HF(aq) + H2O(l) → ← H3O+(aq) + F–(aq)
Ka = [H3O+]e[F–]e [HF]e
Initial0.100 0 0
Change–x +x +x
Equilibrium0.100 – x x x
We can use the pH to find the value of x:
x = [H3O+]e
pH = – log[H3O+]e
or
[H3O+]e = antilog(–pH) = 10–pH = 10–2.08 = 8.3×10–3 M
[F–]e = [H3O+]e = 8.3×0–3 M
[HF]e = 0.100 – x = 0.100 – 0.0083 = 0.092 M