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Question: An excess of \(A{{g}_{2}}Cr{{O}_{4}}\)(s) is added to\(5\times {{10}^{-3}}\)M \({{K}_{2}}Cr{{O}_{4}}...

An excess of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}(s) is added to5×10−35\times {{10}^{-3}}M K2CrO4{{K}_{2}}Cr{{O}_{4}}solution. The concentration of Ag+A{{g}^{+}}in the solution is the closest to –
[Solubility product ofAg2CrO4=1.1×10−12A{{g}_{2}}Cr{{O}_{4}}=1.1\times {{10}^{-12}}]
[A]2.2×10−10M2.2\times {{10}^{-10}}M
[B]1.5×10−5M1.5\times {{10}^{-5}}M
[C]1.0×10−6M1.0\times {{10}^{-6}}M
[D]5.0×10−3M5.0\times {{10}^{-3}}M

Explanation

Solution

To solve this, we can write the dissociation equation of Ag2CrO4A{{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 Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}dissociates into a cation and an anion and we can write the equation as-
Ag2CrO4(s)⇌2Ag++CrO42−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=[cation]×[anion]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=[Ag+]2×[CrO42−]Ksp={{\left[ A{{g}^{+}} \right]}^{2}}\times \left[ Cr{{O}_{4}}^{2-} \right]
Solubility product of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}} is equal to the product of the square of concentration of Ag+A{{g}^{+}}ions and the concentration of CrO42−Cr{{O}_{4}}^{2-}ions.
The value of solubility product of Ag2CrO4A{{g}_{2}}Cr{{O}_{4}}is given to us and it is1.1×10−12M1.1\times {{10}^{-12}}M. By putting this value in the above equation we will get,
1.1×10−12M=[Ag+]2×[CrO42−]1.1\times {{10}^{-12}}M={{\left[ A{{g}^{+}} \right]}^{2}}\times \left[ Cr{{O}_{4}}^{2-} \right]
As we know,K2CrO4{{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-
[CrO42−]=5×10−3M\left[ Cr{{O}_{4}}^{2-} \right]=5\times {{10}^{-3}}M
Putting this value in the solubility product equation, we will get-
1.1×10−12M=[Ag+]2×5×10−3M1.1\times {{10}^{-12}}M={{\left[ A{{g}^{+}} \right]}^{2}}\times 5\times {{10}^{-3}}M
1.1×10−12M÷5×10−3M=[Ag+]21.1\times {{10}^{-12}}M\div 5\times {{10}^{-3}}M={{\left[ A{{g}^{+}} \right]}^{2}}
Or, [Ag+]2=2.2×10−10{{\left[ A{{g}^{+}} \right]}^{2}}=2.2\times {{10}^{-10}}
Therefore, [Ag+]=1.4832×10−5≃1.5×10−5\left[ A{{g}^{+}} \right]=1.4832\times {{10}^{-5}}\simeq 1.5\times {{10}^{-5}}
i.e. the concentration of Ag+A{{g}^{+}}ions in the solution is 1.5×10−51.5\times {{10}^{-5}}M

Therefore, the correct answer is option [B] 1.5×10−5M1.5\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 K2CrO4{{K}_{2}}Cr{{O}_{4}}is equal to its value as it is a strong electrolyte and it dissociates completely in the solution.