Question

A weak monobasic acid is 10% dissociated in 0.05 M solution. What is its percentage dissociation in 0.10 M solution?

• A. 5.27%
• B. 7.17%
• C. 10.3%
• D. 4.5%

Hint:

According to Le Chatelier's principle and Ostwald's law, increasing the concentration of a weak electrolyte will always decrease its degree of dissociation. Since the concentration doubled ($0.05\ \text{M} \rightarrow 0.10\ \text{M}$), the dissociation must decrease below 10%, instantly eliminating option (C).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the percentage dissociation of a weak monobasic acid at one concentration. We need to find its percentage dissociation when the concentration is altered to a new value.

Step 2: Key Formula or Approach:
From Ostwald's dilution law for a weak electrolyte, the dissociation constant ($K_a$) is related to the degree of dissociation ($\alpha$) and concentration ($c$) by: $$K_a = \alpha^2 c$$ Since $K_a$ remains constant at a fixed temperature, we can set up an equilibrium ratio for two different concentrations: $$\alpha_1^2 c_1 = \alpha_2^2 c_2 \implies \frac{\alpha_2}{\alpha_1} = \sqrt{\frac{c_1}{c_2}}$$

Step 3: Detailed Explanation:
Given values: Initial concentration, $c_1 = 0.05\ \text{M}$ Initial degree of dissociation, $\alpha_1 = 10\% = 0.10$ Final concentration, $c_2 = 0.10\ \text{M}$ Let's find the new degree of dissociation ($\alpha_2$): $$\alpha_2 = \alpha_1 \times \sqrt{\frac{c_1}{c_2}}$$ $$\alpha_2 = 0.10 \times \sqrt{\frac{0.05}{0.10}} = 0.10 \times \sqrt{\frac{1}{2}} = \frac{0.10}{\sqrt{2}}$$ Since $\sqrt{2} \approx 1.414$: $$\alpha_2 = \frac{0.10}{1.414} \approx 0.0707$$ Converting the degree of dissociation back to percentage dissociation: $$\text{Percentage dissociation} = 0.0707 \times 180\% = 7.07\% \approx 7.17\%$$

Step 4: Final Answer:
The percentage dissociation of the acid in a $0.10\ \text{M}$ solution is approximately 7.17%, which matches option (B).