Chemistry, 20.09.2020 14:01 Silvanade

A 1.41 L buffer solution consists of 0.149 M butanoic acid and 0.288 M sodium butanoate. Calculate the pH of the solution following the addition of 0.061 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

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Answer from: abgriggs22

pH=5.31

Explanation:

Hello,

In this case, by considering the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

We can compute the pH before the addition of the NaOH:

$pH=-log(1.52x10^{-5})+log(\frac{[0.288M}{0.149M} )\\\\pH=5.10$

Nevertheless, if 0.061 moles of NaOH are added, we first need to compute the present moles of butanoic acid and sodium butanoate:

$n_{acid}=0.149mol/L*1.41L=0.210mol\\\\n_{base}=0.288mol/L*1.41L=0.406mol$

So the moles of acid and base after the addition are:

$n_{acid}=0.210mol-0.061mol=0.149mol\\\\n_{base}=0.406mol+0.061mol=0.467mol$

And the concentrations in the same volume:

$[acid]=\frac{0.149mol}{1.41L} =0.106M$

$[base]=\frac{0.467mol}{1.41L} =0.331M$

Thus, the new pH is:

$pH=-log(1.52x10^{-5})+log(\frac{[0.331M}{0.106M} )\\\\pH=5.31$

Which is a difference of pH of 0.21.

Best regards.

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A 1.41 L buffer solution consists of 0.149 M butanoic acid and 0.288 M sodium butanoate. Calculate t...

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