Question

The solubility product of three sparingly soluble salts ${{\text{M}}_{_{_{\text{2}}}}}{\text{X , MX and M}}{{\text{X}}_{\text{3}}}$ are identical. What will be the order of their solubilities?
A. \[{\text{M}}{{\text{X}}_{\text{3}}}{\text{ > }}{{\text{M}}_{\text{2}}}{\text{X > MX}}\]
B. \[{\text{M}}{{\text{X}}_{\text{3}}}{\text{ > MX}} > {{\text{M}}_{\text{2}}}{\text{X}}\]
C. \[{\text{MX}} > {{\text{M}}_{\text{2}}}{\text{X > M}}{{\text{X}}_{\text{3}}}\]
D. \[{\text{MX > M}}{{\text{X}}_{\text{3}}} > {{\text{M}}_{\text{2}}}{\text{X}}\]

Answer 1

Hint: Write the dissociation reaction for all three salts given to us. Write the solubility product expression for three salts. Rearrange equations for solubility and compare the solubilities.

Complete Step by step answer: The three salts given to us are${{\text{M}}_{_{_{\text{2}}}}}{\text{X , MX and M}}{{\text{X}}_{\text{3}}}$.
The dissociation reaction for three salts is as follows:
\[{\text{MX}} \rightleftharpoons {{\text{M}}^{\text{ + }}}{\text{ + }}{{\text{X}}^{\text{ - }}}\]
\[{{\text{M}}_{\text{2}}}{\text{X}} \rightleftharpoons 2{{\text{M}}^{\text{ + }}}{\text{ + }}{{\text{X}}^{{\text{2 - }}}}\]
\[{\text{M}}{{\text{X}}_{\text{3}}} \rightleftharpoons {{\text{M}}^{{\text{3 + }}}}{\text{ + 3}}{{\text{X}}^{\text{ - }}}\]
Now, using the dissociation reaction we can write the solubility product expression for three salts as follows:
The solubility product constant is denoted by notation\[{{\text{K}}_{{\text{sp}}}}\].
So, the solubility product constant expression for \[{\text{MX}}\] salt is :
\[{{\text{K}}_{{\text{sp}}}} = {\text{ [}}{{\text{M}}^{\text{ + }}}{\text{][}}{{\text{X}}^{\text{ - }}}{\text{]}}\]
Let us assume that solubility is ‘S’
So,
\[{\text{[}}{{\text{M}}^{\text{ + }}}{\text{]}} = {\text{S}}\]
\[\Rightarrow {\text{[}}{{\text{X}}^{\text{ - }}}{\text{] = S}}\]
\[\Rightarrow {{\text{K}}_{{\text{sp}}}} = ({\text{S) (S) = }}{{\text{S}}^{\text{2}}}\]
\[\Rightarrow {\text{S = }}{\left( {{{\text{K}}_{{\text{sp}}}}} \right)^{1/2}}\] ... (1)
The solubility product constant expression for \[{{\text{M}}_{\text{2}}}{\text{X}}\] salt is :
\[{{\text{K}}_{{\text{sp}}}} = {\text{ [}}{{\text{M}}^ + }{{\text{]}}^{\text{2}}}{\text{[}}{{\text{X}}^{2 - }}{\text{]}}\]
Let us assume that solubility is ‘S’
So,
\[{\text{[}}{{\text{M}}^ + }{\text{]}} = 2{\text{S}}\]
\[\Rightarrow {\text{[}}{{\text{X}}^{2 - }}{\text{] = S}}\]
\[\Rightarrow {{\text{K}}_{{\text{sp}}}} = {(2{\text{S)}}^{\text{2}}}{\text{ (S) = 4}}{{\text{S}}^3}\]
\[{\text{S = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{4}} \right)^{1/3}}\] ... (2)
The solubility product constant expression for \[{\text{M}}{{\text{X}}_{\text{3}}}\] salt is :
\[{{\text{K}}_{{\text{sp}}}} = {\text{ [}}{{\text{M}}^{3 + }}{\text{][}}{{\text{X}}^ - }{{\text{]}}^{\text{3}}}\]
Let us assume that solubility is ‘S’
So,
\[{\text{[}}{{\text{M}}^{3 + }}{\text{]}} = {\text{S}}\]
\[{\text{[}}{{\text{X}}^ - }{\text{] = 3S}}\]
\[\Rightarrow {{\text{K}}_{{\text{sp}}}} = ({\text{S) (3S}}{{\text{)}}^{{\text{3 }}}}{\text{ = 27}}{{\text{S}}^4}\]
\[{\text{S = }}{\left( {\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{27}}} \right)^{1/4}}\] ... (3)
As we know the sparingly soluble salts are slightly soluble in water so their solubility product constant values have negative power.
We have given that the solubility product of three sparingly soluble salts ${{\text{M}}_{_{_{\text{2}}}}}{\text{X , MX and M}}{{\text{X}}_{\text{3}}}$ are identical.
So, by comparing equations 1,2 and 3 we can say that the decreasing order of solubility of three salts is :
\[{\text{M}}{{\text{X}}_{\text{3}}}{\text{ > }}{{\text{M}}_{\text{2}}}{\text{X > MX}}\]

Thus, the correct option is (A) \[{\text{M}}{{\text{X}}_{\text{3}}}{\text{ > }}{{\text{M}}_{\text{2}}}{\text{X > MX}}\]


Note: Solubility of salt is the amount of salt dissolved in a given amount of solvent. Do not consider that solubility product constant value of sparingly soluble salt has positive power. As solubility products, the constant value of sparingly soluble salt always has negative power. If we consider it as a positive power we will end up with a reverse order of solubility.