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:

Recall the exact Ostwald dilution-law form $K_b=C\alpha^2/(1-\alpha)$. Use the fact that the degree of dissociation of a weak base is very small.

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: \[ K_b = C\alpha^2 \]

Answer 2

Concept:

• Ostwald dilution law relates the dissociation constant to concentration and degree of dissociation.
• Its exact form is $K_b=\dfrac{C\alpha^2}{1-\alpha}$, which simplifies for a weak base because $\alpha$ is small.

Step 1: Use the exact dilution-law form.
For initial concentration $C$ and degree of dissociation $\alpha$,
$K_b=\dfrac{C\alpha^2}{1-\alpha}$

Step 2: Apply the weak-base condition.
A weak base dissociates only slightly, so $\alpha\ll1$.
Therefore, $1-\alpha\approx1$.

Step 3: Simplify the relation.
$K_b=\dfrac{C\alpha^2}{1-\alpha}\approx\dfrac{C\alpha^2}{1}$
Hence, $K_b=C\alpha^2$.

Final Answer: $K_b=C\alpha^2$, option B.