Question

A 1.31 L buffer solution consists of 0.281 M propanoic acid and 0.104 M sodium propanoate. Calculate the pH of the solution following the addition of 0.068 moles of HCl. Assume that any contribution of the HCl to the volume of the solution is negligible. The Ka of propanoic acid is 1.34 × 10-5.

Answer 1

first find the pKa from Ka

pKa = -logKa = -log(1.34 x 10^-5 ) = 4.87

Moles of propanoic acid = Molarity x volume in liters = 0.281 M x 1.31 L = 0.368 mol

Moles of sodium propanoate = 0.104 M x 1.31 L = 0.1362 mol

moles of HCl aadded = 0.068 mol'

0.068 mol of HCl react with 0.068 mol of sodium propanoate and will form the 0.068 mol of propanoic acid

use the hander son equation

pH = pKa + log[sodium propanoate - moles of HCl / propanoic acid + moles of HCl)

pH = 4.87 + log[0.1362 - 0.068 / 0.368 + 0.068]

pH = 4.87 + log[0.0682 / 0.436]

pH = 4.87 +-0.8

pH = 4.06