One litre of an aqueous solution contains 0.1 | vedclass

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(C) The solution is a buffer containing $CH_3COOH$ (acid) and $CH_3COONa$ (conjugate base).Initially,$[Acid] = 0.15 \ mol$ and $[Salt] = 0.15 \ mol$.W

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$5.1$

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(C) The solution is a buffer containing $CH_3COOH$ (acid) and $CH_3COONa$ (conjugate base).Initially,$[Acid] = 0.15 \ mol$ and $[Salt] = 0.15 \ mol$.When $0.05 \ mol$ of $NaOH$ is added,it reacts with the acid $(CH_3COOH)$:$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$New moles of acid = $0.15 - 0.05 = 0.10 \ mol$.New moles of salt = $0.15 + 0.05 = 0.20 \ mol$.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pH = 4.8 + \log \frac{0.20}{0.10}$$pH = 4.8 + \log (2)$$pH = 4.8 + 0.301 = 5.101 \approx 5.1$.

One litre of an aqueous solution contains $0.15 \ mol$ of $CH_3COOH$ $(pK_a = 4.8)$ and $0.15 \ mol$ of $CH_3COONa$. After the addition of $0.05 \ mol$ of solid $NaOH$ to this solution,the $pH$ will be:

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