Question

What is the pH of 0.213 M NaF(aq)?

HF: Ka = 7.2 * 10-4

Answer is 8.24

Answer 1

NaF is a strong salt :

NaF = Na+ + F-

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

Now prepare ICE table:

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

I                       0.213                                0       0    
C                     -x . .. . . . . . .. . . +x. . .. +x
E                      0.213-x. . . . . . . . . x. . . . .x

Kw = Ka*Kb

Kb= Kw/Ka

Here Ka = 7.2 * 10-4

Kb= 1.0 x 10^-14 / 7.2 x 10^-4

=1.39 x 10^-11
F- + H2O <-----> HF + OH-

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

Assume HF]=[OH-]=X
1.39 x 10^-11 = X^2 / 0.213-x

Here Kb is very small so 0.213-X= 0.213

1.39 x 10^-11 = X^2 / 0.213

X^2= 2.96*10^-12

X= 1.72*10^-6 = [OH-]

POH = -log [OH]-
pOH = 5.76

pH =14-pOH
pH = 14-5.76

= 8.24