Q10 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
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
- Correct answerB$0.1$ M
- C$0.01$ M
- D$0.001$ M
Solution
For a weak base, the dissociation constant $K_b$ is given by Ostwald's dilution law:
$K_b = \dfrac{C \alpha^2}{1 - \alpha}$
Since the degree of dissociation $\alpha = 1.3\% = 0.013$ is very small ($\alpha \ll 1$), we can approximate $1 - \alpha \approx 1$. The expression simplifies to:
$K_b = C \alpha^2$
Rearranging for concentration $C$:
$C = \dfrac{K_b}{\alpha^2}$
Substituting the given values $K_b = 1.69 \times 10^{-5}$ and $\alpha = 0.013$:
$C = \dfrac{1.69 \times 10^{-5}}{(0.013)^2}$
$C = \dfrac{1.69 \times 10^{-5}}{1.69 \times 10^{-4}}$
$C = 10^{-1} = 0.1$ M
Answer: $0.1$ M
$K_b = \dfrac{C \alpha^2}{1 - \alpha}$
Since the degree of dissociation $\alpha = 1.3\% = 0.013$ is very small ($\alpha \ll 1$), we can approximate $1 - \alpha \approx 1$. The expression simplifies to:
$K_b = C \alpha^2$
Rearranging for concentration $C$:
$C = \dfrac{K_b}{\alpha^2}$
Substituting the given values $K_b = 1.69 \times 10^{-5}$ and $\alpha = 0.013$:
$C = \dfrac{1.69 \times 10^{-5}}{(0.013)^2}$
$C = \dfrac{1.69 \times 10^{-5}}{1.69 \times 10^{-4}}$
$C = 10^{-1} = 0.1$ M
Answer: $0.1$ M
Hence, the correct answer is option: 'B'.