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: Understanding the Concept:
For a weak base with small degree of dissociation \(\alpha\), the base constant is \(K_b = C\alpha^2\), where \(C\) is the initial concentration.

Step 2: Convert the Data:
\(\alpha = 1.3\% = 0.013\), so \(\alpha^2 = 1.69\times 10^{-4}\).

Step 3: Solve for C:
\[ C = \frac{K_b}{\alpha^2} = \frac{1.69\times 10^{-5}}{1.69\times 10^{-4}} = 0.1\ \text{M} \]

Step 4: Check the Other Options:
If \(C\) were 1 M, then \(\alpha=\sqrt{1.69\times10^{-5}} = 0.0041\), which is 0.41%. For 0.01 M, \(\alpha\) would be 4.1%. For 0.001 M it would be about 13%. Only 0.1 M gives 1.3%. So (B) is correct.

Final Answer:
The concentration of the base solution is 0.1 M, option (B). \[ \boxed{\text{(B) } 0.1\ \text{M}} \]