Question

Let the solubility of an aqueous solution of Mg(OH)₂ be x then its k_sp is
A. 4x³
B. x³
C. 27x⁴
D. 9x

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, solubility (represented as S) of magnesium hydroxide is considered x $\dfrac{mole}{litre}$ . Solubility is defined when a solute dissolves and dissociates in solvent at a given temperature when equilibrium or saturated point reaches. Given compound is of the MX₂ type and so, its ratio is given as 1:2.

Complete Step by Step Answer:
The equation is given as
\[Mg{{(OH)}_{2}}(s)\rightleftarrows Mg{{(OH)}_{2}}(aq)\]
In this reaction, \[Mg{{\left( OH \right)}_{2}}~\]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 \[Mg{{\left( OH \right)}_{2}}~\]will get dissolved in solvent (aq).
Now \[Mg{{\left( OH \right)}_{2}}~\]is a very valuable or strong electrolyte so, it will easily and quickly dissociate into give ions such as
\[Mg{{(OH)}_{2}}\left( aq \right)\rightleftarrows M{{g}^{2+}}+\text{ }2O{{H}^{-}}\]
Now, indirectly equilibrium set between precipitated \[Mg{{\left( OH \right)}_{2}}~\]and ionized \[Mg{{\left( OH \right)}_{2}}~\]at a saturated point such as
\[Mg{{(OH)}_{2}}\left( s \right)\rightleftarrows M{{g}^{2+}}+\text{ }2O{{H}^{-}}\]
Now at equilibrium, solubility product which is represented as \[{{K}_{sp}}\] is equal to the product of concentration of both ions, \[M{{g}^{2+}}\]and \[2O{{H}^{-}}\]And solubility (ability to get dissolve in solvent to make saturated solution) of both ions is S and \[2S\](as per ratio). The relationship between solubility product and solubility is given as
\[{{K}_{sp}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}}\]
\[{{K}_{sp}}=S\text{ }\times \text{ }\left( 2S \right){}^\text{2}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }4{{S}^{2}}\]
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]
Putting the value of S which is given in question we get the required value of solubility product such as
\[{{K}_{sp}}=\text{ }4{{x}^{3}}\]
Thus, the correct option is A.

Note: It is important to note that \[{{K}_{sp}}\], solubility product for the salts of type \[M{{X}_{2}}\]is always same and that is \[{{K}_{sp}}=\text{ }4{{S}^{3}}\] and if a salt is formed with one anion and one cation such as MX, then the solubility product is equal to square of solubility S such as \[{{K}_{sp}}={{S}^{2}}\].