0.1 , M acetic acid solution is titrated agai | vedclass

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(A) For the titration of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the $pH$ at any stage of neutralization is given by the Henderson-Hassel

Vedclass · Accepted Answer

$2 \, \log \, 3$

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(A) For the titration of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the $pH$ at any stage of neutralization is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.At $1/4$ neutralization,the concentration of salt is $1/4$ and the concentration of remaining acid is $3/4$. Thus,$pH_1 = pK_a + \log \frac{1/4}{3/4} = pK_a + \log \frac{1}{3}$.At $3/4$ neutralization,the concentration of salt is $3/4$ and the concentration of remaining acid is $1/4$. Thus,$pH_2 = pK_a + \log \frac{3/4}{1/4} = pK_a + \log 3$.The difference in $pH$ is $pH_2 - pH_1 = (pK_a + \log 3) - (pK_a + \log \frac{1}{3}) = \log 3 - \log \frac{1}{3} = \log 3 - (-\log 3) = 2 \log 3$.

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

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