MHTCET 22 Apr 2025 Shift 1 Paper

Concept:For a weak monoacidic base, the dissociation constant $K_b$ is related to molar concentration $C$ and degree of dissociation $\alpha$ by the relation $K_b = C\alpha^2$.Explanation:Given concentration of the base, $C = 0.1\ \mathrm{M}$.Given degree of dissociation, $\alpha = 2\% = 0.02$.Since the base is weak, $\alpha$ is very small, so the approximate formula $K_b = C\alpha^2$ is valid.Substitute the values into the formula:$$K_b = 0.1 \times (0.02)^2$$First calculate $(0.02)^2 = 0.0004 = 4 \times 10^{-4}$.Then multiply by concentration:$$K_b = 0.1 \times 4 \times 10^{-4} = 4 \times 10^{-5}$$Thus, the dissociation constant of the weak base is $4 \times 10^{-5}$.Answer:$K_b = 4 \times 10^{-5}$Correct option: B. $4 \times 10^{-5}$