Question

At 25 \(^{\circ}\)C, \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \). What is the percentage of ionization of 0.02 M acetic acid at this temperature?

• A. 3
• B. 4.242
• C. 5
• D. 1.414

Hint:

For weak acids, if the ratio of the initial concentration to the acid dissociation constant ($C/K_a$) is greater than 1000, the approximation $1-\alpha \approx 1$ is generally valid. In this case, $C/K_a = (0.02) / (1.8 \times 10^{-5}) \approx 1111$, so the approximation is appropriate. Remember to convert the degree of ionization ($\alpha$) to percentage by multiplying by 100.

Answer 1

The Correct Option is A

Step 1: Write the Ionization Equilibrium for Acetic Acid
Acetic acid (\(\text{CH}_3\text{COOH}\)) is a weak acid that ionizes partially in water. \[ \text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq) \] Step 2: Set Up an ICE Table (Initial, Change, Equilibrium)
Let the initial concentration of acetic acid be $C = 0.02 \text{ M}$. Let $\alpha$ be the degree of ionization.
Step 3: Write the Expression for the Acid Dissociation Constant (\( K_a \))
The acid dissociation constant ($K_a$) for acetic acid is given by: \[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \] Substitute the equilibrium concentrations from the ICE table: \[ K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C^2\alpha^2}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \] Step 4: Apply Approximation for Weak Acids
For weak acids, the degree of ionization ($\alpha$) is very small (typically much less than 1). Therefore, we can approximate $(1-\alpha) \approx 1$. The $K_a$ expression simplifies to: \[ K_a \approx C\alpha^2 \] From this, we can solve for $\alpha$: \[ \alpha^2 = \frac{K_a}{C} \] \[ \alpha = \sqrt{\frac{K_a}{C}} \] Step 5: Substitute Given Values and Calculate \( \alpha \)
Given: \begin{itemize} \item $K_a = 1.8 \times 10^{-5}$ \item $C = 0.02 \text{ M}$ (which is $2 \times 10^{-2} \text{ M}$) \end{itemize} \[ \alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.02}} \] \[ \alpha = \sqrt{\frac{18 \times 10^{-6}}{2 \times 10^{-2}}} \] \[ \alpha = \sqrt{9 \times 10^{-4}} \] \[ \alpha = 3 \times 10^{-2} \] \[ \alpha = 0.03 \] Step 6: Calculate the Percentage of Ionization
Percentage of ionization = $\alpha \times 100%$ \[ \text{Percentage of ionization} = 0.03 \times 100% = 3% \] Step 7: Analyze Options
\begin{itemize} \item Option (1): 3. Correct, as it matches our calculated percentage of ionization. \item Option (2): 4.242. Incorrect. \item Option (3): 5. Incorrect. \item Option (4): 1.414. Incorrect. \end{itemize}