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

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Question

(C) The hydrolysis reaction for $CH_3COONa$ is: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.Given concentration $C = 0.1 \, M$ and degree of

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$10^{-3}, 10^{-11}, 11$

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(C) The hydrolysis reaction for $CH_3COONa$ is: $CH_3COO^- + H_2O ightleftharpoons CH_3COOH + OH^-$.Given concentration $C = 0.1 \, M$ and degree of hydrolysis $h = 1 \% = 0.01$.The concentration of $[OH^-]$ is given by $C 	imes h = 0.1 	imes 0.01 = 10^{-3} \, M$.Using the ionic product of water $K_w = 10^{-14}$,we find $[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, M$.Finally,$pH = -\log[H^+] = -\log(10^{-11}) = 11$.Thus,the values are $[OH^-] = 10^{-3} \, M, [H^+] = 10^{-11} \, M, pH = 11$.

If $0.1 \, M \, CH_3COONa$ undergoes $1 \%$ hydrolysis,then the values of $[OH^{-}]$,$[H^{+}]$,and $pH$ are respectively:

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