0.1 M CH_3COOH is titrated against 0.1 M Na | vedclass

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(D) For a weak acid titration,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.At the $1/4$ stage of

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$2 \log 3$

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(D) For a weak acid titration,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.At the $1/4$ stage of neutralization,$[Salt] = 1/4$ and $[Acid] = 3/4$. So,$pH_1 = pK_a + \log \frac{1/4}{3/4} = pK_a + \log(1/3) = pK_a - \log 3$.At the $3/4$ stage of neutralization,$[Salt] = 3/4$ and $[Acid] = 1/4$. So,$pH_2 = pK_a + \log \frac{3/4}{1/4} = pK_a + \log 3$.The difference in $pH$ is $\Delta pH = pH_2 - pH_1 = (pK_a + \log 3) - (pK_a - \log 3) = 2 \log 3$.

$0.1 \ M$ $CH_3COOH$ is titrated against $0.1 \ M$ $NaOH$. What would be the difference in $pH$ between $1/4$ and $3/4$ stages of neutralization of the acid?

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