Question

A weak monobasic acid is 3.0% dissociated in it's 0.04 M solution. What is the dissociation constant of acid?

• A. $9 \times 10^{-4}$
• B. $3.6 \times 10^{-5}$
• C. $3 \times 10^{-2}$
• D. $4 \times 10^{-2}$

Hint:

To handle the arithmetic quickly, isolate the significant digits from the powers of ten first. The concentration starts with a $4$ and the squared degree of dissociation contains $3^2 = 9$. Multiplying $4 \times 9 = 36$ tells you immediately that the numerical part must begin with $3.6$, which helps isolate option (B) without further work.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the molar concentration ($C = 0.04\text{ M}$) and the percentage dissociation of a weak monobasic acid. We need to calculate its acid dissociation constant ($K_a$).

Step 2: Key Formula or Approach:
According to Ostwald's Dilution Law for a weak electrolyte with a small degree of dissociation ($\alpha \ll 1$), the dissociation constant is expressed as: $$K_a = C\alpha^2$$ where $C$ is the initial molar concentration and $\alpha$ is the fractional degree of dissociation.

Step 3: Detailed Explanation:
1. First, convert the percentage dissociation into the fractional degree of dissociation ($\alpha$): $$\alpha = \frac{3.0\%}{100} = 0.03 = 3 \times 10^{-2}$$ 2. The initial concentration is given as: $$C = 0.04\text{ M} = 4 \times 10^{-2}\text{ mol L}^{-1}$$ 3. Substitute these values into the Ostwald expression: $$K_a = (4 \times 10^{-2}) \times (3 \times 10^{-2})^2$$ 4. Compute the squared term: $$(3 \times 10^{-2})^2 = 9 \times 10^{-4}$$ 5. Complete the multiplication to find $K_a$: $$K_a = 4 \times 10^{-2} \times 9 \times 10^{-4} = 36 \times 10^{-6}$$ 6. Write the final value in standard scientific notation: $$K_a = 3.6 \times 10^{-5}$$

Step 4: Final Answer:
The dissociation constant of the acid is $3.6 \times 10^{-5}$, which matches option (B).