Question

A 1.45 L buffer solution consists of 0.134 M butanoic acid and 0.344 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 Ka of butanoic acid is 1.52 × 10-5.

Answer 1

Numbers of moles of butanoic acid present in 1.45 L solution = 1.45 L * 0.134 M = 0.1943 moles

Numbers of moles of sodium butanoate present in 1.45 L solution = 1.45 L * 0.344 M = 0.4988 moles

When NaOH is added to the solution, it will react with butanoic acid and will form sodium butanoate. As a result concentration of acid will decrease and concentration of salt will increase.

CH₃-CH₂-CH₂-COOH + NaOH ---> CH₃-CH₂-CH₂-COONa + H₂O

Number of moles of butanoic acid after reaction = 0.1943 - 0.063 = 0.1313 moles

concentration of butanoic acid after reaction = 0.1313 moles/1.45 L = 0.0905 M

Number of moles of sodium butanoate after reaction = 0.4988 + 0.063 = 0.5618 moles

concentration of sodium butanoate after reaction = 0.5618 moles/1.45 L = 0.3874 M

pH of an acidic buffer solution

= pK_a + log[salt]/[acid]

= -log(1.52 * 10^-5) + log (0.3874/0.0905)

= 5.45