Question

Calculate the value of dissociation constant of weak monoacidic base if it dissociates to 2% in 0.1 M solution?

• A. \(6 \times 10^{-5}\)
• B. \(4 \times 10^{-5}\)
• C. \(2 \times 10^{-5}\)
• D. \(1 \times 10^{-5}\)

Hint:

When percentage dissociation is small, use: \[ K \approx C\alpha^2 \] after converting percentage into decimal.

Answer 1

The Correct Option is B

Concept:
For a weak monobasic or monoacidic species with degree of dissociation \(\alpha\), the dissociation constant is: \[ K = \frac{C\alpha^2}{1-\alpha} \] Since \(\alpha\) is very small, we can often use: \[ K \approx C\alpha^2 \] ip

Step 1: Write the given values.
Concentration: \[ C = 0.1 \] Degree of dissociation: \[ \alpha = 2% = \frac{2}{100} = 0.02 \] ip

Step 2: Apply the approximate formula.
\[ K \approx C\alpha^2 \] \[ K = 0.1 \times (0.02)^2 \] \[ K = 0.1 \times 0.0004 \] \[ K = 4 \times 10^{-5} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ 4\times10^{-5}} \]