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Question: Equal volume of the following \(C{{a}^{2+}}\) and \({{F}^{-}}\) solutions are mixed. In which soluti...

Equal volume of the following Ca2+C{{a}^{2+}} and F−{{F}^{-}} solutions are mixed. In which solutions will the precipitation occur?
Ksp{{K}_{sp}} of CaF2=1.7×10−10Ca{{F}_{2}}=1.7\times {{10}^{-10}}
i) 10−2M Ca2++10−5M F−{{10}^{-2}}M\text{ }C{{a}^{2+}}+{{10}^{-5}}M\text{ }{{F}^{-}}
ii) 10−3M Ca2++10−3M F−{{10}^{-3}}M\text{ }C{{a}^{2+}}+{{10}^{-3}}M\text{ }{{F}^{-}}
iii) 10−4M Ca2++10−2M F−{{10}^{-4}}M\text{ }C{{a}^{2+}}+{{10}^{-2}}M\text{ }{{F}^{-}}
iv) 10−2M Ca2++10−3M F−{{10}^{-2}}M\text{ }C{{a}^{2+}}+{{10}^{-3}}M\text{ }{{F}^{-}}
Select the correct answer using the codes given below:
A. In (iv) only
B. In (i) and (ii)
C. In (iii) and (iv)
D. In (ii), (iii) and (iv)
E. In all these

Explanation

Solution

To solve this problem, you need to know about the solubility product or Ksp{{K}_{sp}} value that is mentioned in the question. You have to check the product of concentrations of Ca2+C{{a}^{2+}} and F−{{F}^{-}}. Think about how the concentrations of Ca2+C{{a}^{2+}} and F−{{F}^{-}} ions relate to the solubility product.

Complete answer:
When two solutions of equal volumes are mixed, precipitation will occur. The precipitation only occurs when the product is no longer soluble in the solvent. This threshold is given by the Ksp{{K}_{sp}} value.
We know that the solubility product Ksp of CaF2=1.7×10−10{{K}_{sp}}\text{ of }Ca{{F}_{2}}=1.7\times {{10}^{-10}}. The CaF2Ca{{F}_{2}} will precipitate only when the product of concentrations of the ions is greater than the solubility product. [Ca2+][F−]2>Ksp\left[ C{{a}^{2+}} \right]{{\left[ {{F}^{-}} \right]}^{2}}>{{K}_{sp}}. We are squaring the concentration of the F−{{F}^{-}} ions since 2 of them are required to form one molecule of CaF2Ca{{F}_{2}}.
We will now look at the cases given in the question one by one and determine the concentration:
In the first option, [10−2]2×[10−5]22=10−124\dfrac{\left[ {{10}^{-2}} \right]}{2}\times \dfrac{{{\left[ {{10}^{-5}} \right]}^{2}}}{2}=\dfrac{{{10}^{-12}}}{4}which is less thanKsp{{K}_{sp}}. Therefore, no precipitation will occur.
In the second option, [10−3]2×[10−3]22=10−94\dfrac{\left[ {{10}^{-3}} \right]}{2}\times \dfrac{{{\left[ {{10}^{-3}} \right]}^{2}}}{2}=\dfrac{{{10}^{-9}}}{4} which is greater thanKsp{{K}_{sp}}. Therefore, precipitation occurs.
In the third option, [10−4]2×[10−2]22=10−84\dfrac{\left[ {{10}^{-4}} \right]}{2}\times \dfrac{{{\left[ {{10}^{-2}} \right]}^{2}}}{2}=\dfrac{{{10}^{-8}}}{4} which is greater thanKsp{{K}_{sp}}. Therefore, precipitation occurs.
In the last option, [10−2]2×[10−3]22=10−84\dfrac{\left[ {{10}^{-2}} \right]}{2}\times \dfrac{{{\left[ {{10}^{-3}} \right]}^{2}}}{2}=\dfrac{{{10}^{-8}}}{4}
which is greater than Ksp{{K}_{sp}}. Therefore, precipitation occurs.
So, precipitation occurs in the case (ii), (iii) and (iv).
Therefore, the correct option is ‘D’

Note: When two solutions are mixed of equal volumes, the concentration of the resulting solution will be reduced. It will become half. That is why we divided the concentration terms by 2. The two independent entities will combine to form one entity, which is why the concentration reduces by half. If there are more than 2 entities involved in the making of a compound, some other factor corresponding to the number of entities will be used.