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

Final volume = 0.50 L + 5.0 L = 5.50 L

Final [CH3CO2H] = 0.50 L x 0.40 mol / L = 0.20 mol / 5.50 L = 0.04 M

Final [CH3CO2Na] = 5.0 L x 0.40 mol / L = 2.0 mol / 5.50 L = 0.364 M

Solution is a buffer since it contains a weak acid, CH3CO2H and the conjugate base, CH3CO2Na

Use Henderson Hasselbalch equation to find the pH:

pH = pKa + log [CH3CO2Na] / [CH3CO2H] ... looking up pKa of acetic acid (CH3CO2H), we find 4.75

pH = 4.75 + log (0.364 / 0.04) = 4.75 + log 9.09

pH = 4.75 + 0.96

pH = 5.71