The concentrations of various cations in seawater, in moles per liter, are $$ \begin{array}{llllll} \hline \text { Ion } & \mathrm{Na}^{+} & \mathrm{Mg}^{2+} & \mathrm{Ca}^{2+} & \mathrm{A}^{3+} & \mathrm{Fe}^{3+} \\ \text { Molarity }(M) & 0.46 & 0.056 & 0.01 & 4 \times 10^{-7} & 2 \times 10^{-7} \\ \hline \end{array} $$ (a) At what \(\left[\mathrm{OH}^{-}\right]\) does \(\mathrm{Mg}(\mathrm{OH})_{2}\) start to precipitate? (b) At this concentration, will any of the other ions precipitate? (c) If enough \(\mathrm{OH}^{-}\) is added to precipitate \(50 \%\) of the \(\mathrm{Mg}^{2+}\), what percentage of each of the other ions will precipitate? (d) Under the conditions in (c), what mass of precipitate will be ob- tained from one liter of seawater?

The Ksp values for Mg(OH)2 is given as \(K_{sp} = [Mg^{2+}][OH^-]^2\). We are given the concentration of Mg2+ as 0.056 M. To find the concentration of OH- at which Mg(OH)2 starts to precipitate, we need to solve the Ksp expression for [OH-]: $$ [OH^-] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}$$ Using \(K_{sp} = 5.61 \times 10^{-12}\) for Mg(OH)2 and [Mg2+] = 0.056 M: $$ [OH^-] = \sqrt{\frac{5.61 \times 10^{-12}}{0.056}} \approx 3.54 \times 10^{-6} M$$ So, the concentration of OH- at which Mg(OH)2 starts to precipitate is approximately \(3.54 \times 10^{-6}M\). (b)

We need to compare the Ksp values of the precipitates for other ions (assuming they all form hydroxides) with their actual ion product (Q) at the OH- concentration found in part (a). If Q > Ksp, then the ion will precipitate. Using the given concentrations, we can calculate Q for each ion as follows: $$Q_{NaOH} = [Na^+][OH^-] = 0.46\times 3.54\times 10^{-6}$$ $$Q_{Ca(OH)_2} = [Ca^{2+}][OH^-]^2 = 0.01\times (3.54\times 10^{-6})^2$$ $$Q_{A(OH)_3} = [A^{3+}][OH^-]^3 = 4\times 10^{-7}\times (3.54\times 10^{-6})^3$$ $$Q_{Fe(OH)_3} = [Fe^{3+}][OH^-]^3 = 2\times 10^{-7}\times (3.54\times 10^{-6})^3$$ Now, we compare these values with their respective Ksp values, for example: $$K_{sp} ({Ca(OH)_2}) = 6.5\times 10^{-6}$$ $$K_{sp} ({Fe(OH)_3}) = 2.79\times 10^{-39}$$ $$K_{sp} ({A(OH)_3}) = 3\times 10^{-34}$$ The Ksp value for NaOH is not relevant, as it is a strong base and does not have a significant tendency to precipitate. We compare Q with Ksp and find that none of these values exceeds their respective Ksp values, which means that no other ions will precipitate at this OH- concentration. (c)

When enough OH- is added to precipitate 50% of Mg2+, the concentration of Mg2+ in solution will decrease by half: $$[Mg^{2+}]_{new} = \frac{1}{2} \times 0.056 M$$ At this point, we should recalculate Q for each ion with the new concentration of Mg2+ and considering that the [OH-] required to do this transformation is the same as before (from part (a)): $$Q_{NaOH} = [Na^+][OH^-] = 0.46\times 3.54\times 10^{-6}$$ $$Q_{Ca(OH)_2} = [Ca^{2+}][OH^-]^2 = 0.01\times (3.54\times 10^{-6})^2$$ $$Q_{A(OH)_3} = [A^{3+}][OH^-]^3 = 4\times 10^{-7}\times (3.54\times 10^{-6})^3$$ $$Q_{Fe(OH)_3} = [Fe^{3+}][OH^-]^3 = 2\times 10^{-7}\times (3.54\times 10^{-6})^3$$ We find that still none of these values exceeds their respective Ksp values, which means that no other ions will precipitate at this OH- concentration. (d)

At this point, we know that only Mg(OH)2 is precipitating and that 50% of the Mg2+ ions have precipitated. To find the mass of precipitate, we will use the stoichiometry of the reaction. $$Mg^{2+} (aq) + 2OH^{-} (aq) \rightarrow Mg(OH)_2 (s)$$ Moles of Mg2+ ions precipitated = 0.5 × moles of Mg2+ ions in solution $$= 0.5 \times 0.056 \, mol$$ Since the molar mass of Mg(OH)2 is 58.32 g/mol, the mass of precipitate obtained is: $$Mass \, of \, Mg(OH)_2 = 0.5 \times 0.056 \, mol \times 58.32 \, g/mol = 1.63g$$ Therefore, the mass of precipitate obtained from one liter of seawater under the conditions in part (c) is approximately 1.63 grams.