pH of a solution obtained by mixing equal vol | vedclass

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(C) When equal volumes of $0.2 \ M \ NaOH$ and $0.2 \ M \ CH_3COOH$ are mixed,the number of moles of each reactant is equal.$CH_3COOH + NaOH \rightarr

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) When equal volumes of $0.2 \ M \ NaOH$ and $0.2 \ M \ CH_3COOH$ are mixed,the number of moles of each reactant is equal.$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$Since the reaction is stoichiometric,both reactants are completely consumed,resulting in the formation of a salt,$CH_3COONa$.The concentration of the salt formed is $0.1 \ M$ (since volume doubles).$CH_3COONa$ is a salt of a weak acid and a strong base,which undergoes anionic hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.The $pH$ of such a salt solution is given by: $pH = 7 + \frac{1}{2}(pK_a + \log C)$.Given $K_a = 10^{-5}$,so $pK_a = 5$.$C = 0.1 \ M$.$pH = 7 + \frac{1}{2}(5 + \log 0.1) = 7 + \frac{1}{2}(5 - 1) = 7 + \frac{1}{2}(4) = 7 + 2 = 9$.

$pH$ of a solution obtained by mixing equal volume of $0.2 \ M \ NaOH$ and $0.2 \ M \ CH_3COOH$ $(K_a = 10^{-5})$ is

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