Question

A weak base is \(1.3\%\) dissociated in its aqueous solution. If \(K_b\) for weak base is \(1.69\times 10^{-5}\) at \(298\) K. Find the concentration of aqueous solution of weak base.

• A. \(1\) M
• B. \(0.1\) M
• C. \(0.01\) M
• D. \(0.001\) M

Hint:

Use Kb = C alpha squared for a weak base with small dissociation.

Answer 1

The Correct Option is B

Step 1: Ostwald Dilution Law:
For weak electrolytes $\alpha = \sqrt{K_b/C}$. Squaring and rearranging gives $C = K_b/\alpha^2$.

Step 2: Numbers:
$\alpha^2 = (0.013)^2 = 1.69\times10^{-4}$, so $C = 1.69\times10^{-5}/1.69\times10^{-4} = 10^{-1}$.

Step 3: Conclusion:
The solution is 0.1 M, option (B). The nice cancellation of 1.69 confirms the choice.

Final Answer:
Option (B), 0.1 M. \[ \boxed{\text{(B) } 0.1\ \text{M}} \]