Question

The $K_a$ value of acetic acid is $1.8 \times 10^{-5}$. What would be the degree of ionization of 0.02 M acetic acid, containing 0.01 M Sodium acetate in it?}

• A. $1.5 \times 10^{-3}$
• B. $1.8 \times 10^{-3}$
• C. $3 \times 10^{-2}$
• D. $1.5 \times 10^{-2}$

Hint:

In common ion effect problems for weak acids or bases, the concentration of the common ion supplied by the weak electrolyte is almost always negligible compared to the strong electrolyte. Directly using $[Salt]$ for the conjugate ion concentration saves valuable time during exams.

Answer 1

The Correct Option is B

Step 1: Concept:
The question asks for the degree of ionization ($\alpha$) of a weak acid (acetic acid) in a solution that also contains a strong electrolyte (sodium acetate) providing a common ion.
This is a classic application of the Common Ion Effect, which states that the addition of a common ion to a weak electrolyte solution suppresses the ionization of the weak electrolyte.

Step 2: Key Formula or Approach:
The dissociation of acetic acid is given by:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
The acid dissociation constant ($K_a$) is expressed as:
\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]
Let $C_a$ be the initial concentration of the acid and $C_s$ be the concentration of the added salt (which dissociates completely).

Step 3: Step-by-step Explanation:

• Let the initial concentration of acetic acid be $C_a = 0.02$ M.

• The initial concentration of the common ion from sodium acetate is $C_s = 0.01$ M.

• Let $\alpha$ be the degree of ionization of acetic acid. The change in concentration for $CH_3COOH$ is $-C_a\alpha$, and for $H^+$ and $CH_3COO^-$ it is $+C_a\alpha$.

• At equilibrium, the concentrations are:
$[CH_3COOH] = C_a(1 - \alpha) = 0.02(1 - \alpha)$
$[H^+] = C_a\alpha = 0.02\alpha$
$[CH_3COO^-] = C_s + C_a\alpha = 0.01 + 0.02\alpha$

• Since acetic acid is a weak acid and its dissociation is further suppressed by the common ion, $\alpha$ is extremely small ($\alpha \ll 1$).

• We can approximate: $(1 - \alpha) \approx 1$ and $(0.01 + 0.02\alpha) \approx 0.01$.

• Substituting these into the $K_a$ expression:
\[ 1.8 \times 10^{-5} = \frac{(0.01)(0.02\alpha)}{0.02} \]

• Simplifying the equation:
\[ 1.8 \times 10^{-5} = 0.01 \times \alpha \]

• Solving for $\alpha$:
\[ \alpha = \frac{1.8 \times 10^{-5}}{0.01} = 1.8 \times 10^{-3} \]

Step 4: Final Answer:
The degree of ionization is mathematically calculated to be $1.8 \times 10^{-3}$, which matches option (B).