Question

The ${{K}_{sp}}$ of $A{{g}_{2}}Cr{{O}_{4}}$, $AgCl$, $AgBr$ and $AgI$ are respectively, $1.1\times {{10}^{-12}}$, $1.8\times {{10}^{-10}}$, $5.0\times {{10}^{-13}}$, $8.3\times {{10}^{-17}}$. Which one of the following salts will precipitate last if $AgN{{O}_{3}}$ solution is added to the solution containing equal moles of $NaCl$, $NaBr$, $NaI$ and \[N{{a}_{2}}Cr{{O}_{4}}\]?
A. $AgCl$
B. $AgBr$
C. $A{{g}_{2}}Cr{{O}_{4}}$
D. $AgI$

Answer 1

Hint: 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 ${{K}_{sp}}$ 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:
 $A{{g}_{2}}Cr{{O}_{4}}\to 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}$

1	0	0
1-s	2s	s

We know that ${{K}_{sp}}$ can be calculated as:
${{K}_{sp}}=\dfrac{[A{{g}^{2+}}][Cr{{O}^{4-}}]}{A{{g}_{2}}Cr{{O}_{4}}}$
${{K}_{sp}}=\dfrac{{{(2s)}^{2}}s}{1-s}=1.1\times {{10}^{-12}}$ where s<<1
$1.1\times {{10}^{-12}}=4{{s}^{3}}$
$s=6.5\times {{10}^{-5}}$
$AgCl\to A{{g}^{+}}+C{{l}^{-}}$ ${{K}_{sp}} = {{s}^{2}}=1.8\times {{10}^{-10}}$
$s = 1.34\times {{10}^{-5}}$
Solubility value for $AgBr$ and $AgI$ can be calculated in the same manner as in case of $AgCl$ as in this case solution will dissociate into cation and anion in the same manner as in the reaction of $AgCl$, hence the solubility values of $AgBr$ and $AgI$ are $7.1\times {{10}^{-7}}$ and $9\times {{10}^{-9}}$, 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 $A{{g}_{2}}Cr{{O}_{4}}$ 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.