Question

A monobasic weak acid dissociates 2% in its 0.002 M solution. Calculate the dissociation constant of weak acid.

• A. $2 \times 10^{-9}$
• B. $8 \times 10^{-7}$
• C. $6 \times 10^{-7}$
• D. $4 \times 10^{-6}$

Hint:

When $\alpha$ is 5% or less, the approximation $1 - \alpha \approx 1$ holds true, making $K_a = C\alpha^2$ highly accurate. If $\alpha$ is large, you must use the full quadratic form $K_a = \frac{C\alpha^2}{1-\alpha}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a weak acid, the dissociation constant $K_a$ is related to the degree of dissociation ($\alpha$) and concentration ($C$) by Ostwald's Dilution Law: $K_a = C\alpha^2$.

Step 2: Formula Application:
$C = 0.002$ M $= 2 \times 10^{-3}$ M. $\alpha = 2% = 0.02 = 2 \times 10^{-2}$.

Step 3: Explanation:
$K_a = (2 \times 10^{-3}) \times (2 \times 10^{-2})^2$ $K_a = (2 \times 10^{-3}) \times (4 \times 10^{-4})$ $K_a = 8 \times 10^{-7}$.

Step 4: Final Answer:
The dissociation constant is $8 \times 10^{-7}$.