A \(100.0-\mathrm{mL}\) aliquot of \(0.200 M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{~mL}\) of \(0.200 M\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

To determine the balanced chemical equation, first, write down the formulas of the two reactants and then predict the products through double displacement. Potassium hydroxide: \(KOH\), aqueous Magnesium nitrate: \(Mg(NO_3)_2\), aqueous Double displacement: Potassium will swap places with magnesium in each compound. \(KOH(aq) + Mg(NO_3)_2(aq) \longrightarrow Mg(OH)_2(s) + 2KNO_3(aq)\) Now we have a balanced chemical equation for the reaction taking place.

In the balanced chemical equation, we can see that magnesium hydroxide \(( Mg(OH)_2 )\) is one of the products, and it is in the solid phase. Therefore, magnesium hydroxide forms a precipitate.

First, we need to determine the limiting reactant from the given concentrations of the two reactants: Concentration of \(KOH\) = \(0.200 M\) Volume of \(KOH\) = \(100.0 mL\) Concentration of \(Mg(NO_3)_2\) = \(0.200 M\) Volume of \(Mg(NO_3)_2\) = \(100.0 mL\) Convert the volumes to liters: Volume of \(KOH\) = \(0.100 L\) Volume of \(Mg(NO_3)_2\) = \(0.100 L\) Calculate the moles of each reactant: moles of \(KOH = 0.200 M \times 0.100 L = 0.020\; mol\) moles of \(Mg(NO_3)_2 = 0.200 M \times 0.100 L = 0.020\; mol\) Using the stoichiometry of the balanced chemical equation, we can calculate the moles of \(Mg(OH)_2\) formed: \(1 \;mol\; KOH \rightarrow 1 \;mol\; Mg(OH)_2\) moles of \(Mg(OH)_2 = 0.020\; mol\) Now, we can convert moles to grams using molar mass: Molar mass of \(Mg(OH)_2 = 24.3+2(16.0+1.0) = 58.3 g/mol\) Mass of precipitate produced = moles of \(Mg(OH)_2 * Molar mass of Mg(OH)_2\) Mass of precipitate produced = \(0.020\; mol \times 58.3 g/mol = 1.166 g\)

Since all the moles of \(KOH\) reacted with all the moles of \(Mg(NO_3)_2\) and formed a precipitate, there is no excess of any reactant. The remaining ions are those of the soluble salt \(KNO_3\). The balanced equation shows that two moles of \(KNO_3\) form for every mole of \(KOH\), so: moles of \(K^+\) = moles of \(NO_3^-\) = \(2 \times moles\; of\; KOH = 2 \times 0.020\; mol = 0.040\; mol\) Since the volume of the reaction mixture is the sum of the initial volumes: Final volume = \(100.0 mL + 100.0 mL = 200.0 mL = 0.200 L\) Now, we can calculate the concentration of each ion remaining in the solution: Concentration of \(K^+\) = moles of \(K^+\) / total volume Concentration of \(K^+\) = \(0.040\; mol / 0.200 L = 0.200 M\) Concentration of \(NO_3^-\) = moles of \(NO_3^-\) / total volume Concentration of \(NO_3^-\) = \(0.040\; mol / 0.200 L = 0.200 M\) Thus, the concentrations of the remaining ions in the solution are: \(K^+\): \(0.200 M\) \(NO_3^-\): \(0.200 M\)