During the titration of 0.1 , M CH_3COOH and | vedclass

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(C) The titration involves a weak acid $(CH_3COOH)$ and a strong base $(KOH)$.After $40\%$ neutralization,the amount of salt $(CH_3COOK)$ formed is $0

Vedclass · Accepted Answer

$4.52$

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(C) The titration involves a weak acid $(CH_3COOH)$ and a strong base $(KOH)$.After $40\%$ neutralization,the amount of salt $(CH_3COOK)$ formed is $0.4$ equivalents,and the remaining amount of weak acid $(CH_3COOH)$ is $0.6$ equivalents.This forms an acidic buffer solution.The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} ight)$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(2 	imes 10^{-5}) = 5 - \log(2) = 5 - 0.301 = 4.699$.Now,substitute the values into the Henderson-Hasselbalch equation:$pH = 4.699 + \log \left( \frac{0.4}{0.6} ight) = 4.699 + \log \left( \frac{2}{3} ight)$.$pH = 4.699 + (\log 2 - \log 3) = 4.699 + (0.301 - 0.477) = 4.699 - 0.176 = 4.523$.Rounding to two decimal places,we get $pH = 4.52$.

During the titration of $0.1 \, M$ $CH_3COOH$ and $0.1 \, M$ $KOH$ solution,what will be the $pH$ of the solution after $40\%$ neutralization of the acid? [$K_a$ for $CH_3COOH = 2 \times 10^{-5}$]

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