Q. The solubility product of $Hg_{2}I_{2}$ is equal to
A) $\left\lbrack Hg_{2}^{2}\ \ \right\rbrack\lbrack I\ \ \rbrack$
B) $\left\lbrack Hg^{2}\ \ \right\rbrack\lbrack I\ \ \rbrack$
C) $\left\lbrack Hg_{2}^{2}\ \ \right\rbrack\lbrack I\ \ \rbrack^{2}$
D) $\left\lbrack Hg^{2}\ \right\rbrack\lbrack I\ \ \rbrack^{2}$
Explanation
(c)
$Hg_{2}I_{2} \rightleftharpoons {Hg}_{2}^{2 +} + 2I^{-}$
$K_{sp} = \left\lbrack Hg_{2}^{2 +} \right\rbrack\left\lbrack I^{-} \right\rbrack^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.