what is the pH of the buffer?

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 0.060 mol of NaOH were added?

Answer 1

The added NaOH (OH^-) will react with the acid (HF) to form H₂O and F^-.

HF + OH^- ===> F^- + H₂O

0.300....0.060..........0.200...........Initial

-0.06.....-0.06..........+0.06...........Change

0.240.......0...............0.26..........Equilibrium

So, after the addition of 0.060 mols NaOH, we have 0.240 M HF and 0.26 M F^-

We can now use the Henderson Hasselbalch equation to calculate the pH:

pH = pKa + log [F^-] / [HF] and pKa = -log Ka = -log 6.8x10^-4 = 3.17

pH = 3.17 + log (0.26/0.24)

pH = 3.17 + log 1.08

pH = 3.17 + 0.0348

pH = 3.20