Question

What is the pH of a solution that is 0.300M in NaF? The pKa of HF is 3.455

​Please walk me through all the steps to figuring this one out. For some reason it is really not making sense to me at all.

Answer 1

1st find kb of F-

use:

pKa = -log Ka

3.455 = -log Ka

Ka = 3.508*10^-4

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/3.508*10^-4

Kb = 2.851*10^-11

F- dissociates as

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

0.3 0 0

0.3-x x x

Kb = [HF][OH-]/[F-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.851*10^-11)*0.3) = 2.925*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.925*10^-6 M

So,

[OH-] = 2.925*10^-6 M

use:

pOH = -log [OH-]

= -log (2.925*10^-6)

= 5.53

use:

PH = 14 - pOH

= 14 - 5.53

= 8.47

Answer: 8.47