Question

${{K}_{sp}}$ of \[Mg{{(OH)}_{2}}\] is $1X{{10}^{12}},$ 0.01M $MgC{{l}_{2}}$ will be precipitating at the limiting pH:
(A) 8
(B) 9
(C) 10
(D) 12

Answer 1

Hint: Solubility product ${{K}_{sp}}$ describes the concentration of ions for a saturated solution, which means the solution at maximum equilibrium with molecules and ions of solution. This equilibrium constant ${{K}_{sp}}$measures the maximum amount of ions and it changes with temperature.

Complete step by step solution:
However, the concentration of ions in a solution may be diluted, saturated, or supersaturated. So, the ionic product which measures the maximum amount of ions in a solution and also varies on the concentration of ions dissolved.
If the solution is diluted, saturated, or supersaturated depends on the relation between ionic product and solubility product values.
If the ionic product of a solution is equal to the solubility product, then the solution is saturation without any precipitation. After this saturation point, the solution is unable to add any ions, because of the actual amount of ions equal to the maximum amount of ions.
When the actual amount of ions less than the maximum amount of ions. i.e, the ionic product is less than the solubility product. Then the solution is diluted but no precipitate will form. There is a possibility of adding more ions in the dilute solution.
If the ionic product greater than the solubility product means that the actual amount of ions is more than the maximum amount of ions. The resultant solution will be supersaturated.
$Mg{{(OH)}_{2}}\rightleftharpoons M{{g}^{+2}}+2O{{H}^{-}}$
Then solubility product is, ${{K}_{sp}}=[M{{g}^{+2}}]{{[O{{H}^{-}}]}^{2}}$ -- (1)
Given the value of ${{K}_{sp}}$ \[Mg{{(OH)}_{2}}\] = $1X{{10}^{-12}}$
The concentration of $[M{{g}^{+2}}]=0.01M$ , because due to common ion in $[MgC{{l}_{2}}]=0.01M$
Substitute the above values in the above equation (1),
Then, $1X{{10}^{-12}}=0.01X{{[O{{H}^{-}}]}^{2}}$
$[O{{H}^{-}}]={{10}^{-5}}M$
Ionic product ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
Substitute the $[O{{H}^{-}}]$ value in the above equation,
Then, $[{{H}^{=}}]=\dfrac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}}M$
Since the if ionic product exceeds the solubility product then precipitation occurs at limiting pH = 9.

Hence the correct answer to this question is option (B).

Note: The ionic product of water means concentrations of both ions in water at a particular temperature because water undergoes self ionization to some extent. Because water is a weak electrolyte. If both ions concentration is equal then water is a neutral medium and acidity or basicity depends on the concentration of ions in water.