# Question 5cf09

Mar 12, 2017

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

${\text{CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

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.

$\left[{\text{CH"_3"COO"^(-)] = ["H"_3"O}}^{+}\right]$

Now, you know that in a $\text{0.1 M}$ acetic acid solution, 10% of the acid is ionized. This means that 10%, or $\frac{1}{10} \text{^"th}$, 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}}}}$

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.