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(C) For precipitation to occur,the ionic product must exceed the solubility product $(K_{sp})$.For $Ag_{2}CrO_{4}$:Ionic Product $(IP) = [Ag^{+}]^{2}[

Vedclass · Accepted Answer

Silver chloride gets precipitated first

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(C) For precipitation to occur,the ionic product must exceed the solubility product $(K_{sp})$.For $Ag_{2}CrO_{4}$:Ionic Product $(IP) = [Ag^{+}]^{2}[CrO_{4}^{2-}] = (10^{-4})^{2}(10^{-5}) = 10^{-13}$.Given $K_{sp}(Ag_{2}CrO_{4}) = 4 	imes 10^{-12}$.Since $IP For $AgCl$:Ionic Product $(IP) = [Ag^{+}][Cl^{-}] = (10^{-4})(10^{-5}) = 10^{-9}$.Given $K_{sp}(AgCl) = 1 	imes 10^{-10}$.Since $IP > K_{sp}$,$AgCl$ will precipitate.Therefore,silver chloride gets precipitated first.

$1 \ dm^{3}$ solution containing $10^{-5} \ mol$ each of $Cl^{-}$ ions and $CrO_{4}^{2-}$ ions is treated with $10^{-4} \ mol$ of silver nitrate. Which one of the following observations is made?
$[K_{sp} \ Ag_{2}CrO_{4} = 4 \times 10^{-12}]$
$[K_{sp} \ AgCl = 1 \times 10^{-10}]$

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$1 \ dm^{3}$ solution containing $10^{-5} \ mol$ each of $Cl^{-}$ ions and $CrO_{4}^{2-}$ ions is treated with $10^{-4} \ mol$ of silver nitrate. Which one of the following observations is made?$[K_{sp} \ Ag_{2}CrO_{4} = 4 \times 10^{-12}]$$[K_{sp} \ AgCl = 1 \times 10^{-10}]$