What is the pH at the equivalence point when 95.0 mL of a 0.175 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?

Question

What is the pH at the equivalence point when 95.0 mL of a 0.175 M  solution of acetic acid (CH₃COOH) is titrated with 0.100 M  NaOH to its end point?

Answer 1

The pH at the equivalence point between a weak acid (CH3COOH) and a strong base (NaOH) will be greater than 7. We will determine the pH by looking at the hydrolysis of the conjugate base (CH3COO-) since that is the species present at the equivalence point (moles acid = moles base)

CH3COOH + NaOH ==> CH3COONa + H2O

CH3COOH + OH- ==> CH3COO- + H2O ... net ionic equation

moles CH3COOH present = 95.0 ml x 1 L / 1000 ml x 0.175 mol/L = 0.0166 moles CH3COOH

moles NaOH needed for equivalence = 0.0166 moles NaOH

Volume NaOH needed for 0.0166 moles = 0.0166 mols NaOH x 1 L / 0.100 mols = 0.166 L

Final volume of solution = 0.0950 L + 0.166 L = 0.261 L

Moles CH3COO- formed = 0.0166 moles (see stoichiometry of net ionic equation)

Final [CH3COO-] = 0.0166 mols / 0.261 L = 0.0636 M

CH3COO- + H2O ==> CH3COOH + OH-

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

Kb = 1x10^-14 / 1.76x10^-5 = 5.68x10^-10 = (x)(x) / 0.0636 - x and assume x is small relative to 0.0636

5.68x10^-10 = x² / 0.0636

x = [OH-] = 6.01x10^-6 M

pOH = -log 6.01x10^-6 = 5.221

pH = 14 - pOH

pH = 8.779