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Question: The solubility product of \({K_{sp}}\) of \(Mg{\left( {OH} \right)_2}\) is \(9.0 \times {10^{ - 12}}...

The solubility product of Ksp{K_{sp}} of Mg(OH)2Mg{\left( {OH} \right)_2} is 9.0×10−129.0 \times {10^{ - 12}} . If a solution is 0.010 M with respect to Mg2+M{g^{2 + }} ion. What is the maximum hydroxide ion concentration which could be present without causing the precipitation of Mg(OH)2Mg{\left( {OH} \right)_2}?

Explanation

Solution

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: Ksp=[A]p[B]q{K_{sp}} = {\left[ A \right]^p}{\left[ B \right]^q}
where [A]\left[ A \right] and [B]\left[ B \right] are the concentrations of A and B in the solution, Ksp{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,
ApBq⇄pA+qB{A_p}{B_q} \rightleftarrows pA + qB
the solubility product (Ksp)\left( {{K_{sp}}} \right) for the compound ApBq{A_p}{B_q} is defined as follows Ksp=[A]p[B]q{K_{sp}} = {\left[ A \right]^p}{\left[ B \right]^q}
where[A]\left[ A \right] and [B]\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 Ksp{K_{sp}} has the dimension of (concentration)p+q{\left( {concentration} \right)^{p + q}} .
In a chemical equilibrium,
Mg(OH)2⇄1Mg+2OHMg{\left( {OH} \right)_2} \rightleftarrows 1Mg + 2OH
the value of Ksp=9×10−12M3{K_{sp}} = 9 \times {10^{ - 12}}{M^3}
and also
Ksp=[Mg2+]1[OH−]2{K_{sp}} = {\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}
where [Mg2+]\left[ {M{g^{2 + }}} \right] and [OH−]\left[ {O{H^ - }} \right] represents concentrations of Mg and OH.
Ksp=[Mg2+]1[OH−]2{K_{sp}} = {\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}
9×10−12=[0.010]1[OH−]29 \times {10^{ - 12}} = {\left[ {0.010} \right]^1}{\left[ {O{H^ - }} \right]^2}
9×10−12=10−2[OH−]29 \times {10^{ - 12}} = {10^{ - 2}}{\left[ {O{H^ - }} \right]^2}
9×10−1210−2=[OH−]2\dfrac{{9 \times {{10}^{ - 12}}}}{{{{10}^{ - 2}}}} = {\left[ {O{H^ - }} \right]^2}
9×10−10=[OH−]29 \times {10^{ - 10}} = {\left[ {O{H^ - }} \right]^2}
[OH−]=9×10−10\left[ {O{H^ - }} \right] = \sqrt {9 \times {{10}^{ - 10}}}
[OH−]=3×10−5\left[ {O{H^ - }} \right] = 3 \times {10^{ - 5}} M
Therefore, maximum hydroxide concentration which could be present without causing the precipitation of Mg(OH)2Mg{\left( {OH} \right)_2} is 3×10−53 \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{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.