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