Question

Calculate the percentage ionization of 0.01 M acetic acid in 0.1 M HCl. ${K_a}$ of acetic acid is $1.8 \times {10^{ - 5}}$.
A.) 0.18%
B.) 0.018%
C.) 1.8%
D.) 18%

Answer 1

Hint: HCl is a strong acid. It dissociates completely into hydrogen ion and chlorine ion. However, acetic acid is a weak acid. It undergoes a common ion effect by the hydrogen ions of HCl and its ionization decreases.

Complete step by step answer:
A weak acid is an acid that ionizes only slightly in an aqueous solution. An example of a weak acid is acetic acid. The extent of ionization of weak acids varies but is generally less than 10%. Weak acids, like strong acids, ionize to yield the ${H^ + }$ ion and a conjugate base. Also, the ${H^ + }$ ions of acetic acid undergo the common ion effect by the ${H^ + }$ ions of HCl. The common-ion effect is the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. It is a consequence of Le Chatelier’s principle for the equilibrium reaction of the ionic association/dissociation.
In the given question, acetic acid decomposes according to the reaction:
$\mathop {C{H_3}COOH}\limits_{0.01 - 0.01\alpha } \to \mathop {C{H_3}CO{O^ - }}\limits_{0.01\alpha } + \mathop {{H^ + }}\limits_{0.01\alpha + 0.1}$
The equilibrium equation is:
${K_a} = \dfrac{{[C{H_3}CO{O^ - }][{H^ + }]}}{{[C{H_3}COOH]}}$
$1.8 \times {10^{ - 5}} = \dfrac{{(0.01\alpha )\left( {0.01\alpha + 0.1} \right)}}{{0.01 - 0.01\alpha }}$
As <<< 1, 0.01 <<< 0.01
$\therefore 1.8 \times {10^{ - 5}} = \dfrac{{(0.01\alpha )(0.1)}}{{0.01}}$
$\therefore \alpha = 1.8 \times {10^{ - 4}}$
Percentage dissociation of acetic acid = 100%
= 0.018%

Hence, the correct answer is (B) 0.018%.

Note: Remember that the common ion effect is by the strong acid on the weak acid. It hinders its dissociation if they have any ion in common.