Question #823c6

The pH of your solution will be equal to 8.90.

Sodium fluoride, `NaF`, is a soluble salt that dissociates completely in aqueous solution to give sodium cations, `Na^(+)`, and fluoride anions, `F^(-)`.

`NaF_((s)) -> Na_((aq))^(+) + F_((aq))^(-)`

Since 1 mole of sodium fluoride produces 1 mole of fluoride anions, the concentration of the fluoride ions will be equal to that of the salt.

`[F^(-)] = [NaF] = "0.22 M"`

The fluoride ions will react with water to form hydrofluoric acid, a weak acid, and hydroxide ions, `OH^(-)`, which is an indicator that the pH of the solution will be greater than 7.

Use an ICE table to determine the concentration of the hydroxide ions

`" "F_((aq))^(-) + H_2O_((l)) rightleftharpoons HF_((aq)) + OH_((aq))^(-)`
I....0.22...............................0..................0
C....(-x)...............................(+x)..............(+x)
E...0.22-x............................x...................x

The base dissociation constant, `K_b`, will be equal to

`K_b = K_w/K_a = 10^(-14)/(3.5 * 10^(-5)) = 2.86 * 10^(-10)`

This means that you'll get

`K_b = ([HF] * [OH^(-)])/([F^(-)]) = (x * x)/(0.22 - x) = x^2/(0.22-x)`

Since `K_b` is so small, you can approximate `"(0.22-x)"` with `0.22` to get

`K_b = x^2/0.22 = 2.86 * 10^(-10)`

Therefore,

`x = sqrt(0.22 * 2.86 * 10^(-10)) = 7.9 * 10^(-6)`

You can now determine the pOH of the solution

`pOH = -log([OH^(-)])`

`pOH = -log(7.9 * 10^(-6)) = 5.10`

As a result, the solution's pH will be

`pH_"sol" = 14 - pOH = 14 - 5.10 = color(green)("8.90")`