What will be the unit in which the solubility product of $C{a_3}{(P{O_4})_2}$ is expressed?
A) $mol\;d{m^{ - 3}}$
B) $mo{l^2}\;d{m^{ - 6}}$
C) $mo{l^3}\;d{m^{ - 9}}$
D) $mo{l^5}\;d{m^{ - 15}}$

Hint: To answer the unit in which the solubility product of the given salt can be expressed, you need to calculate the solubility product of $C{a_3}{(P{O_4})_2}$. You should also know that $C{a_3}{(P{O_4})_2}$ dissociates into 3 calcium ions having +2 charge and two phosphate ions having -3 charge.

Complete Solution :
- We are given salt and this is calcium phosphate, $C{a_3}{(P{O_4})_2}$. Let us write the ionisation reaction of given salt, $C{a_3}{(P{O_4})_2}$ when it is dissolved in water. The equilibrium reaction can be represented as:
- $C{a_3}{(P{O_4})_2} \rightleftharpoons 3C{a^{2 + }} + 2PO_4^{3 - }$
- Thus, 1 mole of calcium phosphate dissociates into 3 moles of calcium ions and two moles of phosphate ions dissolving in water.
- Now, the solubility product, denoted by ${K_{sp}}$, is the equilibrium constant for the dissolution of a solid substance into an aqueous solution.
- To solve ${K_{sp}}$ of a salt, we need to take the product of the concentrations of the products, each raise to the power of their stoichiometric coefficient and also multiply concentration by their respective stoichiometric coefficient.
- Thus, ${K_{sp}}$ for $C{a_3}{(P{O_4})_2}$ = ${[C{a^{2 + }}]^3}{[PO_4^{3 - }]^2}$
- Now, let us consider the molar solubility of $C{a_3}{(P{O_4})_2}$ is S. Molar solubility of the compound is expressed in units $mol\;d{m^{ - 3}}$. The concentrations of the ions are equal to the molar solubility of the salt. If molar solubility is S, then
${K_{sp}}$ = ${(3S)^3}{(2S)^2} = 108{S^5}\;mo{l^5}d{m^{ - 15}}$

- Therefore, the unit in which the solubility product of $C{a_3}{(P{O_4})_2}$ is expressed is $mo{l^5}\;d{m^{ - 15}}$.
So, the correct answer is “Option D”.

Note: When the equilibrium is established between the undissolved solid and the ions in the solution, the solution is said to be a saturated solution. The concentration of the solute or salt in a saturated solution is nothing but molar solubility or solubility. Units of solubility are expressed as mass per unit volume i.e. $mol\;d{m^{ - 3}}$ or $mol\;{L^{ - 1}}$.