A 100.0-mL aliquot of 0.200 M aqueous potassium hydroxide is mixed with 100.0 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.

The reaction between potassium hydroxide (KOH) and magnesium nitrate [Mg(NO₃)₂] can be written as: KOH (aq) + Mg(NO₃)₂ (aq) → ? To create a balanced chemical equation, we must first predict the products formed. According to the solubility rules, alkali metal ions (Group 1 elements) and nitrate ions are soluble, so we expect potassium nitrate (KNO₃) to be soluble. Magnesium hydroxide (Mg(OH)₂), on the other hand, is slightly soluble in water. Therefore, our balanced chemical equation should look like this: 2 KOH (aq) + Mg(NO₃)₂ (aq) → 2 KNO₃ (aq) + Mg(OH)₂ (s)

In this reaction, the insoluble product formed is magnesium hydroxide (Mg(OH)₂). It forms as a solid precipitate, so we indicate this in our balanced chemical equation with the (s) state symbol: 2 KOH (aq) + Mg(NO₃)₂ (aq) → 2 KNO₃ (aq) + Mg(OH)₂ (s)

To calculate the mass of Mg(OH)₂ produced, we must identify the limiting reactant and use stoichiometry. We have 100.0 mL of 0.200 M KOH and 100.0 mL of 0.200 M Mg(NO₃)₂ as our starting amounts. First convert the volumes to moles: moles KOH = volume × concentration moles KOH = (100.0 mL)(0.200 mol/L) = 0.0200 mol moles Mg(NO₃)₂ = volume × concentration moles Mg(NO₃)₂ = (100.0 mL)(0.200 mol/L) = 0.0200 mol Now, use stoichiometry to find the moles of Mg(OH)₂ formed: moles Mg(OH)₂ = moles Mg(NO₃)₂ × (1 mol Mg(OH)₂ / 1 mol Mg(NO₃)₂) moles Mg(OH)₂ = 0.0200 mol Mg(NO₃)₂ × (1 mol Mg(OH)₂ / 1 mol Mg(NO₃)₂) = 0.0200 mol Mg(OH)₂ Next, convert moles to mass: mass Mg(OH)₂ = moles × molar mass mass Mg(OH)₂ = (0.0200 mol)(58.32 g/mol) = 1.1664 g Therefore, 1.1664 g of Mg(OH)₂ (s) precipitate is produced.

Since Mg(OH)₂ is the limiting reactant, 1 mole of Mg(NO₃)₂ will react with 2 moles of KOH. After precipitation is complete, all moles of Mg(NO₃)₂ will have reacted. However, we will have an excess of KOH remaining. Calculate the moles of KOH remaining: moles KOH remaining = moles KOH initially - (moles Mg(NO₃)₂ × 2) moles KOH remaining = 0.0200 mol - (0.0200 mol × 2) = 0.0200 mol - 0.0400 mol = -0.0200 mol Since the remaining moles of KOH cannot be negative, the limiting reactant is KOH instead of Mg(NO₃)₂. moles KOH reacted = moles KOH initially / 2 moles reacted = 0.0200 mol / 2 = 0.0100 mol Now, find the moles of KNO₃ formed and the moles of Mg(NO₃)₂ remaining: moles KNO₃ = moles KOH reacted × 2 moles KNO₃ = 0.0100 mol × 2 = 0.0200 mol KNO₃ moles Mg(NO₃)₂ remaining = moles Mg(NO₃)₂ initially - moles reacted moles Mg(NO₃)₂ remaining = 0.0200 mol - 0.0100 mol = 0.0100 mol Finally, calculate the concentration of each ion in solution. The total volume of the solution is 200.0 mL. Concentration of K⁺: \[c(K^+) = \frac{moles \, KNO_3}{total \, volume} = \frac{0.0200 \, mol \, KNO_3}{0.200 \, L} = 0.100 \, M\] Concentration of Mg²⁺: \[c(Mg^{2+}) = \frac{moles \, Mg(NO_3)_2}{total \, volume} = \frac{0.0100 \, mol \, Mg(NO_3)_2}{0.200 \, L} = 0.0500 \, M\] Concentration of NO₃⁻: \[c(NO_3^-) = \frac{moles \, Mg(NO_3)_2 \times 2 + moles \, KNO_3}{total \, volume} = \frac{(0.0100 \, mol \times 2) + 0.0200 \, mol}{0.200 \, L} = \frac{0.0400 \, mol}{0.200 \, L} = 0.200 \, M\] So, after precipitation is complete, the concentration of K⁺ is 0.100 M, Mg²⁺ is 0.0500 M, and NO₃⁻ is 0.200 M.