Calculate the pH and the concentrations of all species present, "H"_3"O"^(+), "F"^(-), "HF", and "OH"^(-), in "0.05 M HF" ?
1 Answer
Here's what I got.
Explanation:
For starters, make sure that you have the value of the acid dissociation constant for hydrofluoric acid
K_a = 7.2 * 10^(-4)
So, hydrofluoric acid is a weak acid that only partially ionizes in aqueous solution to produce hydronium cations and fluoride anions
"HF"_ ((aq)) + "H"_ 2"O" _ ((l)) rightleftharpoons "H"_ 3 "O"_ ((aq))^(+) + "F"_ ((aq))^(-)
Notice that the ionization equilibrium produces
If you take
["H"_3"O"^(+)] = ["F"^(-) ] =x "M " -> the two ions are produced in a1:1 mole ratio
["HF"] = ["HF"]_0 - x -> the concentration of the acid decreases byx "M"
In your case
["HF"]_0 = "0.05 M"
and so the equilibrium concentration of the acid, i.e. what remains unionized in solution, will be
["HF"] = (0.05 - x) "M"
By definition, the acid dissociation constant is equal to
K_a = (["H"_3"O"^(+)] * ["F"^(-)])/(["HF"])
In your case, you have
7.2 * 10^(-4) = (x * x)/(0.05 - x) = x^2/(0.05 - x)
Rearrange to quadratic equation form
x^2 + 7.2 * 10^(-4) * x - 0.05 * 7.2 * 10^(-4) = 0
SIDE NOTE: You cannot use the approximation
0.05 - x ~~ 0.05
because you will have
7.2 * 10^(-4) = x^2/0.05
which will get you
x = 0.006
Remember that the approximation holds only if
x/(["HF"]_0) xx 100% < 5%
In this case, the approximation does not hold, since
(0.006 color(red)(cancel(color(black)("M"))))/(0.005color(red)(cancel(color(black)("M")))) xx 100% = 12% > 5%
This quadratic equation will produce two solutions, one positive and one negative. Since
Therefore, you will have
x = 0.00656
This means that at equilibrium, you will have
["H"_ 3"O"^(+)] = "0.00656 M"
["F"^(-)] = "0.00656 M"
["HF"] = "0.05 M" - "0.00565 M" = "0.0444 M"
As you know, an aqueous solution at room temperature has
color(blue)(ul(color(black)(["H"_ 3"O"^(+)] * ["OH"^(-)] = 10^(-14))))
This means that you will have
["OH"^(-)] = 10^(-14)/(["H"_3"O"^(+)])
which gets you
["OH"^(-)] = 10^(-14)/(0.00565) = 1.77 * 10^(-12) "M"
Finally, the pH of the solution
color(blue)(ul(color(black)("pH" = - log( ["H"_3"O"^(+)]))))
will be equal to
"pH" = - log(0.00565) = 2.25
I'll leave the concentrations rounded to three sig figs and the pH rounded to two decimal places, but keep in mind that you only have one significant figure for the initial concentration of the acid.