Question #5cf09

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.