Question

An excess of $A{{g}_{2}}Cr{{O}_{4}}$(s) is added to$5\times {{10}^{-3}}$M ${{K}_{2}}Cr{{O}_{4}}$solution. The concentration of $A{{g}^{+}}$in the solution is the closest to –
[Solubility product of$A{{g}_{2}}Cr{{O}_{4}}=1.1\times {{10}^{-12}}$]
[A]$2.2\times {{10}^{-10}}M$
[B]$1.5\times {{10}^{-5}}M$
[C]$1.0\times {{10}^{-6}}M$
[D]$5.0\times {{10}^{-3}}M$

Answer 1

Hint: To solve this, we can write the dissociation equation of $A{{g}_{2}}Cr{{O}_{4}}$ which will give us the ions it is dissociated into. We can write the equation for the solubility product using that and the concentration of anion will be equal to the electrolyte as the given electrolyte is strong. The value of solubility product is also given to us and we can put the values of the concentration of anion and the solubility product to get the concentration of the cation.

Complete step by step solution:
According to the given question, we can write that $A{{g}_{2}}Cr{{O}_{4}}$dissociates into a cation and an anion and we can write the equation as-
     $A{{g}_{2}}Cr{{O}_{4}}(s)\rightleftharpoons 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}$
Ksp is the solubility product of a reaction. It is the measure for the degree of dissociation of the ions in a solution. It is written in terms of concentration of the ions-
     \[Ksp=\left[ cation \right]\times \left[ anion \right]\]
I.e. Ksp is the product of concentration of anions and cations present in the solution.
Therefore, for the above reaction we can write that-
     \[Ksp={{\left[ A{{g}^{+}} \right]}^{2}}\times \left[ Cr{{O}_{4}}^{2-} \right]\]
Solubility product of $A{{g}_{2}}Cr{{O}_{4}}$ is equal to the product of the square of concentration of $A{{g}^{+}}$ions and the concentration of $Cr{{O}_{4}}^{2-}$ions.
The value of solubility product of $A{{g}_{2}}Cr{{O}_{4}}$is given to us and it is$1.1\times {{10}^{-12}}M$. By putting this value in the above equation we will get,
     \[1.1\times {{10}^{-12}}M={{\left[ A{{g}^{+}} \right]}^{2}}\times \left[ Cr{{O}_{4}}^{2-} \right]\]
As we know,${{K}_{2}}Cr{{O}_{4}}$is a strong electrolyte i.e. it dissociates readily in the solution. Therefore, the concentration of the anion will be equal to the electrolyte itself.
Therefore, we can write that-
     \[\left[ Cr{{O}_{4}}^{2-} \right]=5\times {{10}^{-3}}M\]
Putting this value in the solubility product equation, we will get-
     \[1.1\times {{10}^{-12}}M={{\left[ A{{g}^{+}} \right]}^{2}}\times 5\times {{10}^{-3}}M\]
     \[1.1\times {{10}^{-12}}M\div 5\times {{10}^{-3}}M={{\left[ A{{g}^{+}} \right]}^{2}}\]
Or, \[{{\left[ A{{g}^{+}} \right]}^{2}}=2.2\times {{10}^{-10}}\]
Therefore, \[\left[ A{{g}^{+}} \right]=1.4832\times {{10}^{-5}}\simeq 1.5\times {{10}^{-5}}\]
i.e. the concentration of $A{{g}^{+}}$ions in the solution is \[1.5\times {{10}^{-5}}\]M

Therefore, the correct answer is option [B] $1.5\times {{10}^{-5}}M$

Note: The solubility product can be calculated by using the concentration of the ions of the dissociated salt. It is dependent on other surrounding factors like temperature, pressure and nature of the electrolyte.
Here, it is important to remember that the concentration of the anion of ${{K}_{2}}Cr{{O}_{4}}$is equal to its value as it is a strong electrolyte and it dissociates completely in the solution.