Question

The dissociation constant of weak monobasic acid is $2.7 \times 10^{-5}$. If degree of dissociation of acid is $3 \times 10^{-2}$, what is the concentration of acid?

• A. $0.24\ \mathrm{M}$
• B. $0.03\ \mathrm{M}$
• C. $0.3\ \mathrm{M}$
• D. $0.11\ \mathrm{M}$

Hint:

To avoid power sign confusion during multi-step division, write out numbers in direct scientific notation first. Squaring $3 \times 10^{-2}$ gives $9 \times 10^{-4}$. Since $2.7 / 9 = 0.3$, shifting the decimal balance leaves you immediately with $0.03\ \mathrm{M}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the acid dissociation constant ($K_a = 2.7 \times 10^{-5}$) and the degree of dissociation ($\alpha = 3 \times 10^{-2}$) for a weak monobasic acid. We need to calculate its initial molar concentration ($C$).

Step 2: Key Formula or Approach:
According to Ostwald's Dilution Law for a weak monobasic acid where the value of $\alpha$ is small ($\alpha \ll 1$), the dissociation constant is linked to concentration by the formula:
$$K_a = C \cdot \alpha^2$$ Rearranging this formula to calculate concentration ($C$) gives:
$$C = \frac{K_a}{\alpha^2}$$

Step 3: Detailed Explanation:
Substitute the given values into the rearranged equation:
$$K_a = 2.7 \times 10^{-5}$$ $$\alpha = 3 \times 10^{-2} \implies \alpha^2 = (3 \times 10^{-2})^2 = 9 \times 10^{-4}$$ Now calculate the concentration $C$:
$$C = \frac{2.7 \times 10^{-5}}{9 \times 10^{-4}}$$ $$C = \frac{2.7}{9} \times 10^{-5 - (-4)}$$ $$C = 0.3 \times 10^{-1}$$ $$C = 0.03\ \mathrm{M}$$

Step 4: Final Answer:
The initial concentration of the weak monobasic acid is $0.03\ \mathrm{M}$, matching option (B).