Question

The solubility product of ${K_{sp}}$ of $Mg{\left( {OH} \right)_2}$ is $9.0 \times {10^{ - 12}}$ . If a solution is 0.010 M with respect to $M{g^{2 + }}$ ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of $Mg{\left( {OH} \right)_2}$?

Answer 1

Hint: The solubility product is the product of each concentration and each concentration is raised to the subscript of each element as a power as given in its compound. Then, we’ll put the values of solubility product and concentration of Mg ions in the formula and concentration of OH ions can be calculated.

Formula used: ${K_{sp}} = {\left[ A \right]^p}{\left[ B \right]^q}$
where $\left[ A \right]$ and $\left[ B \right]$ are the concentrations of A and B in the solution, ${K_{sp}}$ is the solubility product and p, q are the subscripts of A and B in the formula of its compound.

 Complete step by step answer:
Solubility equilibrium exists when a chemical compound in the solid state is in chemical equilibrium with a solution containing the compound. In each case i.e., in saturated, supersaturated, etc, equilibrium constant can be specified as a quotient of activities (measure of the effective concentration of the substance).
 In a chemical equilibrium,
 ${A_p}{B_q} \rightleftarrows pA + qB$
 the solubility product $\left( {{K_{sp}}} \right)$ for the compound ${A_p}{B_q}$ is defined as follows ${K_{sp}} = {\left[ A \right]^p}{\left[ B \right]^q}$
where$\left[ A \right]$ and $\left[ B \right]$are the concentrations of A and B in a saturated solution (solution in which no more solute can be dissolved). A solubility product has a similar function to an equilibrium constant but ${K_{sp}}$ has the dimension of ${\left( {concentration} \right)^{p + q}}$ .
In a chemical equilibrium,
 $Mg{\left( {OH} \right)_2} \rightleftarrows 1Mg + 2OH$
 the value of ${K_{sp}} = 9 \times {10^{ - 12}}{M^3}$
 and also
 ${K_{sp}} = {\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}$
where $\left[ {M{g^{2 + }}} \right]$ and $\left[ {O{H^ - }} \right]$ represents concentrations of Mg and OH.
${K_{sp}} = {\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}$
$9 \times {10^{ - 12}} = {\left[ {0.010} \right]^1}{\left[ {O{H^ - }} \right]^2}$
$9 \times {10^{ - 12}} = {10^{ - 2}}{\left[ {O{H^ - }} \right]^2}$
$\dfrac{{9 \times {{10}^{ - 12}}}}{{{{10}^{ - 2}}}} = {\left[ {O{H^ - }} \right]^2}$
$9 \times {10^{ - 10}} = {\left[ {O{H^ - }} \right]^2}$
$\left[ {O{H^ - }} \right] = \sqrt {9 \times {{10}^{ - 10}}} $
$\left[ {O{H^ - }} \right] = 3 \times {10^{ - 5}}$ M
Therefore, maximum hydroxide concentration which could be present without causing the precipitation of $Mg{\left( {OH} \right)_2}$ is $3 \times {10^{ - 5}}$ M or 0.00003M.

Note: Kindly be careful while doing the calculations as the concentration of Mg is given as 0.010 M which means ${10^{ - 2}}$ M as the zero after 1 doesn’t signify anything. Remember the formula of Solubility product and it has the same dimension or unit as concentration.