A 450.0 mL sample of 0.20 M HF is titrated with 0.10 M NaOH. Determine the pH of the solution after the addition of 450.0 mL of NaOH. The Ka of HF is 6.8 × 10-4.

Answer 1

HF is a weak acid. Titrating with NaOH produces NaF + H2O and NaF (or F^-) is the conjugate base. When you have a weak acid and its conjugate base, you have a buffer.

mols HF = 0.450 L x 0.20 mol/L = 0.090 mols

mols OH- = 0.450 L x 0.10 mol/L = 0.045 mols

HF + OH^- ==> F^- + H2O

0.09.....0.045......0............Initial

-0.0045..-0.045...+0.045...Change

0.045........0........0.045......Equilibrium

So @ equilib. you have equal mols of weak acid and conjugate base.

At this point, pH = pKa (see Henderson Hasselbalch equation below)**

pH = - log 6.8x10-4

pH = 3.16

**HH equation

pH = pKa + log [conj. base] / [acid]

pH = 3.16 + log [0.045] / [0.045]

pH = 3.16