Q. Solubility product of a salt $AB\ is\ 1 \times 10^{- 8}\ M^{2}$ in a solution in which the concentration of $A^{+}$ ions is $10^{- 3}$ M. The salt will precipitate when the concentration of $B^{-}$ ions is kept
A) Between $10^{- 8}\ to\ 10^{- 7}$ M
B) Between $10^{- 7}M\ to\ 10^{- 8}$ M
C) $> 10^{- 5}$ M
D) $< 10^{- 8}$ M
Explanation
(c)
A salt is precipitated only when the product of ionic concentration is more than its solubility product.
$K_{sp} = 1 \times 10^{- 8}$
$\left\lbrack A^{+} \right\rbrack = 10^{- 3}$ M
$\left\lbrack B^{-} \right\rbrack = \frac{1 \times 10^{- 8}}{10^{- 3}} = 10^{- 5}$ M
So, $AB$ will be more precipitated only when the concentration of $\lbrack B^{-}\rbrack$ is more than $10^{- 5}$ M.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.