Question

A saturated solution of $ Mg{\left( {OH} \right)_2} $ ​ in water has pH =10.32. How do you calculate the Ksp of $ Mg{\left( {OH} \right)_2} $ ​ ?

Answer 1

Hint : To solve this question, we must first know the basic concepts about the solubility product constant Ksp, then you need to know about the formulae and methods to calculate Ksp of when the pH is being given.

Complete Step By Step Answer:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
Ksp (Solubility Product Constant) is the equilibrium constant for a solid substance or solute which is dissolved in an aqueous solution. It represents the level at which a solute gets dissolved in solution. The more the Ksp value is the higher the solubility of that substance is.
Step 1: As we know that:
 $ Mg{\left( {OH} \right)_2} \rightleftharpoons M{g^{2 + }} + {\text{ }}2O{H^ - } $
Step 2: And also, $ pH{\text{ }} + {\text{ }}pOH{\text{ }} = {\text{ }}14 $
 $ \begin{array}{*{20}{l}}
  { \Rightarrow pOH{\text{ }} = {\text{ }}14-pH{\text{ }} = {\text{ }}14-10.32} \\
  {\therefore - log\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}3.68} \\
  { \Rightarrow \left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}2{\text{ }} \times {{10}^{ - 4}}mol{\text{ }}{l^{ - 1}}}
\end{array} $
Step 3: From the reaction in the step 1 we can that the concentration of $ \left[ {M{g^{2 + }}} \right] $ will be half of $ \left[ {O{H^ - }} \right] $ :
 $ \therefore \left[ {M{g^{2 + }}} \right]{\text{ }} = {\text{ }}\dfrac{{\left( {2{\text{ }} \times {\text{ }}{{10}^{ - 4}}} \right)}}{2}{\text{ }} = {\text{ }}{10^{ - 4}}mol{\text{ }}{l^{ - 1}} $
Step 4: Now the Ksp will be given by:
 $ \begin{array}{*{20}{l}}
  {Ksp{\text{ }} = {\text{ }}\left[ {M{g^{2 + }}} \right]{{\left[ {O{H^ - }} \right]}^2}} \\
  { \Rightarrow Ksp{\text{ }} = {\text{ }}{{10}^{ - 4}} \times {\text{ }}{{\left( {2{\text{ }} \times {\text{ }}{{10}^{ - 4}}} \right)}^2} = 4{\text{ }} \times {{10}^{ - 12}}mol{\text{ }}{l^ - }^3}
\end{array} $
Hence, $ Ksp{\text{ }} = 4{\text{ }} \times {10^{ - 12}}mol{\text{ }}{{\text{l}}^ - }^{\text{3}} $ is the required answer.

Note :
The total amount of the solute that can be dissolved in the solvent at equilibrium is called the solubility of a substance in a solvent. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.