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

1 Answer
Oct 4, 2015

Answer:

#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)#