Question

The solubility product K sp of $Mg{(OH)_2}$ is 9.0$ \times {10^{ - 12}}$. If a solution is 0.0100 M with respect to $M{g^{2 + }}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg{(OH)_2}$?
A. $1.5 \times {10^{ - 7}}$ M
B. $3.0 \times {10^{ - 7}}$ M
C. $1.5 \times {10^{ - 5}}$ M
D. $3.0 \times {10^{ - 5}}$ M

Answer 1

Hint: the question is based on ionic equilibrium. The solubility product depends on the concentration of ions dissolved in the solution.

Complete step by step solution: the maximum number of moles of a salt that can be dissolved per litre of a solution at a given temperature is called the solubility of the given salt at the given temperature. Let us consider the salt of magnesium hydroxide that is dissolved in water. The salt starts ionizing and reaction can proceed in both the directions and a equilibrium can be written as
     ${K_{sp}} = [M{g^{2 + }}][O{H^ - }]$
     $Mg{(OH)_2} \to M{g^{2 + }} + 2O{H^ - }$
(K sp) solubility product of an electrolyte at a specified temperature may be defined as the molar concentration of its ions in a saturated solution, each concentration raised to the power of equal to the number of ions produced on dissociation of one molecule of the electrolyte. Substituting the values in the formula from the question
      $
  9 \times {10^{ - 12}} = 0.01 \times {[O{H^ - }]^2} \\
  {[O{H^ - }]^2} = \dfrac{{9 \times {{10}^{ - 12}}}}{{{{10}^{ - 2}}}} = 9 \times {10^{ - 10}} = \sqrt {9 \times {{10}^{ - 10}}} = 3 \times {10^{ - 5}} \\
$
Hence the correct option is option D.

Additional information: If an ionic product is less than Ksp more of the salt dissolves. If the ionic product is equal to the K sp then the solution is saturated.
Note: write a balanced equation.