Question

The solubility of \[Ca{{F}_{2}}\] is \[2\times {{10}^{-4}}~\] \[mole/litre\]. Its solubility product is:
A. \[2.0\text{ }\times \text{ }{{10}^{-4}}\]
B. \[4.0\text{ }\times \text{ }{{10}^{-8}}\]
C. \[4.0\text{ }\times \text{ }8.0\text{ }\times {{10}^{-8}}\]
D. \[3.2\text{ }\times \text{ }{{10}^{-11}}\]

Answer 1

Hint: Solubility product is the product of the concentration of ions produced in equilibrium or saturated solution and it is represented as \[{{K}_{sp}}\]. In the given question, solubility (represented as S) of calcium hydroxide is \[2\times {{10}^{-4}}~\]\[mole/litre\]. Solubility is defined when a solute dissolves and dissociates in solvent at a given temperature when equilibrium or saturated point reaches.

Complete Step by Step Answer:
The equation is given as,
\[Ca{{F}_{2}}\left( s \right)~\rightleftarrows ~Ca{{F}_{2}}\left( aq \right)\]

In this reaction, \[Ca{{F}_{2}}\] is dissolved in a solvent and as it is a slightly soluble salt thus, most of its part will get precipitate (in solid form) and a very less amount of \[Ca{{F}_{2}}\]will get dissolved in solvent (aq).

Now \[Ca{{F}_{2}}\]is a very valuable electrolyte so, it will easily and quickly dissociate into give ions such as,
\[Ca{{F}_{2}}\left( aq \right)\rightleftarrows C{{a}^{2+}}+\text{ }2{{F}^{-}}\]

Now, indirectly equilibrium set between precipitated \[Ca{{F}_{2}}\] and ionised \[Ca{{F}_{2}}\] at a saturated point such as,
\[Ca{{F}_{2}}\left( s \right)\rightleftarrows C{{a}^{2+}}+\text{ }2{{F}^{-}}\]

Now at equilibrium, solubility product which is represented as \[{{K}_{sp}}\] is equal to the product of concentration of both ions, \[C{{a}^{2+}}\]and \[2{{F}^{-}}\]. And solubility (ability to get dissolve in solvent to make saturated solution) of both ions is S and \[2S\].The relationship between solubility product and solubility is given as,
\[{{K}_{sp}}=[C{{a}^{2+}}]{{[{{F}^{-}}]}^{2}}\]
\[{{K}_{sp}}=S\text{ }\times \text{ }\left( 2S \right){}^\text{2}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }4{{S}^{2}}\]
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]

Putting the value of S which is given in question we get the required value of solubility product such as
\[{{K}_{sp}}=\text{ }4{{(2\times {{10}^{-4}})}^{3}}\]
\[{{K}_{sp}}=\text{ }4\times 2\times {{10}^{-4}}\times 2\times {{10}^{-4}}\times 2\times {{10}^{-4}}\]
\[{{K}_{sp}}=\text{ 4}\times \text{8}\times \text{1}{{\text{0}}^{-12}}\]
Thus, the correct option is C.

Note: It is important to note that \[{{K}_{sp}}\], solubility product for the salts of type \[M{{X}_{2}}\] is always same and that is \[{{K}_{sp}}=\text{ }4{{S}^{3}}\] and if a salt is formed with one anion and one cation such as MX, then the solubility product is equal to square of solubility S such as \[{{K}_{sp}}={{S}^{2}}\].