Question

For a weak base of concentration \(C\) and degree of dissociation \(\alpha\), what is the correct relation between \(K_b\) and \(C\)?

• A. \(K_b = C\alpha\)
• B. \(K_b = C\alpha^2\)
• C. \(K_b = \dfrac{\alpha}{C}\)
• D. \(K_b = \dfrac{C}{\alpha^2}\)

Hint:

For weak electrolytes (weak acids or weak bases), when the degree of dissociation \(\alpha\) is very small, we use the approximation \(1-\alpha \approx 1\). Thus, the relation becomes: \(K = C\alpha^2\).

Answer 1

The Correct Option is B

Concept: A weak base partially dissociates in water. If the base is represented as \(B\), the dissociation equilibrium can be written as: \[ B + H_2O \rightleftharpoons BH^+ + OH^- \] The equilibrium constant for a weak base is called the base dissociation constant \(K_b\). \[ K_b = \frac{[BH^+][OH^-]}{[B]} \] If the initial concentration of the base is \(C\) and the degree of dissociation is \(\alpha\):
• Concentration of dissociated base = \(C\alpha\)
• Concentration of \(BH^+\) formed = \(C\alpha\)
• Concentration of \(OH^-\) formed = \(C\alpha\)
• Remaining base concentration = \(C(1-\alpha)\) Since weak bases dissociate very slightly, we approximate: \[ 1-\alpha \approx 1 \] This simplifies the calculation of \(K_b\).

Step 1: Write the equilibrium expression for \(K_b\). \[ K_b = \frac{[BH^+][OH^-]}{[B]} \]

Step 2: Substitute the concentrations using degree of dissociation. \[ K_b = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} \]

Step 3: Apply the approximation for weak bases. Since \(\alpha\) is very small: \[ 1-\alpha \approx 1 \] Thus, \[ K_b = \frac{C^2\alpha^2}{C} \] \[ K_b = C\alpha^2 \]

Step 4: Identify the correct option. Therefore, the correct relation between \(K_b\) and \(C\) is: \[ K_b = C\alpha^2 \]