Buffer Problem (Help needed!!)

Question

Calculate the pH after 0.18 mol of NaOH is added to 1.00 L of the solution that is 0.79 M HF and 1.00 M KF.

Answer 1

Moles HF = 0.79 M x 1.00 L = 0.79

Moles F- = 1.00 M x 1.00 L = 1.00

HF + OH- >> F- + H2O

the effect of the added 0.18 mol of OH- is to convert HF to F-

Moles HF = 0.79 - 0.18 = 0.61

Moles F- = 1.00 + 0.18 = 1.18

V = 1.00 L

[HF] = 0.61 mol / 1.00 L = 0.61M

[F-] = 1.18 mol / 1.00 L = 1.18 M

pH = pKa + log [F-] /[HF]

pKa = 3.15

pH = 3.15 + log 1.18 / 0.61 =  3.44