Q. The solubility of $Sb_{2}S_{3}\ in\ water\ is\ 1.0 \times 10^{- 5}\ \frac{mol}{L}$ at 298 K. What will be its solubility product?
A) $108 \times 10^{- 25}$
B) $1.0 \times 10^{- 25}$
C) $144 \times 10^{- 25}$
D) $126 \times 10^{- 24}$
Explanation
(a)
$Sb_{2}S_{3} \rightleftharpoons 2Sb^{3 +} + 3S^{2 -}$
$s$ mol/L $2s$ $3s$
Solubility product $\left( K_{sp} \right) = \left\lbrack Sb^{3 +} \right\rbrack^{2}\left\lbrack S^{2 -} \right\rbrack^{3}$
$\ = (2s)^{2}(3s)^{3} = 108s^{5}$
$= 108 \times \left( 1.0 \times 10^{- 5} \right)^{5} = 108 \times 10^{- 25}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.