Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 ∘C to 27.7 ∘C. C: Write complete equation for the reaction that occurs when the two solutions are mixed. D: Write net ionic equation for the reaction that occurs when the two solutions are mixed. E: From the calorimetric data, calculate ΔH for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water.
The Correct Answer and Explanation is:
C. Complete Equation:
When copper(II) sulfate (CuSO₄) and potassium hydroxide (KOH) are mixed, a double displacement (precipitation) reaction occurs:
CuSO₄(aq) + 2KOH(aq) → Cu(OH)₂(s) + K₂SO₄(aq)
D. Net Ionic Equation:
Remove the spectator ions (K⁺ and SO₄²⁻) to focus on the actual chemical change:
Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
This is the net ionic equation for the formation of solid copper(II) hydroxide from its ions in solution.
E. Calculation of ΔH (Enthalpy Change):
To calculate ΔH, we follow these steps:
Step 1: Determine the heat released (q):
We use the formula:
q = m × c × ΔT
- m = mass of the solution = 100.0 mL × 1.00 g/mL = 100.0 g
- c = specific heat of water = 4.18 J/g·°C
- ΔT = change in temperature = 27.7°C – 21.5°C = 6.2°C
So,
q = 100.0 g × 4.18 J/g·°C × 6.2°C = 2,591.6 J = 2.5916 kJ
This is the heat released by the reaction. Since the reaction releases heat, ΔH is negative:
q = –2.5916 kJ
Step 2: Determine moles of limiting reactant:
From the balanced net ionic equation, 1 mol Cu²⁺ reacts with 2 mol OH⁻.
- CuSO₄: 50.0 mL × 1.00 mol/L = 0.0500 mol Cu²⁺
- KOH: 50.0 mL × 2.00 mol/L = 0.100 mol OH⁻
The molar ratio is exactly 1:2, so both are completely consumed. The limiting reagent is Cu²⁺, with 0.0500 mol reacting.
Step 3: Calculate ΔH per mole of Cu²⁺:
ΔH = q / mol of reaction =
ΔH = (–2.5916 kJ) / (0.0500 mol) = –51.83 kJ/mol
Final Answer:
- Complete equation: CuSO₄(aq) + 2KOH(aq) → Cu(OH)₂(s) + K₂SO₄(aq)
- Net ionic equation: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
- ΔH = –51.8 kJ/mol (to 3 significant figures)
Explanation
When two aqueous solutions—copper(II) sulfate and potassium hydroxide—are combined, a chemical reaction takes place in which a solid precipitate forms and heat is released. This is a typical double displacement reaction, where Cu²⁺ cations from CuSO₄ react with OH⁻ anions from KOH to form insoluble copper(II) hydroxide (Cu(OH)₂). The remaining K⁺ and SO₄²⁻ ions remain dissolved in solution and do not participate in the reaction. The complete and net ionic equations reflect this process.
The rise in temperature indicates the reaction is exothermic, meaning it releases energy into the surroundings—in this case, the solution. The amount of heat released is determined using the formula q = m × c × ΔT, where m is the mass of the solution (assumed equal to its volume in grams), c is the specific heat of water, and ΔT is the change in temperature. The calculated heat released is about 2.59 kJ.
To find the enthalpy change per mole of reaction (ΔH), the heat released is divided by the moles of limiting reagent, in this case, Cu²⁺, since it completely reacts with the available hydroxide. This gives an enthalpy change of –51.8 kJ/mol, which represents the energy released when 1 mole of copper(II) ions reacts with hydroxide ions to form the precipitate. This value helps chemists understand the energy profile of the reaction and classify it as energetically favorable due to the exothermic nature.
