K_a for CH_3COOH is 1.9 times 10^{-5}. Calcul | vedclass

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(A) At the equivalence point,$CH_3COOH$ reacts with $NaOH$ to form $CH_3COONa$ and $H_2O$. Since the initial concentrations are $0.1 \ M$ each,the vol

Vedclass · Accepted Answer

$8.71$

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(A) At the equivalence point,$CH_3COOH$ reacts with $NaOH$ to form $CH_3COONa$ and $H_2O$. Since the initial concentrations are $0.1 \ M$ each,the volume doubles,so the concentration of the salt $CH_3COONa$ is $C = 0.05 \ M$. The salt undergoes hydrolysis: $CH_3COO^- + H_2O ightleftharpoons CH_3COOH + OH^-$. 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 $K_a = 1.9 	imes 10^{-5}$,then $pK_a = -\log(1.9 	imes 10^{-5}) \approx 4.72$. $pH = \frac{1}{2} [14 + 4.72 + \log(0.05)]$. $pH = \frac{1}{2} [18.72 - 1.30] = \frac{1}{2} [17.42] = 8.71$.

$K_a$ for $CH_3COOH$ is $1.9 \times 10^{-5}$. Calculate the $pH$ at the equivalence point in the titration of $0.1 \ M$ $CH_3COOH$ with $0.1 \ M$ $NaOH$.

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