Gen Chem 2 Buffer pH

Question

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 0.160 mol of NaOH were added?

Answer 1

Henderson Hasselbalch equation:

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

Initial mols HF = 0.300 mol

Initial mols F- = 0.200 mol

Initial mols OH- = 0.160 mol

pKa = -log Ka = -log 6.8x10^-4 = 3.17

HF + OH- ==> F- + H2O

0.300....0.200......0.200..........Initial

-0.2......-0.2..........+0.2............Change

0.100......0...........0.400..........Equilibrium

pH = 3.17 + log (0.400/0.100)

pH = 3.17 + log 4

pH = 3.17 + 0.602

pH = 3.772