Question

If solubility of calcium hydroxide is $\sqrt{3}$, then its solubility product will be
A 27
B 3
C 9
D $12\sqrt{3}$

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 $\sqrt{3}$. 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 Solution:
The equation is given as:
\[Ca{{\left( OH \right)}_{2}}\left( s \right)~\rightleftarrows ~Ca{{\left( OH \right)}_{2}}\left( aq \right)\]

In this reaction, \[Ca{{\left( OH \right)}_{2}}\]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 \[Ca{{\left( OH \right)}_{2}}\]will get dissolved in solvent (aq).

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

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

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 \[2O{{H}^{-}}\]. 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
\[{{K}_{sp}}=[C{{a}^{2+}}]{{[O{{H}^{-}}]}^{2}}\]
\[{{K}_{sp}}=S\text{ }\times \text{ }\left( 2S \right){}^\text{2}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }4{{S}^{2}}\]
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]

Putting the value of S which is given in question we get the required value of solubility product such as:
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]
\[{{K}_{sp}}=\text{ }4{{(\sqrt{3})}^{3}}\]
\[{{K}_{sp}}=\text{ }4\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}\]
\[{{K}_{sp}}=\text{ 12}\sqrt{3}\]
Thus, the correct option is D.

Note: It is important to note that \[{{K}_{sp}}\], solubility product for the salts of type MX2 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 \[{{K}_{sp}}\rightleftarrows {{S}^{2}}\].