What is the pH of a solution prepared by mixing 0.50 L of 0.40 M CH3CO2H with 5.0 L of 0.40 M CH3CO2Na?

Question

What is the pH of a solution prepared by mixing 0.50 L of 0.40 M CH₃CO₂H with 5.0 L of 0.40 M CH₃CO₂Na?

Answer 1

Since we have an acid (CH₃CO₂H) and a base (CH₃CO₂Na), this is a buffer problem and we can use the Henderson-Hasselbach equation to solve for the pH of the solution.

1) First, lets find the number of moles of acid and base in solution by using the molarity and volume.

Moles of acid: 0.50 L CH₃CO₂H x 0.40 M CH₃CO₂H = 0.20 mol CH₃CO₂H

Moles of base: 5.0 L CH₃CO₂Na x 0.40 M CH₃CO₂Na = 2.0 mol CH₃CO₂Na

2) Now, lets find the concentration of each of these in the solution. Since we’re mixing the two solutions, our total volume is 0.50 L + 5.0 L = 5.5 L. 

concentration of acid: 0.20 mol CH₃CO₂H / 5.5 L = 0.036 M CH₃CO₂H

concentration of base: 2.0 mol CH₃CO₂Na / 5.5 L = 0.36 M CH₃CO₂Na

3) Using the Henderson-Hasselbach equation, which says pH = pKa + log([base]/[acid]), we can solve for the pH of the solution. From the pKa table, the pKa of CH₃CO₂H is 4.76.

We then have pH = 4.76 + log(0.36/0.036) = 4.76 + 1.00 = 5.76