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 .

Explanation:

Hello,

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

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

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

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

And the concentrations in the same volume:

Thus, the new pH is:

Which is a difference of pH of 0.21.

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