Question

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH₃CO₂H?

CH₃CO₂H(aq) + H₂O(l)⇌H₃O⁺(aq) + CH₃CO₂⁻(aq) Ka = 1.8 × 10−5

(Hint: Determine [CH₃CO₂⁻] at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or

[CH₃CO₂⁻] / [CH₃CO₂H]_initial × 100.

Short answer

The percentage ionization of a week acid can be explained as ratio of H3O+ at equilibrium to the initial HA concentration increased by 100%

Step-by-step solution

calculating Kc

Acetic acid reacts with water to give the following reaction

Given:

\(\begin{array}{l}C{H_3}COO{H_{(eq)}} + {H_2}O\rightleftharpoons {H_3}{O^ + }_{(aq)} + C{H_3}CO{O^ - }\\c = 0 \times 100M\\{k_a} = 1.8 \times 1{0^{ - 5}}\\\left[ {{H_3}{O^ + }} \right] = \sqrt {Kq.C} = \sqrt {1.8 \times 1{0^{ - 5}} \times 0.100} \\\left[ {{H_3}{O^ + }} \right] = 1.34 \times 1{0^{ - 3}}\;M\end{array}\)

\(\begin{array}{l}\% lonisation = \frac{{\left[ {{H_3}{O^ + }} \right] \times 100}}{C} = \frac{{1.34 \times 1{0^{ - 3}}}}{{0.100}} \times 100\\\% lonisation = 1.34\% \end{array}\)

Final answer

\(\% lonisation = 1.34\% \)

Thus the percentage of ionization is 1.34%