_a = 1.8 \times 10^{-5}$ for $CH_3COOH$)

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