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 Question:
We are given the percentage dissociation ($% \alpha$) and the molar concentration ($C$) of a weak monobasic acid.
We need to calculate the acid dissociation constant ($K_a$).

Step 2: Key Formula or Approach:
For a weak monobasic acid $HA \rightleftharpoons H^+ + A^-$, the dissociation constant $K_a$ is defined by Ostwald's Dilution Law.
When the degree of dissociation ($\alpha$) is very small ($\alpha < 0.05$), the equation approximates to:
$$K_a \approx C \alpha^2$$
Where $C$ is the initial concentration and $\alpha$ is the fractional degree of dissociation.

Step 3: Detailed Explanation:
First, convert the percentage dissociation into the fractional degree of dissociation ($\alpha$):
$$\alpha = \frac{2%}{100} = 0.02 = 2 \times 10^{-2}$$
Now, write down the concentration in scientific notation for easier multiplication:
$$C = 0.002 \text{ M} = 2 \times 10^{-3} \text{ M}$$
Substitute these values into the approximation formula:
$$K_a = (2 \times 10^{-3}) \times (2 \times 10^{-2})^2$$
Calculate the squared term first:
$$(2 \times 10^{-2})^2 = 4 \times 10^{-4}$$
Multiply by the concentration:
$$K_a = (2 \times 10^{-3}) \times (4 \times 10^{-4})$$
$$K_a = 8 \times 10^{-7}$$

Step 4: Final Answer:
The dissociation constant is $8 \times 10^{-7}$, which matches option (b).