Hydrofluoric acid, HF, has an acid dissociation constant of #6.8 * 10^(-4)# at 25 °C. At equilibrium the concentration of HF is 0.025M. a. What is the equilibrium constant expression for this dissociation? b. What is the pH of this solution?
1 Answer
Explanation:
Hydrofluoric acid,
More specifically, the acid is partially ionized in aqueous solution, the extent of ionization depending on the value of the acid dissociation constant,
The equilibrium that is established when hydrofluoric acid ionizes looks like this
#"HF"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "F"_((aq))^(-)#
By definition, the acid dissociation constant for this equilibrium will be
#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_a = (["F"^(-)] * ["H"_3"O"^(+)])/(["HF"]))color(white)(a/a)|)))#
Keep in mind that the expression of the acid dissociation constant uses equilibrium concentrations.
Now, you know that the equilibrium concentration of hydrofluoric acid is equal to
Notice that every mole of hydrofluoric acid that dissociates produces one mole of hydronium cations and one mole of fluoride anions.
This means that if you take
This means that you have
#["F"^(-)] = x" "# and#" " ["H"_3"O"^(+)] = x#
Use the expression of the acid dissociation constant to find the value of
#K_a = (x * x)/0.025 = x^2/0.025#
This will get you
#x = sqrt(0.025 * K_a)#
#x = sqrt(0.025 * 6.8 * 10^(-4)) = 4.12 * 10^(-3)#
Since
#["H"_3"O"^(+)] = 4.12 * 10^(-3)"M"#
The pH of the solution is defined as
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
In your case, you will end up with
#"pH" = - log(4.12 * 10^(-3)) = color(green)(|bar(ul(color(white)(a/a)2.39color(white)(a/a)|)))#