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(A) The solution is a buffer containing a weak acid $(CH_3COOH)$ and its salt $((CH_3COO)_2Ba)$.Using the Henderson-Hasselbalch equation: $pH = pK_a +

Vedclass · Accepted Answer

$5.046$

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(A) The solution is a buffer containing a weak acid $(CH_3COOH)$ and its salt $((CH_3COO)_2Ba)$.Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.First,calculate $pK_a = -\log(1.8 	imes 10^{-5}) = 4.7447$.Since $(CH_3COO)_2Ba$ dissociates as $(CH_3COO)_2Ba ightarrow 2CH_3COO^- + Ba^{2+}$,the concentration of the conjugate base $[CH_3COO^-] = 2 	imes 0.1 \, M = 0.2 \, M$.The concentration of the acid $[CH_3COOH] = 0.1 \, M$.Substituting the values: $pH = 4.7447 + \log \frac{0.2}{0.1} = 4.7447 + \log(2) = 4.7447 + 0.301 = 5.0457 \approx 5.046$.

Calculate the $pH$ of a $2 \, L$ solution containing $0.1 \, M \, CH_3COOH$ and $0.1 \, M \, (CH_3COO)_2Ba$. Given $K_a (CH_3COOH) = 1.8 \times 10^{-5}$.

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