Question

Sodium fluoride is added to many municipal water supplies to reduce tooth decay. Calculate the pH of a 0.00335 M solution of NaF, given that the Ka of HF= 6.80x10^-4 at 25 degrees celcius.

Answer 1

NaF is a strong electrolyte so it will be 100% dissociated. The anion is the ion of a weak acid, however so it will undergo hydrolysis according to the reaction:

F^-_(aq) + H2O_(l) HF + OH^-

This is base hydrolysis.

Kb = 1.47 * 10^-11

From the equilibrium expression:

[HF] = [OH^-] = x; and [F^-] =0.00335 - x

or

from the quadratic equation we can get x

Solving this we get

x = 2.23*10^-7 M
Since x=[OH-], [OH-] = 2.23*10^-7 M

[H⁺][OH^-] = 1.00 x 10^-14

[H⁺] = 4.48*10^-8

pH = -log [H⁺]

     = -log (4.48*10^-8)

pH = 7.34