Question #5cf09

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Question #5cf09

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Answer 1 · 2017-03-12T01:45:41

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

${CH}_{3} {COOH}_{(a q)} + H_{2} O_{(l)} ⇌ {CH}_{3} {COO}_{(a q)}^{-} + H_{3} O_{(a q)}^{+}$ $"CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)$

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

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

$[{CH}_{3} {COO}^{-}] = [H_{3} O^{+}]$ $["CH"_3"COO"^(-)] = ["H"_3"O"^(+)]$

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

$[H_{3} O^{+}] = 0.1 M \cdot \frac{1}{10} = 0.01 M = 1 \cdot 10^{- 2} M$ $["H"_3"O"^(+)] = "0.1 M" * 1/10 = "0.01 M" = 1 * 10^(-2)"M"$

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.

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Question #5cf09

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