Question

Calculate the pH of a .003 M solution of NaF, given that the Ka of HF = 6.8 x 10^-4 at 25 degrees celcius.

Answer 1

Okay, so the first thing that happens it that the NaF completely dissociates:
NaF ---> Na+ + F-

Cool. Now, the Na+ is just floating around, so we can ignore it. Let's deal with the F-. The F- is in the water, so it's going to do this:
F- + H2O <---> HF + OH-
Note that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.

But first, because I'm dealing with OH- ions, I'll need a Kb value. Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:
(1E-14)/(6.8E-4)= Kb = 1.47E-11

Now,
Kb=[OH-][HF]/[F-]
or
1.47E-11 = x^2/(.003-x)
solve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):
x=2.2E-7 M
Since x=[OH-], [OH-]=2.2E-7 M

We can calculate pOH by taking the negative log of [OH-]:
-log 2.23E-7 M = 6.65
Since pH + pOH = 14.00:
pH + 6.65 = 14.00
pH = 14.00-6.65
pH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy. Yay!