Determine the pH of a .22-M ##NaF## solution at ##25^@"C"##? The ##K_a## of ##HF## is ##3.5xx10^(-5)##

The 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 of the solution will be greater than 7.

Use an ICE table to determine the of the hydroxide ions

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