Question

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?

Answer 1

`"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)|)))`