If 12.5 mL of 0.1 M KOH is added into 50 m | vedclass

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(D) The reaction forms an acidic buffer solution.$pK_a = -\log(K_a) = -\log(2 	imes 10^{-5}) = 5 - \log(2) = 5 - 0.3 = 4.7$.$CH_3COOH + KOH ightarr

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$4.22$

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(D) The reaction forms an acidic buffer solution.$pK_a = -\log(K_a) = -\log(2 	imes 10^{-5}) = 5 - \log(2) = 5 - 0.3 = 4.7$.$CH_3COOH + KOH ightarrow CH_3COOK + H_2O$.Initial moles of $CH_3COOH = 50 	imes 0.1 = 5 \ mmol$.Initial moles of $KOH = 12.5 	imes 0.1 = 1.25 \ mmol$.After reaction,remaining $CH_3COOH = 5 - 1.25 = 3.75 \ mmol$.Formed $CH_3COOK = 1.25 \ mmol$.Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.$pH = 4.7 + \log \frac{1.25}{3.75} = 4.7 + \log(\frac{1}{3})$.$pH = 4.7 - 0.477 \approx 4.22$.

If $12.5 \ mL$ of $0.1 \ M$ $KOH$ is added into $50 \ mL$ of $0.1 \ M$ $CH_3COOH$ solution,then what would be the final $pH$? $(K_a(CH_3COOH) = 2 \times 10^{-5})$

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