Question

The solubility product of silver sulphide is \[3.2\times {{10}^{-11}}\]. Its solubility at the experimental temperature is:
(A) \[2\times {{10}^{-4}}moles\text{ }per\text{ }litre\]
(B) \[6\times {{10}^{-6}}moles\text{ }per\text{ }litre\]
(C) \[1.2\times {{10}^{-5}}moles\text{ }per\text{ }litre\]
(D) \[8\times {{10}^{-4}}moles\text{ }per\text{ }litre\]

Answer 1

Hint: In the given question the solubility product of \[A{{g}_{2}}S\]is given \[3.2\text{ }\times {{10}^{-11}}\]. When \[A{{g}_{2}}S\]is dissolved, it will dissociate into \[2A{{g}^{+}}\]ions and\[~{{S}^{2-}}\]ions in a saturated solution. The solubility of both ions is S.

Complete Step by Step Solution:
The equation is given as:
\[A{{g}_{2}}S(s)\rightleftarrows A{{g}_{2}}S(aq)\] \[A{{g}_{2}}S(s)\rightleftarrows A{{g}_{2}}S(aq)\]

In this reaction, \[A{{g}_{2}}S\]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 \[A{{g}_{2}}S\] will get dissolved in solvent (aq).

Now, \[A{{g}_{2}}S\] is a good electrolyte because it is an ionic compound so, it will dissociate to give ions in water such as
\[A{{g}_{2}}S\left( aq \right)\rightleftarrows 2A{{g}^{+}}+\text{ }{{S}^{2-}}\]

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

As per hint, the solubility product of \[A{{g}_{2}}S\]is given (3.2 ×10-11) in water which is equal to the product of the concentration of both ions produced in saturated solution such as
\[{{K}_{ps}}={{[2A{{g}^{+}}]}^{2}}[{{S}^{2-}}]\]

The solubility (dissociate to give ion in a saturated solution at a given temperature) of \[2A{{g}^{+}}\]and \[~{{S}^{2-}}\]produced in saturated solution after dissociation is 2S and S
such as:
\[{{K}_{ps}}={{[2A{{g}^{+}}]}^{2}}[{{S}^{2-}}]\]
\[{{K}_{sp}}=\text{ }{{\left( 2S \right)}^{2}}\times \text{ }S\]
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]
\[S\text{ }=\text{ 3}\sqrt{{{K}_{sp}}/4}\]
\[S\text{ }=\text{ 3}\sqrt{3.2\times {{10}^{-11}}/4}\]

Putting the value of the solubility product (\[3.2\times {{10}^{-11}}\]) in this given equation we get
\[S\text{ }=\text{ 3}\sqrt{3.2\times {{10}^{-11}}/4}\]
\[S=2\times {{10}^{-4}}\]mol/litre \[S=2\times {{10}^{-4}}\]
Thus, the correct option is A.

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 \[{{M}_{2}}X\]\[(A{{g}_{2}}S)\], solubility product will remain same as above, \[{{K}_{sp}}=\text{ }4{{S}^{3}}\].