Q. pH of a saturated solution of $Ca(OH)_2$ is 9. The solubility product ($K_{sp}$) of $Ca(OH)_2$ is :
A) $0.5 \times 10^{-15}$
B) $0.25 \times 10^{-10}$
C) $0.125 \times 10^{-15}$
D) $0.5 \times 10^{-10}$
Explanation
[Solution] pOH = 5, $[OH^-] = 10^{-5}$, $[Ca^{2+}] = 0.5 \times 10^{-5}$; $K_{sp} = 0.5 \times 10^{-5} \times (10^{-5})^2 = 0.5 \times 10^{-15}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.