Question

What is the relationship between the degree of dissociation of a weak base and its concentration?

• A. The degree of dissociation of a weak base is directly proportional to square root of its concentration.
• B. The degree of dissociation of a weak base is inversely proportional to square root of its concentration.
• C. The degree of dissociation of a weak base doesn't depend on concentration.
• D. The degree of dissociation of a weak base is directly proportional to its concentration

Hint:

Use Ostwald dilution law: the degree of dissociation varies as the inverse square root of concentration.

Answer 1

The Correct Option is B

Step 1: Start from the equilibrium.
BOH $\rightleftharpoons$ B$^+$ + OH$^-$. With initial concentration $C$ and degree $\alpha$: $[\text{B}^+] = [\text{OH}^-] = C\alpha$ and $[\text{BOH}] = C(1-\alpha)$.

Step 2: Write the constant.
\[ K_b = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \]

Step 3: Approximate.
For a weak base $\alpha$ is tiny, so $1-\alpha \approx 1$ and $K_b = C\alpha^2$, giving $\alpha = \sqrt{K_b/C}$.

Step 4: Read the dependence.
$K_b$ is fixed at a given temperature, so $\alpha$ falls when $C$ rises. This is an inverse square root relation, which rules out (A), (C) and (D).

Final Answer:
Option (B) is correct. \[ \boxed{\alpha = \sqrt{K_b/C}} \]