0.1 M formic acid solution is titrated again | vedclass

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(D) When formic acid is titrated with $NaOH$,it forms a buffer solution.For an acidic buffer,the Henderson-Hasselbalch equation is: $pH = pK_a + \log

Vedclass · Accepted Answer

$2 \log \ 4$

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(D) When formic acid is titrated with $NaOH$,it forms a buffer solution.For an acidic buffer,the Henderson-Hasselbalch equation is: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$At $1/5$ neutralization:$[Salt] = 1/5$ and $[Acid] = 4/5$$pH_1 = pK_a + \log \frac{1/5}{4/5} = pK_a + \log(1/4)$At $4/5$ neutralization:$[Salt] = 4/5$ and $[Acid] = 1/5$$pH_2 = pK_a + \log \frac{4/5}{1/5} = pK_a + \log(4)$Difference in $pH = pH_2 - pH_1 = (pK_a + \log 4) - (pK_a + \log(1/4))$$Difference = \log 4 - \log(1/4) = \log 4 - (-\log 4) = 2 \log 4$

$0.1 \ M$ formic acid solution is titrated against $0.1 \ M \ NaOH$ solution. What would be the difference in $pH$ between $1/5$ and $4/5$ stages of neutralization of acid?

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