Question

If the solubility product of \[BaS{{O}_{4}}\]is \[1.5\times {{10}^{-10}}\]in water. Its solubility, in moles per litre, is.
A \[1.5\text{ }\times \text{ }{{10}^{-9}}\]
B \[3.9\text{ }\times \text{ }{{10}^{-5}}\]
C \[7.5\text{ }\times \text{ }{{10}^{-5}}\]
D \[1.2\text{ }\times \text{ }{{10}^{-5}}\]

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 the solubility product of \[BaS{{O}_{4}}\]is given \[1.5\times {{10}^{-10}}\]. When \[BaS{{O}_{4}}\]is dissolved, it will dissociate into \[B{{a}^{+2}}\] ions and \[S{{O}_{4}}^{-2}\]ions in a saturated solution. The solubility of both ions is S.

Complete Step by Step Solution:
The reaction is given as
\[BaS{{O}_{4}}(s)\rightleftarrows BaS{{O}_{4}}(aq)\]

In this reaction, \[BaS{{O}_{4}}\] 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 \[BaS{{O}_{4}}\] will get dissolved in solvent (aq).

Now \[BaS{{O}_{4}}\] is a weak electrolyte so, it will dissociate little to give ions in water such as
\[BaS{{O}_{4}}(aq)\rightleftarrows B{{a}^{2+}}+S{{O}_{4}}^{2-}\]

Now, indirectly equilibrium set between precipitated \[BaS{{O}_{4}}\] and ionised \[BaS{{O}_{4}}\] at a saturated point such as
\[BaS{{O}_{4}}(s)\rightleftarrows B{{a}^{2+}}+S{{O}_{4}}^{2-}\]

As per hint, the solubility product \[({{K}_{sp}})\]of \[BaS{{O}_{4}}\] is given (\[1.5\times {{10}^{-10}}\]) in water which is equal to the product of the concentration of both ions produced in saturated solution such as
\[({{K}_{sp}})=[B{{a}^{2+}}][S{{O}_{4}}^{2-}]\]

The solubility (dissociate to give ion in a saturated solution at a given temperature) of both the ion produced in saturated solution after dissociation is S such as
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }S\]
\[{{K}_{sp}}=\text{ }{{S}^{2}}\]
\[S\text{ }=\text{ }\surd {{K}_{sp}}\]
Putting the value of the solubility product in this given equation we get
\[S=\sqrt{1.5\times {{10}^{-10}}}\] \[S=\sqrt{1.5\times {{10}^{-10}}}\]
\[S=1.2\times {{10}^{-5}}M\]
Thus, the correct option is D.

Note: It is important to note that solubility is defined in two ways. The first one when solute is taken 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, M.