Q. Solubility of $AgCl\ at\ 20℃\ is\ 1.435 \times 10^{- 3}\ \frac{g}{L}$ . The solubility product of $AgCl$ is
A) $1 \times 10^{- 5}$
B) $1 \times 10^{- 10}$
C) $1.435 \times 10^{- 5}$
D) $108 \times 10^{- 3}$
Explanation
(b)
$AgCl \rightleftharpoons {Ag}^{+} + {Cl}^{-}$
$K_{sp} = \left\lbrack {Ag}^{+} \right\rbrack\left\lbrack {Cl}^{-} \right\rbrack$
$K_{sp} = S^{2}$ ($S$=solubility in mol/L)
$S = \frac{1.435 \times 10^{- 3}\ \frac{g}{L}\ }{143.5} = 1 \times 10^{- 10}$ mol/L
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.