Question

Sodiumfluoride, NaF,is added to many municipal water supplies to reduce tooth decay in the population. Calculate the pH of a 0.30M solution of NaF when 0.10 M of NaOH is added to the solution, Kb=1.5×10-11.

Answer 1

NaF will ionize as

NaF ---> Na⁺ + F^-

Flouride ion will undergo hydrolysis as
                  F^- + H₂O   ---> HF + OH-

Initial           0.3                  0      0

Chnage      -x                    +x   +x

Equilibr     0.3-x                  x        x

Kb = [HF] [ OH^-] / [F^-] = 1.5 X 10^-11
1.5 X 10^-11 = x² / (0.3-x)

x can be ignored in denominator as Kb is very low

1.5 X 10^-11 = x² / 0.3

4.5 X 10^-12 = x²

x = 2.12 X X 10^-6 = [OH^-]

Now on adding NaOH it will also furnish OH- = 0.1 M

So total [OH-] = [OH-]_NaF + [OH-]_NaOH =2.12 X X 10^-6 + 0.1 = 0.10000212 M

pOH = -log[OH-] = 0.99

pH= 14- pOH = 14- 0.99 = 13 (approx)