Question

Calculate the pH of a 3.66×10-3 M solution of NaF, given that the Ka of HF = 6.80 x 10-4 at 25°C.

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)/6.8*10^-4

Kb = 1.471*10^-11

F- dissociates as

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

0.0037 0 0

0.0037-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 ((1.471*10^-11)*3.66*10^-3) = 2.32*10^-7

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

so, x = 2.32*10^-7 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.32*10^-7)

= 6.63

we have below equation to be used:

PH = 14 - pOH

= 14 - 6.63

= 7.37

Answer: 7.37