Question

At $27^\circ C$, the $K_a$ of acetic acid is $1.8 \times 10^{-5}$. What is the percentage of ionization of 0.02M of this acid at $27^\circ C$?

• A. 6
• B. 3
• C. 2
• D. 1.5

Hint:

Always check if the approximation $\alpha \ll 1$ holds. Here, $0.03$ is indeed much less than $1$, so the simplified formula is valid.

Answer 1

The Correct Option is B

Step 1: Formula for Degree of Ionization ($\alpha$):
For a weak monobasic acid like acetic acid ($CH_3COOH$), the degree of ionization $\alpha$ is related to the acid dissociation constant $K_a$ and concentration $C$ by Ostwald's dilution law: \[ K_a = C\alpha^2 \] (Assuming $\alpha \ll 1$, so $1-\alpha \approx 1$). \[ \alpha = \sqrt{\frac{K_a}{C}} \]
Step 2: Substitution and Calculation:
Given: $K_a = 1.8 \times 10^{-5}$ $C = 0.02 M = 2 \times 10^{-2} M$ \[ \alpha = \sqrt{\frac{1.8 \times 10^{-5}}{2 \times 10^{-2}}} \] \[ \alpha = \sqrt{0.9 \times 10^{-3}} \] Convert to even power of 10 for easier root: \[ \alpha = \sqrt{9 \times 10^{-4}} \] \[ \alpha = 3 \times 10^{-2} \] \[ \alpha = 0.03 \]
Step 3: Convert to Percentage:
Percentage Ionization = $\alpha \times 100$. \[ % = 0.03 \times 100 = 3% \]
Step 4: Final Answer:
The percentage of ionization is 3%.