Question

What is the pH of a 2.75 x 10^-3 M solution of potassium fluoride, KF, in water? The K_a for hydrofluoric acid is 3.55 x 10^-5.

please show steps!!

the answer is 7.94

Answer 1

KF is a weak acid... therefore

In solution:

KF -> K+ and F-

Also, there could be an acid forming H+ and F- => HF

The F- ion will "steal" H+ ions from the water! therefore the OH- concentration will be greater than H+

Remember: KF is a basic salt since F- is the conjugate base of the weak acid HF

F- + H2O ---> HF + OH-

We really need Kb not Ka! Kw = 10^-14

Kb = Kw / Ka = (10^-14 / 3.55*10^-5) = 2.8*10^-10

LET [HF] = [OH-] = x

and [F-] = M - x = 2.75*10^-3 - x

substitute in Kb value

Kb = [HF][OH-] / [F-] (by definition)

substitute everything

Kb = x^2 / (2.75*10^-3 - x)

You could ignore -x because this is so small (if you don't want to ignore it its ok, just solve for x with a quadratic formula)

Kb = x^2 / 2.75*10^-3 = 2.8*10^-10

x^2 = 2.1 x 10^-12
x^2 = 7.7*10^-13

x = sqrt (7.7*10^-13)

x= 8.8*10^-7

x= [OH-] = 8.8*10^-7
pOH = -log[OH-] = -log(8.8*10^-7) = 6.05
pH = 14 - pOH = 14 - 6.05 = 7.95

pH= 7.95 which is what you proposed at first!