Q. The solubility of $BaSO_4$ in water is $2.42 \times 10^{-3}$ $gL^{-1}$ at 298 K. The value of its solubility product ($K_{sp}$) will be (Given molar mass of $BaSO_4$ = 233 g $mol^{-1}$)
A) $1.08 \times 10^{-14}$ $mol^2L^{-2}$
B) $1.08 \times 10^{-12}$ $mol^2L^{-2}$
C) $1.08 \times 10^{-10}$ $mol^2L^{-2}$
D) $1.08 \times 10^{-8}$ $mol^2L^{-2}$
Explanation
[Solution] s = 2.42 × $10^{-3}$/233 = 1.04 × $10^{-5}$ M; $K_{sp} = s^2 = 1.08 \times 10^{-10}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.