Question

Question #fcf52

Answer 1

`7.2 * 10^(-4)`

Explanation:

The first thing to note here is that the concentration of the acid is said to be `"0.1 F"`, or `"0.1 formal"`.

Formality is simply a measure of concentration that does not take into account the form in which the solute exists in solution.

In this case, you can say that a `"0.1-F"` solution will be equivalent to a `"0.1-M"` solution.

Now, hydrofluoric acid, `"HF"`, is a weak acid that does not ionize completely in aqueous solution. The problem tells you that in your solution, only `8.13%` of the acid is ionized.

If you think about this at a molecular level, a `8.13%` ionization basically tells you that out of `10000`, or `10^4`, molecules of hydrofluoric acid present in solution, only `813` will ionize, i.e. give off their proton.

The equilibrium reaction for the ionization of hydrofluoric acid looks like this

`color(red)("H")"F"_text((aq]) + "H"_2"O"_text((l]) rightleftharpoons "F"_text((aq])^(-) + "H"_3"O"_text((aq])^(color(red)(+)`

Notice that every molecule of `"HF"` that ionizes forms one molecule of fluoride anions, `"F"^(-)`, the conjugate base of the acid, and one molecule of hydronium ions, `"H"_3"O"^(+)`.

By definition, the acid dissociation constant, `K_a`, will be equal to

`K_a = (["F"^(-)] * ["H"_3"O"^(+)])/(["HF"])`

Now, if you start with `"0.1 M"` of hydrofluoric acid, and you know that `8.13%` of the molecules ionize, you can calculate the equilibrium concentration of the fluoride and hydronium ions by using a proportion

`["F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * ["HF"]`

In this case, you will have

`["F"^(-)] = ["H"_3"O"^(+)] = 8.13/100 * "0.1 M" = 8.13 * 10^(-3)"M"`

But remember, if `8.13 * 10^(-3)"M"` is ionized, that must mean that `8.13 * 10^(-3)` of the acid is no longer in molecular form. Therefore, the equilibrium concentration of the acid will be

`overbrace("0.1 M")^(color(purple)("what you start with")) - overbrace(8.13 * 10^(-3)"M")^(color(blue)("what ionizes")) ~~ "0.09187 M"`

Now you're ready to plug in your values into the equation for `K_a`

`K_a = (8.13 * 10^(-3) * 8.13 * 10^(-3))/0.09187 = color(green)(|bar(ul(color(white)(a/a)7.2 * 10^(-4)color(white)(a/a)|)))`

The actual value for the acid dissociation constant of hydrofluoric acid is `7.2 * 10^(-4)`, so this is an excellent result!

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf