
Let the solubility of an aqueous solution of Mg(OH)2 be x then its ksp is
A. 4x3
B. x3
C. 27x4
D. 9x
Answer
164.7k+ views
Hint: Solubility product is the product of the concentration of ions produced in equilibrium or saturated solution and it is represented as ksp. 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 \[{{K}_{sp}}={{S}^{2}}\].
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 \[{{K}_{sp}}={{S}^{2}}\].
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