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Chapter 4: Problem 87

A solution of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{KOH}\) is mixed with a solution of 200.0 \(\mathrm{mL}\) of 0.150 \(\mathrm{M} \mathrm{MiSO}_{4}\) . (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

Short Answer

Expert verified
The balanced chemical equation for the reaction between KOH and MgSO4 is: \(2 KOH(aq) + MgSO_4(aq) \rightarrow Mg(OH)_2(s) + 2K^{+}(aq) + SO_{4}^{2-}(aq)\). The formed precipitate is Mg(OH)2 (magnesium hydroxide). The limiting reactant is KOH. Approximately 0.1166 grams of Mg(OH)2 precipitate will form. The concentrations of the ions remaining in the solution are: Mg2+ (0.0933 M), K+ (0.0267 M), and SO42- (0.0933 M).

Step by step solution

01

1. Balanced Chemical Equation

: The equation for the reaction between KOH (potassium hydroxide) and MiSO4 (magnesium sulfate) is: \(2 KOH(aq) + MgSO_4(aq) \rightarrow Mg(OH)_2(s) + 2K^{+}(aq) + SO_{4}^{2-}(aq)\)
02

2. Precipitate Formation

: The precipitate that forms is Mg(OH)2 (magnesium hydroxide), which is an insoluble solid.
03

3. Limiting Reactant

: First, we need to determine which reactant is limiting. We can calculate the moles of KOH and MgSO4: moles of KOH = volume x concentration = 100.0 mL x 0.200 M = 0.02 L x 0.200 mol/L = 0.004 mol moles of MgSO4 = volume x concentration = 200.0 mL x 0.150 M = 0.2 L x 0.150 mol/L = 0.03 mol Next, divide the moles of each reactant by their respective stoichiometric coefficients: KOH: 0.004 mol / 2 = 0.002 MgSO4: 0.03 mol / 1 = 0.03 Since the ratio of KOH is lower, it is the limiting reactant.
04

4. Calculating the Precipitate Mass

: We can now use the limiting reactant (KOH) to calculate the moles of Mg(OH)2 formed: moles of Mg(OH)2 = (0.004 mol KOH) x (1 mol Mg(OH)2 / 2 mol KOH) = 0.002 mol Now, we convert moles of Mg(OH)2 to grams: mass of Mg(OH)2 = moles x molar mass = 0.002 mol x 58.319 g/mol ≈ 0.1166 g So, 0.1166 grams of Mg(OH)2 precipitate will form.
05

5. Concentration of Ions Remaining

: First, we can calculate the moles of the reactants and products after the reaction: moles of MgSO4 remaining = moles(initial) - moles(reacted) = 0.03 - 0.002 = 0.028 mol moles of K+ formed = 2 x moles of KOH reacted = 2 x 0.004 = 0.008 mol Now, calculate the total volume after mixing: total volume = 100 mL + 200 mL = 300 mL = 0.3 L Finally, calculate the concentration of each ion remaining in the solution: conc. Mg2+ = moles(Mg2+) / total volume = 0.028 mol / 0.3 L ≈ 0.0933 M conc. K+ = moles(K+) / total volume = 0.008 mol / 0.3 L ≈ 0.0267 M conc. SO42- = moles(SO42-) / total volume = 0.028 mol / 0.3 L ≈ 0.0933 M The concentrations of the ions remaining in the solution are: Mg2+ (0.0933 M), K+ (0.0267 M), and SO42- (0.0933 M).

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