What is the pH at the equivalence point when 25.0 mL of a 0.275 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?

Answer 1

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

Recall that the equivalence point is when the moles of acetic acid equals the moles of NaOH.

moles CH3COOH present = 25.0 ml x 1 L / 1000 ml x 0.275 mols / L = 6.875x10^-3 mols

for equivalence, we need to provide 6.875x10^-3 mols NaOH

volume of NaOH to provide 6.875x10^-3 mols NaOH = 6.875x10^-3 mols x 1 L / 0.100 mol = 0.06875 L

To find the final pH of solution, we will have to look at the hydrolysis of CH3COONa, since it is now the only species present. To do this, we will need to know the final volume of solution, the final concentration of CH3COONa and the Ka of CH3COOH, or the Kb of CH3COO^-.

Final volume of solution (L) = 0.025 L + 0.06875 L = 0.09375 L

Final [CH3COONa] = 6.875x10^-3 mols / 0.09375 L = 0.0733 M

Hydrolysis reaction:

CH3COO^- + H2O ==> CH3COOH + OH- (Na omitted as it is a spectator ion)

CH3COO^- is acting as a base, so we need the Kb ...

Looking of Ka CH3COOH, we find it to be 1.76x10^-5

Kb = 1x10^-14 / 1.76x10^-5 = 5.68x10^-10

Write the Kb expression, and solve for [OH-]...

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

5.68x10^-10 = (x)(x) / 0.0733 - x (assume x is small and ignore it in denominator)

5.68x10^-10 = x² / 0.0733

x² = 4.16x10^-11

x = 6.45x10^-6 (note this is very small relative to 0.0733 so above assumption was valid)

This is the [OH-], so we can now calculate pOH and then pH ...

pOH = -log [OH-] = -log 6.45x10^-6

pOH = 5.19

pH = 14 - 5.19

pH = 8.81