A 1.50 L buffer solution consists of 0.118 M butanoic acid and 0.318 M sodium butanoate. Calculate the pH of the solution following the addition of 0.063 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The K a of butanoic acid is 1.52 × 10 − 5 .

pH = 5.493

Explanation:

It is possible to find pH of a buffer using H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of sodium butanoate and [HA] is concentration of butanoic acid.

pKa is -log Ka. As Ka of butanoic acid is 1.52×10⁻⁵, pKa is 4.818

Before reaction, moles of butanoic acid and sodium butanoate are:

butanoic acid: 1.50L × (0.118mol / L) = 0.177 moles butanoic acid

sodium butanoate: 1.50L × (0.318mol / L) = 0.477 moles sodium butanoate

NaOH reacts with butanoic acid thus:

NaOH + butanoic acid → sodium butanoate + water.

Butanoic acid decreases concentration and sodium butanoate increases concentration

Thus, after reaction, moles of butanoic acid and sodium butanoate are:

butanoic acid: 0.177mol - 0.063mol = 0.114mol

sodium butanoate: 0.477mol + 0.063mol = 0.540mol

As total volume is 1.50L, concentrations are:

[A⁻] [sodium butanoate] = 0.540mol / 1.50L = 0.360M

[HA] [butanoic acid] = 0.114mol / 1.50L = 0.076M

Replacing in H-H equation:

pH = 4.818 + log₁₀ [0.360M] / [0.076M]

pH = 5.493

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