Q. The solubility of $CaF_{2}$ is $2 \times 10^{- 4}$ mol/L. Its solubility product $(K_{sp})$ is
A) $2.0 \times 10^{- 4}$
B) $4.0 \times 10^{- 3}$
C) $8.0 \times 10^{- 12}$
D) $3.2 \times 10^{- 11}$
Explanation
The correct answer is Option D: $3.2 \times 10^{- 11}$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.