The dissociation constants of CH_3COOH and HC | vedclass

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(A) The salts are formed from a weak acid and a strong base.The degree of hydrolysis ($h$ or $x$) is given by the formula: $x = \sqrt{\frac{K_h}{C}} =

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$x_1 > x_2$

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(A) The salts are formed from a weak acid and a strong base.The degree of hydrolysis ($h$ or $x$) is given by the formula: $x = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a 	imes C}}$.Assuming the concentration $(C)$ is the same for both salts,the relationship is $x \propto \sqrt{\frac{1}{K_a}}$.For $KCN$,the acid is $HCN$ $(K_a = 7.2 	imes 10^{-10})$.For $CH_3COOK$,the acid is $CH_3COOH$ $(K_a = 1.8 	imes 10^{-5})$.Since $K_a(CH_3COOH) > K_a(HCN)$,it follows that $x_1 > x_2$ because $x$ is inversely proportional to the square root of $K_a$.

The dissociation constants of $CH_3COOH$ and $HCN$ are $1.8 \times 10^{-5}$ and $7.2 \times 10^{-10}$ respectively. What is the relationship between the degree of hydrolysis of $KCN$ $(x_1)$ and $CH_3COOK$ $(x_2)$?

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