Question

a solution that is made by combining 55 mL of 6.0×10−2 M hydrofluoric acid with 125 mL of 0.12 M sodium fluoride I kept getting around 7.7 for the pH. I found the ml of HF and NaF to be (3.3*10^-3 mol) and (1.5*10^-2) respectively. The total volume to be 0.180L. and pKa value of 3.17 with a Ka value of (6.8*10^-4). The answer given was pH = 3.83 but I can't seem to get the same answer.

Answer 1

HF -----> H+ + F-

Ka = [H+][F-] / [HF] = 6.8 x 10^-4

After mixing, the concentrations are:
[HF] = 0.0600 M x 55/ 180 = 0.0183 M
[F-] = 0.12 M x 125/180 = 0.0833 M

Let s = amount of HF that ionizes (or dissociates)
[HF] = 0.0183 - s
[H+] = s
[F-] = 0.0833 + s

Ka = [H+][F-] / [HF]
6.8 x 10^-4 = (s)(0.0833 + s) / (0.0183 - s)
0.00068 = (0.0833 s + s^2) / (0.0183 - s)
0.00068 (0.0183 - s ) = 0.0833 s + s^2
0.00001244 - 0.00068 s = 0.0833 s + s^2
s^2 + 0.0839 s - 0.00001244 = 0

Solve the quadratic equation for s.

s = 0.000148 M = [H+]+
pH = - log (0.000148) = 3.83

F- is a common ion in this solution. The presence of F- causes the equilibrium to shift to the left which reduces the amount of ionization of the HF, thus reducing the H+ and increasing the pH.