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")#