Will a precipitate form when 100.0 \(\mathrm{mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to 100.0 \(\mathrm{mL}\) of \(2.0 \times 10^{-4} \mathrm{MNaOH}\)?

The balanced chemical equation for the reaction between Mg(NO₃)₂ and NaOH can be written as: \[Mg(NO_3)_2(aq) + 2NaOH(aq) \rightarrow Mg(OH)_2(s) + 2NaNO_3(aq)\] This equation indicates that one mole of magnesium nitrate reacts with two moles of sodium hydroxide to produce one mole of magnesium hydroxide precipitate and two moles of sodium nitrate in the solution.

We need to calculate the millimoles (mmol) of each ion in the mixed solution. For magnesium nitrate: Volume = 100.0 mL = 0.100 L Concentration = \(4.0 \times 10^{-4}\,M\) mmol of \(Mg^{2+}\) = Volume × Concentration = 0.100 L ×\(4.0 \times 10^{-4}\,M\) = 0.040 mmol For sodium hydroxide: Volume = 100.0 mL = 0.100 L Concentration = \(2.0 \times 10^{-4}\,M\) mmol of \(OH^-\) = Volume × Concentration = 0.100 L ×\(2.0 \times 10^{-4}\,M\) = 0.020 mmol Next, we calculate the initial concentrations of \(Mg^{2+}\) and \(OH^-\) ions in the mixed solution: Volume of the mixed solution = 100.0 mL + 100.0 mL = 200.0 mL = 0.200 L Initial concentration of \(Mg^{2+}\) = (0.040 mmol) / (0.200 L) = \(2.0 \times 10^{-4} M\) Initial concentration of \(OH^-\) = (0.020 mmol) / (0.200 L) = \(1.0 \times 10^{-4} M\)

To calculate the reaction quotient, Q, we need to use the initial concentrations of \(Mg^{2+}\) and \(OH^-\) ions: \[Q = [Mg^{2+}][OH^-]^{2}\] Q = \( (2.0 \times 10^{-4})(1.0 \times 10^{-4})^{2}\) = \(2.0 \times 10^{-12}\) Now, we need to compare Q with the solubility product constant, Ksp, for magnesium hydroxide to predict if a precipitate would form. The Ksp of \(Mg(OH)_2\) is \(1.0 \times 10^{-11}\). Since Q < Ksp (\(2.0 \times 10^{-12} < 1.0 \times 10^{-11}\)), no precipitate will form. In conclusion, no precipitate will form when 100.0 mL of \(4.0 \times 10^{-4} M Mg(NO_3)_2\) is added to 100.0 mL of \(2.0 \times 10^{-4} M NaOH\).