Determine the [OH-] of a solution that is 0.100 M in F-? Determine the pH of this solution?

1 Answer
Oct 4, 2015

pH_"sol" = 8.07

Explanation:

You actually need the base dissociation constant, K_b, of the fluoride anion, "F"^(-), in order to be able to calculate the concentration of hydroxide ions it produces in aqueous solution.

Since you did not provide that value, I will use the acid dissociaation constant, K_a, of hydrofluoric acid, "HF", to calculate the K_b of "F"^(-).

The K_a of hydrofluoric acid is equal to 7.2 * 10^(-4). You know that water's self-ionization constant, K_W, is euqal to 10^(-14) at room temperature.

Moreover, you know that

K_W = K_a xx K_b

This means that K_b will be equal to

K_b = K_W/K_a = 10^(-14)/(7.2 * 10^(-4)) = 1.39 * 10^(-11)

So, the fluoride anion will actually react with water to form hydrofluoric acid. Use an ICE table to help you determine the equilibrium concentration of the hydroxide ions

"F"_text((aq])^(-) + "H"_2"O"_text((l]) -> "HF"_text((aq]) + "OH"_text((aq])^(-)

color(purple)("I")" " " "0.100" " " " " " " " " " "0" " " " " " " "0
color(purple)("C")" " " "(-x)" " " " " " " "(+x)" " "(+x)
color(purple)("E")" "(0.100-x)" " " " " "x" " " " " " "x

By definition, the base dissociaation constant will be equal to

K_b = (["HF"] * ["OH"^(-)])/(["F"^(-)]) = (x * x)/(0.100 - x)

Because K_b is so small, you can say that (0.100-x) ~~ 0.100, which means that you have

K_b = x^2/0.100 = 1.39 * 10^(-11)

x = sqrt(0.100 * 1.39 * 10^(-11)) = 1.18 * 10^(-6)

The equilibrium concentration of hydroxide ions will thus be

x = ["OH"^(-)] = 1.18 * 10^(-6)"M"

The pOH of the solution will be

pOH = -log(["OH"^(-)]) = - log(1.18 * 10^(-6)) = 5.93

Therefore, the pH of the solution will be

pH_"sol" = 14 - pOH = 14 - 5.93 = color(green)(8.07)