What is the pH?

Question

Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 M CH3COOH with 0.300 M NaOH. The value Ka for CH3COOH is 1.8x10^-5

Answer 1

CH3COOH + NaOH ==> H2O + CH3COONa ... balance equation

moles CH3COOH = 50.0 ml x 1 L /1000 ml x 0.300 mol/L = 0.015 moles CH3COOH

At equivalence, moles of NaOH = moles CH3COOH

moles NaOH = 0.015 mols NaOH

Volume NaOH = 0.015 mols x 1 L / 0.300 mols = 0.05 L

Total volume = 0.05 L NaOH + 0.05 L CH3COOH = 0.1 L

At this point, all of the CH3COOH has been neutralized and converted to CH3COO-

Now, we look at the hydrolysis of CH3COO-

CH3COO- +H2O ==> CH3COOH + OH-

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

Kb = 1x10^-14 / 1.8x10^-5 = 5.56x10^-10

[CH3COO-] = 0.015 moles / 0.1 L = 0.15 M

Kb = (x)(x) / 0.15

5.56x10^-10 = x² / 0.15

x = 9.13x10^-6 M = [OH-]

pOH = -log 9.13x10^-6

pOH = 5.04

pH = 14 - 5.04

pH = 8.96