3 g of acetic acid is added to 250 mL of 0.1 M HCL and the solution made up to 500 mL. To 20 mL of this solutions 1/2 mL of 5 M NaOH is added. The pH of the solution is __________. (Round off to the nearest integer)
[Given : pKa of acetic acid = 4.75, molar mass of acetic of acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.

Solution

Chemistry ma’am · 4 steps

Let NaOH finish off the HCl first; what is left over turns acetic acid into an acetate buffer.

1. Count the acids. Acetic acid: 3/60 = 0.05 mol, and HCl: 0.250 × 0.1 = 0.025 mol, both in 500 mL. So a 20 mL portion holds 2 mmol of CH₃COOH and 1 mmol of HCl.
2. Strong acid first. The base added is 0.5 mL × 5 M = 2.5 mmol of NaOH. It neutralises all 1 mmol of HCl first, leaving 1.5 mmol of NaOH.
3. Then the weak acid. The remaining 1.5 mmol of NaOH converts 1.5 mmol of CH₃COOH into CH₃COO⁻. That leaves 0.5 mmol of acid with 1.5 mmol of its salt, which is a buffer.

   pH = pKₐ + log[CH₃COO⁻]/[CH₃COOH] = 4.75 + log 1.5/0.5 = 4.75 + 0.4771

4. So. pH = 5.227, which rounds to 5.

The common slip: 4. Putting acid over salt in the log gives 4.75 - 0.477 = 4.27, which rounds to 4. The salt goes on top.

The idea to keep. When a strong and a weak acid share a beaker, added base always eats the strong acid first. Only the leftover base builds the buffer, so do the bookkeeping in millimoles before Henderson.

Answer from the NTA final answer key. Solution by PrepWiser.