Question

Question #5cf09

Answer 1

Here's what I got.

Explanation:

Acetic acid is a weak acid, which means that an ionization equilibrium exists in aqueous solutions of acetic acid

`"CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)`

Notice that every mole of acetic acid that ionizes in solution produces `1` mole of acetate anions and `1` mole of hydronium cations.

This implies that the equilibrium concentrations of the two ions will be equal.

`["CH"_3"COO"^(-)] = ["H"_3"O"^(+)]`

Now, you know that in a `"0.1 M"` acetic acid solution, `10%` of the acid is ionized. This means that `10%`, or `1/10 ""^"th"`, of the initial concentration of the acid ionized to form acetate anions and hydronium cations.

In other words, you know that at equilibrium, the solution will contain

`["H"_3"O"^(+)] = "0.1 M" * 1/10 = "0.01 M" = 1 * 10^(-2)"M"`

As you know, an aqueous solution at room temperature has

`color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14))))`

This means that the concentration of hydroxide anions will be equal to

`["OH"^(-)] = 10^(-14)/(1 * 10^(-2)) = color(darkgreen)(ul(color(black)(1 * 10^(-12)"M")))`

Both values are rounded to one significant figure, the number of sig figs you have for the molarity of the acetic acid.