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 0.090 mol of HCl were added?

Answer 1

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

F- + H+ ===> HF

0.2.....0.09.......0.3......Initial

-0.09...-0.09...+0.09...Change

0.11......0.......0.39......Equilibrium

Henderson Hasselbalch equation:

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

pH = 3.17 + log (0.11 / 0.39)

pH = 3.17 + (-0.55)

pH = 2.62