One litre of an aqueous solution contains 0.1 | vedclass

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

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$5.1$

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(C) The given solution is a buffer solution containing $CH_3COOH$ (acid) and $CH_3COONa$ (conjugate base).Initially,$[CH_3COOH] = 0.15 \ mol$ and $[CH_3COO^-] = 0.15 \ mol$.When $0.05 \ mol$ of $NaOH$ (a strong base) is added,it reacts with the acid $(CH_3COOH)$ to form more conjugate base $(CH_3COO^-)$.The new concentrations are:$[CH_3COOH]_{new} = 0.15 - 0.05 = 0.10 \ mol$$[CH_3COO^-]_{new} = 0.15 + 0.05 = 0.20 \ mol$Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}$$pH = 4.8 + \log \frac{0.20}{0.10}$$pH = 4.8 + \log(2)$Since $\log(2) \approx 0.3$,$pH = 4.8 + 0.3 = 5.1$

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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