If 0.1 , M , CH_3COONa undergoes 1 % hydrolys | vedclass

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(A) Given: Concentration $C = 0.1 \, M$,Degree of hydrolysis $h = 1 \% = 0.01 = 10^{-2}$.For the salt of a weak acid and strong base,the hydrolysis co

Vedclass · Accepted Answer

$10^{-5}, 10^{-9}, 10^{-3}$

Vedclass · Answer

(A) Given: Concentration $C = 0.1 \, M$,Degree of hydrolysis $h = 1 \% = 0.01 = 10^{-2}$.For the salt of a weak acid and strong base,the hydrolysis constant $K_h$ is given by $K_h = C h^2$.$K_h = 0.1 	imes (10^{-2})^2 = 0.1 	imes 10^{-4} = 10^{-5}$.We know $K_h = \frac{K_w}{K_a}$,so $K_a = \frac{K_w}{K_h} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.The concentration of $CH_3COOH$ formed is $[CH_3COOH] = C 	imes h = 0.1 	imes 10^{-2} = 10^{-3} \, M$.Thus,the values are $10^{-5}, 10^{-9}, 10^{-3}$.

If $0.1 \, M \, CH_3COONa$ undergoes $1 \%$ hydrolysis,then the values of $K_h$,$K_a$ of the acid,and the concentration of $CH_3COOH$ are respectively:

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