Question

What is the maximum pH of a solution 0.10M in $M{g^{2 + }}$ from which $Mg{(OH)_2}$ will not precipitate? Given, ${K_{sp}}\left[ {Mg{{(OH)}_2}} \right] = 1.2 \times {10^{ - 11}}$.

Answer 1

Hint: For precipitation to occur the ionic product (i.e ${K_{ip}}$) should be greater than the solubility product(i.e ${K_{sp}}$ ) or ${K_{ip}}$> ${K_{sp}}$, and ${K_{ip}}$< ${K_{sp}}$ for precipitation to not occur. Here we are given the condition of no precipitation so will use the later mentioned formula.

Complete step-by-step answer:
First let us understand what an ionic and solubility product is.
The solubility product (${K_{sp}}$) is defined as the product of solubilities of the ions in moles per liter.
Whereas ionic product (${K_{ip}}$) is defined as the product of the concentrations of ions, each raised to the power determined by their stoichiometric coefficient in a solution of a salt.
Ideally, both ionic product and solubility product represent the product of concentrations of the ions in the solution. The term ionic product has a broader definition since it is applicable to all types of solutions(both saturated and unsaturated)
Whereas, we can apply solubility products only to a saturated solution where there exists a dynamic equilibrium between the ions and the undissolved salt present in the solution.
Therefore the solubility product can also be defined as the ionic product for a saturated solution at a constant temperature.
Now coming back to the question, we want $Mg{(OH)_2}$ to not precipitate, so we will use the condition : ${K_{ip}}$< ${K_{sp}}$ i.e the ionic product must be lesser than the solubility product.
$Mg{(OH)_2} \rightleftarrows M{g^{2 + }} + 2O{H^ - }$
Now, the ionic product ${K_{ip}}$is: $\left[ {M{g^{2 + }}} \right]\,{\left[ {O{H^ - }} \right]^2}$
Since ${K_{ip}}$< ${K_{sp}}$and it is given that: ${K_{sp}}\left[ {Mg{{(OH)}_2}} \right] = 1.2 \times {10^{ - 11}}$.
$ \Leftrightarrow $ $\left[ {M{g^{2 + }}} \right]\,{\left[ {O{H^ - }} \right]^2}\, < \,1.2 \times {10^{ - 11}}$
$
\left[ {O{H^ - }} \right]\, < \,{\left[ {\dfrac{{{K_{sp}}\,(Mg{{(OH)}_2})}}{{\left[ {M{g^{2 + }}} \right]}}} \right]^{\dfrac{1}{2}}} \\
\left[ {O{H^ - }} \right]\, < \,{\left[ {\dfrac{{1.2 \times {{10}^{ - 11}}{M^3}}}{{0.10M}}} \right]^{\dfrac{1}{2}}} \\
 $
On solving we get:
$\left[ {O{H^ - }} \right]\, < \,1.035\, \times {10^{ - 11}}$
Now to find pH, we must first find the pOH of the solution.
So taking negative logarithm both sides:
$ - \log \left[ {O{H^ - }} \right]\, < \, - \log (1.035 \times {10^{ - 5}})$
Upon calculating we get : pOH < 4.36
As, we know pH+pOH=14
pH = 14-pOH
pH > 14-4.36 (Inequality altered due to negative sign)
pH > 9.04
Therefore, the maximum pH at which there will be no precipitation is: 9.04

Note: A saturated solution can be defined as a chemical solution that contains the maximum concentration of a solute dissolved in solvent. Any amount of additional solute will not dissolve in a saturated solution. An unsaturated solution is termed as a chemical solution in which the concentration of solute is lower than its equilibrium solubility.