Q18 / 150 · MHT CET 2026 (13 April Shift 1) · Chemistry

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.
  • Correct answer
    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

Solution

For a weak base $BOH$, the dissociation reaction is $BOH \rightleftharpoons B^+ + OH^-$.

Let the initial concentration be $C$ and the degree of dissociation be $\alpha$.

At equilibrium, the concentrations are $[BOH] = C(1-\alpha)$, $[B^+] = C\alpha$, and $[OH^-] = C\alpha$.

The base dissociation constant $K_b$ is given by $K_b = \dfrac{[B^+][OH^-]}{[BOH]} = \dfrac{C\alpha \cdot C\alpha}{C(1-\alpha)} = \dfrac{C\alpha^2}{1-\alpha}$.

For a weak base, $\alpha$ is very small compared to $1$, so $1-\alpha \approx 1$.

Thus, $K_b \approx C\alpha^2 \Rightarrow \alpha = \sqrt{\dfrac{K_b}{C}}$.

Therefore, the degree of dissociation of a weak base is inversely proportional to the square root of its concentration.

Answer: The degree of dissociation of a weak base is inversely proportional to square root of its concentration.

Hence, the correct answer is option: 'B'.

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