Question: Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from seawater by the following sequence of reactions:

\(\begin{array}{*{20}{c}}{M{g^{2 + }}(aq) + Ca{{(OH)}_2}(aq) \to Mg{{(OH)}_2}(s) + C{a^{2 + }}(aq)}\\{Mg{{(OH)}_2}(s) + 2HCl(aq) \to MgC{l_2}(s) + 2{H_2}O(l)}\end{array}\)

\(MgC{l_2}(l)\mathop \to \limits^{\;electrolysis\;} Mg(s) + C{l_2}(g)\)

Sea water has a density of 1.026 g/cm3 and contains 1272 parts per million of magnesium \(M{g^{2 + }}(aq)\)by mass. What mass, in kilograms, \(Ca{(OH)_2}\)is required to precipitate 99.9% of the magnesium in 1.00 × 103 L of seawater?

The mass of magnesium,

\(m(Mg) = V \times \rho = 1.0 \times {10^3}L \times {\rm{c}}{{\rm{m}}^3}/L \times 1.026{\rm{g}}/{\rm{c}}{{\rm{m}}^3} \times 1272{\rm{ppm}} \times {10^{ - 6}}{\rm{pp}}{{\rm{m}}^{ - 1}}\)

\(m(Mg) = 1.305 \times {10^3}g\)

The mass concentration of magnesium equals to \(1.305{\rm{g}}/{\rm{L}}\).

99.9% will be predicted as \(1.305g/L \times 0.999 = 1.304g/L\)

\(\begin{array}{l}c(Mg) = \frac{{1.304{\rm{g}}/{\rm{L}}}}{{24.305{\rm{gol}}}}\\ = 0.05365{\rm{M}}\end{array}\)

One mole of \({\rm{Ca}}{({\rm{OH}})_2}\)reacts with one mole of \({M^{2 + }}\)so the amount of \({\rm{Ca}}{({\rm{OH}})_2}\) required to precipitate 99.9% of the \(M{g^{2 + }}\)in 1L is,

\(\begin{array}{l}\gamma = 0.05365M \times 74.09{\rm{g}}/{\rm{mol}}\\ = 3.97{\rm{g}}/{\rm{L}}\end{array}\)

\(O{H^ - }\)ions need to be added for equilibrium. The added amount of \(O{H^ - }\) required is found from the solubility product

\(\begin{array}{*{20}{c}}{{K_{sp}} = \left( {M{g^{2 + }}} \right){{\left( {O{H^ - }} \right)}^2}}\\{}\end{array}\)

\(\begin{array}{l}\left( {O{H^ - }} \right) = \sqrt {\frac{{{K_{sp}}}}{{\left( {M{g^{2 + }}} \right)}}} \\ = \sqrt {\frac{{1.5 \times {{10}^{ - 11}}}}{{5.369 \times {{10}^{ - 5}}}}} \\ = 5.29 \times {10^{ - 4}}\end{array}\)

\({\rm{Ca}}{({\rm{OH}})_2}{\rm{\;in\;}}1{\rm{L}} = 2.69 \times {10^{ - 4}}{\rm{mol}}\)

\(\begin{array}{l}{\rm{m}}\left( {{\rm{Ca}}{{({\rm{OH}})}_2}} \right) = 2.65 \times {10^{ - 4}}{\rm{mol}} \times 1.0 \times 1000{\rm{L}} \times 74.0964{\rm{g}}/{\rm{mol}}\\ = 20{\rm{g}}\end{array}\)

So, the total mass is \(3.97{\rm{kg}} + 0.02{\rm{kg}} = 3.99{\rm{kg}}\)