Question

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

Answer 1

`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)`