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

"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)|)))