When a 0.1 , M acetic acid solution is titrat | vedclass

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(C) The reaction between acetic acid and sodium hydroxide is: $CH_3COOH + NaOH \rightleftharpoons CH_3COONa + H_2O$ At the equivalence point,the total

Vedclass · Accepted Answer

$8.71$

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(C) The reaction between acetic acid and sodium hydroxide is: $CH_3COOH + NaOH ightleftharpoons CH_3COONa + H_2O$ At the equivalence point,the total volume doubles,so the concentration of the salt $(CH_3COONa)$ becomes $C = \frac{0.1 	imes V}{2V} = 0.05 \, M$. The $pH$ of a salt of a weak acid and a strong base is given by: $pH = \frac{1}{2} [pK_w + pK_a + \log C]$ Given $K_w = 10^{-14}$,$K_a = 1.9 	imes 10^{-5}$,and $C = 0.05 \, M$: $pK_w = 14$ $pK_a = -\log(1.9 	imes 10^{-5}) = 5 - 0.2788 = 4.7212$ $pH = \frac{1}{2} [14 + 4.7212 + \log(0.05)]$ $pH = \frac{1}{2} [18.7212 + (-1.301)]$ $pH = \frac{1}{2} [17.4202] = 8.7101 \approx 8.71$

When a $0.1 \, M$ acetic acid solution is titrated with a $0.1 \, M$ $NaOH$ solution,calculate the $pH$ at the equivalence point ($K_a$ for acetic acid = $1.9 \times 10^{-5}$).

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