What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.100 mol of NaOH were added?

Answer 1

HF + NaOH ==> NaF + H2O

0.3......0.1.............0.2............Initial

-0.1...-0.1..............+0.1........Change

0.2.....0..................0.3..........Equilibrium

pH = pKa + log [NaF]/[HF]

pKa = -log Ka = 3.17

pH = 3.17 + log (0.3/0.2) = 3.17 + log 1.5

pH = 3.17 + 0.18

pH = 3.35