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
May 6, 2016

Answer:

#"pH" = 2.39#

Explanation:

Hydrofluoric acid, #"HF"#, is a weak acid that does not ionize completely in aqueous solution to form hydronium cations, #"H"_3"O"^(+)#, and fluoride anions, #"F"^(-)#.

More specifically, the acid is partially ionized in aqueous solution, the extent of ionization depending on the value of the acid dissociation constant, #K_a#.

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 #"0.025 M"#.

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 #x# to be the concentration of hydrofluoric acid that ionizes, you can say that this concentration will produce a concentration of #x# of hydronium cations and a concentration of #x# of fluoride anions.

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 #x#

#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 #x# represents the equilibrium concentration of hydronium cations, you will have

#["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)|)))#