Chemistry HW help

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 0.110 mol of HCI were added?

Answer 1

NaF + HCl ==> HF + NaCl

0.200.....0.11........0.300...............Initial

-0.11.....-0.11.......+0.11................Change

0.09.........0...........0.41.................Equilibrium

We can use the Henderson Hasselbalch equation

pH = pKa + log [NaF] / HF]

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

pH = 3.17 + log (0.09 / 0.41) = 3.17 + log 0.220

pH = 3.17 - 0.66

pH = 2.51