Calculate the equilibrium constant for the following reaction of a weak acid and base HF + NH3 <=> NH4+ + F-

Question

given Ka (HF) = 6.9 x 10^-4 and Kb (NH3) = 1.8 x 10^-5 and Kw = 1.0 x 10^-14

Answer 1

I would look at this.
Write Ka expression for HF.
Write Kb expression for NH3(aq)
Write Kw expression for H2O.

Now, Write Keq expression for the reaction. Substitute for (NH4^+) from the Kb expression, substitute for the (F^-) from the Ka expression and cancel similar terms. I think that will get Keq in terms of Kw, Ka, and Kb. You will need to remember that (H^+) = (F^-) for the last step.

Answer 2

much thanks!!!