Q. $1\ dm^{3}$ solution containing $10^{\ 5}$ moles each of $Cl$ ions and $CrO_{4}^{2}$ ions is treated with $10^{\ 4}$ moles of silver nitrate. Which one of the following observation is made? $\lbrack K_{sp}Ag_{2}CrO_{4}\ \ \ \ \ \ \ \ 4\ \ \ \ \ \ \ \ \ \ \ 10^{\ 12}\rbrack$ $\lbrack K_{sp}AgCl\ \ \ \ \ \ 1\ \ \ \ \ \ \ \ \ \ \ 10^{\ 10}\rbrack$
Explanation
(c)
For precipitation,
ionic product > solubility product $(K_{sp})$
For, $Ag_{2}CrO_{4}$
ionic product =$\ \left\lbrack Ag^{+} \right\rbrack^{2}\left\lbrack CrO_{4}^{-} \right\rbrack$
$= \left( 10^{- 4} \right)^{2}\left( 10^{- 5} \right) = 10^{- 13}$
$K_{sp}\ of\ Ag_{2}CrO_{4} = 4 \times 10^{- 12}$
Here, $K_{sp} > IP$
This, no precipitate is obtained.
For $AgCl,\ ionic\ product = \left\lbrack {Ag}^{+} \right\rbrack\left\lbrack {Cl}^{-} \right\rbrack = \left\lbrack 10^{- 4} \right\rbrack\left\lbrack 10^{- 5} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{- 9}$
$K_{sp}(AgCl\ ) = 1 \times 10^{- 10}$
Here, $IP > K_{sp}$
So, precipitate will form.
Thus, silver chloride gets precipitated first.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.