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Question: The \({{K}_{sp}}\) of \(A{{g}_{2}}Cr{{O}_{4}}\), \(AgCl\), \(AgBr\) and \(AgI\) are respectively, \(...

The Ksp{{K}_{sp}} of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}, AgClAgCl, AgBrAgBr and AgIAgI are respectively, 1.1×10−121.1\times {{10}^{-12}}, 1.8×10−101.8\times {{10}^{-10}}, 5.0×10−135.0\times {{10}^{-13}}, 8.3×10−178.3\times {{10}^{-17}}. Which one of the following salts will precipitate last if AgNO3AgN{{O}_{3}} solution is added to the solution containing equal moles of NaClNaCl, NaBrNaBr, NaINaI and Na2CrO4N{{a}_{2}}Cr{{O}_{4}}?
A. AgClAgCl
B. AgBrAgBr
C. Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}
D. AgIAgI

Explanation

Solution

The solubility product constant defined in the terms of an equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is represented by the symbol Ksp{{K}_{sp}} and its value generally depends on the temperature.

Complete Solution : To find out which salt is precipitate last we have to find out the solubility product of the given compounds:
Ag2CrO4→2Ag++CrO42−A{{g}_{2}}Cr{{O}_{4}}\to 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}

100
1-s2ss

We know that Ksp{{K}_{sp}} can be calculated as:
Ksp=[Ag2+][CrO4−]Ag2CrO4{{K}_{sp}}=\dfrac{[A{{g}^{2+}}][Cr{{O}^{4-}}]}{A{{g}_{2}}Cr{{O}_{4}}}
Ksp=(2s)2s1−s=1.1×10−12{{K}_{sp}}=\dfrac{{{(2s)}^{2}}s}{1-s}=1.1\times {{10}^{-12}} where s<<1
1.1×10−12=4s31.1\times {{10}^{-12}}=4{{s}^{3}}
s=6.5×10−5s=6.5\times {{10}^{-5}}
AgCl→Ag++Cl−AgCl\to A{{g}^{+}}+C{{l}^{-}} Ksp=s2=1.8×10−10{{K}_{sp}} = {{s}^{2}}=1.8\times {{10}^{-10}}
s=1.34×10−5s = 1.34\times {{10}^{-5}}
Solubility value for AgBrAgBr and AgIAgI can be calculated in the same manner as in case of AgClAgCl as in this case solution will dissociate into cation and anion in the same manner as in the reaction of AgClAgCl, hence the solubility values of AgBrAgBr and AgIAgI are 7.1×10−77.1\times {{10}^{-7}} and 9×10−99\times {{10}^{-9}}, respectively.
Now we know that lesser the solubility of a substance longer time the precipitates last in the solution hence we can say that out of these we find that Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} have lesser solubility so its salt will remain last. So, the correct answer is “Option C”.

Note: Solubility is defined as that property of a substance by which a solute is dissolved in a solvent in order to form a solution. Solubility products generally increase with an increase in temperature due to increase in the solubility.