Question

I'll become a fan of anyone who can explain this.
How would I answer the following:
What is the pH of .5 M NaF? the Ka of HF is 4.5e-4.

Answer 1

pH = -log\([H^+]\)

Answer 2

p = P - x H+ y.

Answer 3

what equation is that?

Answer 4

ill give you a hint:

when you put NaF in water it will dissociate into Na+ and F-, some of the F- will steal H's from water making OH- in the process

F- + H2O <-> HF + OH-

the Ka given is for the dissociation of HF into H+ and F-

HF <-> H+ + F-
Ka = [H+][F-]/[HF]

Answer 5

what would I do next?

Answer 6

you have to use the Kb value for HF and set up an equilibrium expression of this equation: F- + H2O <-> HF + OH-

Answer 7

ps. Kw=Ka*Kb

Answer 8

So eventually I got the Kb to be 2.2e-11

Answer 9

If I take the -log of that and then subtract that value from 14, is that the pH?

Answer 10

@aaronq

Answer 11

you have to find OH- from the equilibrium expression you set up

Answer 12

Kb does not go into -log[H+]

Answer 13

can you show me how?

Answer 14

so pH does not equal pKa?

Answer 15

nope, pH is related to the hydronium concentration, pKa is the point where half of the acid is in protonated and half in deprotonated forms

Answer 16

F- + H2O <-> HF + OH-

Kb=products/reactants

Answer 17

?