Question

Solubility of \[Mg{\left( {OH} \right)_2}\] is \[1.6 \times {10^{ - 4}}mol{L^{ - 1}}\] at \[298K\] . What is its solubility product?

Answer 1

Hint: Magnesium hydroxide is a compound with molecular formula of \[Mg{\left( {OH} \right)_2}\] . solubility is given which can be represented as \[s\] the compound dissociates into magnesium and hydroxide ions. The concentration can be determined from the number of moles and volume which is one litre. The solubility product can be calculated from the product of the concentrations of the ions.

Complete answer:
Given that solubility of magnesium hydroxide is \[1.6 \times {10^{ - 4}}mol{L^{ - 1}}\] . Magnesium hydroxide has the molecular formula of \[Mg{\left( {OH} \right)_2}\]
It can be dissociated partially in presence of water as follows:
\[Mg{\left( {OH} \right)_2} \rightleftharpoons M{g^{2 + }} + 2O{H^ - }\]
Given that the solubility is \[1.6 \times {10^{ - 4}}mol{L^{ - 1}}\] , the number of moles produced from one moles of magnesium hydroxide is one mole of magnesium and two moles of hydroxide ions.
The number of moles of each ion will be
\[{n_{M{g^{ + 2}}}} = 1 \times 1.6 \times {10^{ - 4}}mol = 1.6 \times {10^{ - 4}}mol\]
\[{n_{M{g^{ + 2}}}} = 2 \times 1.6 \times {10^{ - 4}}mol = 3.2 \times {10^{ - 4}}mol\]
Thus, the number of moles of the two ions that were dissociated were obtained.
The volume of solution is one litre the molarity or concentration will be equal to the number of moles only.
\[\left[ {M{g^{ + 2}}} \right] = 1.6 \times {10^{ - 4}}M\]
\[\left[ {O{H^ - }} \right] = 3.2 \times {10^{ - 4}}M\]
The solubility product of magnesium hydroxide is the product of the concentration of magnesium and hydroxide ions.
\[{K_{sp}} = \left[ {M{g^{ + 2}}} \right] \times {\left[ {O{H^ - }} \right]^2}\]
By substituting the values of concentration of ions,
\[{K_{sp}} = 1.6 \times {10^{ - 4}} \times {\left( {3.2 \times {{10}^{ - 4}}} \right)^2} = 1.6 \times {10^{ - 11}}\]
Solubility of \[Mg{\left( {OH} \right)_2}\] is \[1.6 \times {10^{ - 4}}mol{L^{ - 1}}\] at \[298K\] . The solubility product is \[1.6 \times {10^{ - 11}}\]

Note:
The given compound partially dissolves in water. Thus, it can dissolve only \[1.6 \times {10^{ - 4}}mol{L^{ - 1}}\] moles per one litre of solution. Which will be the molar solubility and the concentration. As the volume of solution is litre, the molarity is same as the number of moles of each ion dissociated.