Question

What is the pH of a 0.895M sodium fluoride solution (NaF) knowing that the K_​a of hydrofluoric acid is 6.6 x 10^-4 ​?

Answer 1

we have below equation to be used:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

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

Kb = (1.0*10^-14)/6.6*10^-4

Kb = 1.515*10^-11

F- dissociates as

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

0.895 0 0

0.895-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.515*10^-11)*0.895) = 3.682*10^-6

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

so, x = 3.682*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.682*10^-6)

= 5.43

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.43

= 8.57

Answer: 8.57