NaF____-->____Na+ +___ F-
5x10^-2M ____0_______0_________all’inizio
____0_____5x10^-2M __5x10^-2M _alla fine

F- + H2O ____<-- --> HF + OH-
5x10^-2M ________0_____0_
5x10^-2M -X_______X____X_

Con

Ki = Kw/Ka = 10^-14 / 6.8x10^-4 =1,4*10^-11

Ki = [HF]*[OH-]/[F-] = X^2 / (5x10^-2 - X)
1,4*10^-11= X^2 / (5x10^-2 - X)
Risolviamo:
X^2 + 1,4*10^-11 X – 7*10^-13= 0
Si resolve e:

X = [OH-] = 7,9*10^-7

pOH = -Log(7,9*10^-7) = 6,1

pH = 14 – 6,1 = 7,9