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Question

A 1.42 L buffer solution consists of 0.310 M propanoic acid and 0.199 M sodium propanoate. Calculate the pH of the solution following the addition of 0.076 mol 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.

pH=?

Answer 1

We can use the Henderson Hasselbalch equation:

pH = pKa + log [salt] / [acid]

The pKa = -log Ka = -log 1.34x10-5 = 4.87

Next, we must find [sodium propionate] and the [propionic acid]

Let HA = propionic acid

NaA = sodium propionate

Initial [HA] = 0.310 M

Initial [NaA] =0.199 M

Inital [HCl] = 0.076 mol / 1.42 L = 0.0535 M

The added HCl will react with the NaA to produce HA + NaCl

NaA + HCl ==> HA + NaCl

Final [NaA] = 0.199 M - 0.0535 M = 0.1455 M

Final [HA] = 0.310 M + 0.0535 M = 0.3635 M

Substituting into the Henderson Hasselbalch equation and solving for pH:

pH = 4.87 + log (0.1455 / 0.3635) = 4.87 - 0.398

pH = 4.47