What is the PH of a solution of 25mL of 0.100M CH3CO2H titrated with 25mL of 0.100M NaOH?

Question

I calculated the moles of OH- in solution from the NaOH to be 0.0025 moles and 0.00003354 moles from acetate, the conjugate base. I also calculated the H+ to be 0.00003354 moles from the acetic acid. I tried to subtract the H+ from the total OH- and then divide those moles by 0.05 L and use negative log to get the PH from that molarity, but am getting the wrong answer. I don't know if the above calculations are correct or if I am approaching this in the wrong way. The answer is 8.72

Answer 1

Ok, Katie. I think you took the wrong approach to answering this problem. As stated in my comment, you have equal moles of NaOH and CH3COOH, so all of the CH3COOH is converted to CH3COONa, or simply CH3COO^-. See the equation below.

CH3COOH + NaOH ==> CH3COONa + H2O

moles CH3COOH = 25 ml x 1 L / 1000 ml x 0.100 mol/L = 0.0025 moles

moles NaOH = 25 ml x 1 L / 1000 ml x 0.100 mol/L = 0.0025 moles

Equal moles of acid and base means you're at the equivalence point so you end up with

0.0025 moles CH3COO^- in 50 ml or in 0.05 L

Final concentration of CH3COO^- = 0.0025 moles / 0.05 L = 0.05 M

CH3COO^- + H2O = CH3COOH + OH^-

Kb = [CH3COOH][OH^-] / [CH3COO^-]

Looking up the Kb for acetate, or the Ka for acetic acid and calculating the Kb for acetate, I get

Kb = 5.68x10^-10 = (x)(x) / 0.05

x² = 2.84x10^-11

x = 5.33x10^-6 M = [OH^-]

pOH = -log 5.33x10^-6 = 5.27

pH = 14 - 5.27 = 8.72

I hope the makes sense to you, and I hope it helps.