Determine the [OH-] of a solution that is 0.100 M in F-? Determine the pH of this solution?
1 Answer
Explanation:
You actually need the base dissociation constant,
Since you did not provide that value, I will use the acid dissociaation constant,
The
Moreover, you know that
K_W = K_a xx K_b
This means that
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])^(-)
By definition, the base dissociaation constant will be equal to
K_b = (["HF"] * ["OH"^(-)])/(["F"^(-)]) = (x * x)/(0.100 - x)
Because
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 = -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)