Question

The solubility of\[CaC{{O}_{3}}~\] in water is\[3.05\times {{10}^{-4}}\]\[moles/litre\]. Its solubility product will be
(A) \[3.05\times {{10}^{-4}}\]
(B) \[10\]
(C) \[6.1\times {{10}^{-4}}\]
(D) \[9.3\times {{10}^{-8}}\]

Answer 1

Hint: Solubility (represented as S) of CaCO₃ is \[3.05\times {{10}^{-4}}\]\[moles/litre\]. Solubility is defined as the ability of solute to dissolve and dissociate in solvent at a given temperature when equilibrium or saturated point reaches.

Complete Step by Step Solution:
The equation is given as:
\[CaC{{O}_{3}}\left( s \right)~\rightleftarrows CaC{{O}_{3}}\left( aq \right)\]

In this reaction, \[CaC{{O}_{3}}~\] 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 CaCO3 will get dissolved in solvent (aq). And \[CaC{{O}_{3}}~\] is a very good electrolyte so, it will easily and quickly dissociate to give ions such as
\[CaC{{O}_{3}}\left( aq \right)~\rightleftarrows C{{a}^{2+}}+\text{ }C{{O}_{3}}^{2-}\]

Now, indirectly equilibrium set between precipitated \[CaC{{O}_{3}}~\] and ionised \[CaC{{O}_{3}}~\]at a saturated point such as
\[CaC{{O}_{3}}\left( s \right)~\rightleftarrows C{{a}^{2+}}+\text{ }C{{O}_{3}}^{2-}\]

Now at equilibrium, solubility product which is represented as \[{{K}_{sp}}\]of \[CaC{{O}_{3}}~\]is equal to the product of concentration of both ions, \[C{{a}^{2+}}\]and\[C{{O}_{3}}^{2-}\]produced in saturated solution. And solubility of both ions is S and S. The relationship between solubility product and solubility is given as
\[{{K}_{sp}}=[C{{a}^{2+}}][C{{O}_{3}}^{2-}]\] \[{{K}_{sp}}=[C{{a}^{2+}}][C{{O}_{3}}^{2-}]\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }S\]
\[{{K}_{sp}}=\text{ }{{S}^{2}}\]

Putting the value of S which is given in question \[(3.05\times {{10}^{-4}})\]we get the required value of solubility product such as
\[{{K}_{sp}}={{(3.05\times {{10}^{-4}})}^{2}}\]
Thus, the correct option is D.

Note: It is important to note that solubility is defined in two ways. The first one when solute in gram is dissolved in litre solvent (gram solubility) and the other is when mol of solute is dissolved in litre solution (molar solubility). In this question, we need to express solubility as mol per litre. Also for the salt of type MX \[(CaC{{O}_{3}}~)\]solubility product will remain same as above, \[{{K}_{sp}}=\text{ }{{S}^{2}}\].