Question

Which formula co-relates degree of dissociation and concentration of electrolyte?

• A. \( c = \sqrt{\frac{K_a}{\alpha}} \)
• B. \( \alpha = \sqrt{\frac{K_a}{c}} \)
• C. \( c = \sqrt{K_a \alpha} \)
• D. \( c = \sqrt{\frac{\alpha}{K_a}} \)

Hint:

Ostwald's Dilution Law is a cornerstone for understanding weak electrolytes. The key approximation is \(1-\alpha \approx 1\), which simplifies the math and leads directly to the relationship \( \alpha = \sqrt{K_a/c} \).

Answer 1

The Correct Option is B

Step 1: State Ostwald's Dilution Law. This law relates the dissociation constant (\(K_a\)) of a weak electrolyte, its degree of dissociation (\(\alpha\)), and its concentration (\(c\)).
Step 2: Derive the relationship.
Consider a weak monoprotic acid HA dissociating in solution: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] Initial concentration: \(c\) \quad 0 \quad 0 Equilibrium concentration: \(c(1-\alpha)\) \quad \(c\alpha\) \quad \(c\alpha\)
The acid dissociation constant, \(K_a\), is given by: \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha} \] For a weak electrolyte, the degree of dissociation \(\alpha\) is very small, so we can approximate \(1-\alpha \approx 1\). \[ K_a \approx c\alpha^2 \]
Step 3: Rearrange the formula to solve for \(\alpha\). \[ \alpha^2 = \frac{K_a}{c} \quad \Rightarrow \quad \alpha = \sqrt{\frac{K_a}{c}} \] This shows that the degree of dissociation is inversely proportional to the square root of the concentration.