Q. Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$.
A) $2 \times 10^{-8}$ M
B) $1 \times 10^{-13}$ M
C) $1 \times 10^8$ M
D) $2 \times 10^{-13}$ M
Explanation
[Solution] $s = K_{sp}/[OH^-]^2 = 2 \times 10^{-15}/(0.1)^2 = 2 \times 10^{-13}$ M. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.