Problem 87 - Chemistry The Central Science

Question

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?

Step-by-Step Solution

Answer

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 1: 1. Balanced Chemical Equation

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

Step 2: 2. Precipitate Formation

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

Step 3: 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.

Step 4: 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.

Step 5: 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).

Key Concepts

Chemical Equations

Chemical equations are like recipes for chemical reactions. They show reactants transforming into products. For each reaction, the reactants are written on the left side, while products are on the right. Between them is an arrow, indicating the reaction process.
In the given reaction, potassium hydroxide (KOH) reacts with magnesium sulfate (MgSO₄). You write the chemical equation as:

• Reactants: 2 KOH(aq) + MgSO₄(aq)
• Products: Mg(OH)₂(s) + 2K⁺(aq) + SO₄²⁻(aq)

The equation needs balancing. This means ensuring the same type and number of atoms are on both sides. Here, we need two moles of KOH to balance with one mole of MgSO₄.
Balancing these equations not only keeps things neat but helps in calculating the reactions at a deeper level, such as in stoichiometry.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that runs out first, thus stopping the reaction. It's like having just enough tickets for a train ride; once they're gone, no more passengers can board.
To find the limiting reactant, compare the mole ratios of your reactants to the coefficients in your balanced equation. It's important to calculate the moles for each based on their respective volumes and molarities.
In the exercise:

• KOH: Calculate 0.004 moles using its concentration and volume.
• MgSO₄: Calculate 0.03 moles.

When compared with the reaction stoichiometry, it's clear that KOH limits the reaction. Because of its lower ratio (0.002), KOH gets used up first, dictating the reaction's progression.

Precipitate Formation

One fascinating aspect of some reactions is precipitate formation. A precipitate is a solid formed from a solution during a chemical reaction and is insoluble in water.
In the mixture of KOH and MgSO₄, a white, murky solid called magnesium hydroxide (Mg(OH)₂) forms. It's the result of these ions coming together and having low solubility in the aqueous solution.
For any precipitate reaction, you apply solubility rules. These rules help predict the occurrence of a precipitate.

• Example Rule: Hydroxides are generally insoluble in water, except those of alkali metals and a few others.

Here, Mg(OH)₂ remains as a solid, illustrating precipitate formation, which can indicate that a chemical change has occurred.

Ion Concentration

After a reaction completes, it's essential to calculate the concentration of ions that remain in the solution. These concentrations reveal how much of each ion is still present, which helps in determining the reaction's final state.
For the exercise, you'll need to figure out the remaining moles of ions post-reaction:

• Mg²⁺ and SO₄²⁻ both continue in solution, as their initial concentration was higher than changes made by the reaction.
• K⁺ formed when KOH dissolved, and its concentration depends on the amount of KOH initially present.

Using the total volume of the solution, which is 300 mL or 0.3 L, calculate the concentration for each: - Mg²⁺: 0.0933 M - K⁺: 0.0267 M - SO₄²⁻: 0.0933 M
These concentrations tell us how dilute or concentrated the solution becomes, informing further chemical interactions or applications.