Consider two solutions, the first being \(50.0 \mathrm{~mL}\) of $1.00 \mathrm{M} \mathrm{CuSO}_{4}\( and the second \)50.0 \mathrm{~mL}\( of \)2.00 \mathrm{M} \mathrm{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 to \(27.7^{\circ} \mathrm{C} .(\mathbf{a})\) Before mixing, how many grams of Cu are present in the solution of \(\mathrm{CuSO}_{4}\) ? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. \((\mathbf{d})\) From the calorimetric data, calculate \(\Delta H\) 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 \(100.0 \mathrm{~mL},\) and that the specific heat and density of the solution after mixing are the same as those of pure water.

Step by step solution

01

Calculate grams of Cu in CuSO4 solution

To find the grams of Cu in the CuSO4 solution, we first need to determine the moles of CuSO4. Volume of CuSO4 solution: \(50.0\,mL\) or \(0.0500\,L\) Concentration of CuSO4: \(1.00\,M\) Using the formula moles = molarity × volume, we have: Moles of CuSO4: \((1.00\,M) (0.0500\,L) = 0.0500\,mol\) Now, we find the grams of Cu in the CuSO4 solution using the molar mass of Cu: Molar mass of Cu = \(63.55\,g/mol\) Grams of Cu = moles × molar mass = \(0.0500\,mol × 63.55\,g/mol = 3.18\,g\) So, there are \(3.18\,g\) of Cu present in the CuSO4 solution.

02

Identify the precipitate formed

When CuSO4 and KOH are mixed, the following double displacement reaction occurs: CuSO4 (aq) + 2 KOH(aq) → Cu(OH)2 (s) + K2SO4 (aq) In this reaction, Cu(OH)2 is the precipitate formed.

03

Write complete and net ionic equations for the reaction

Complete ionic equation: Cu2+ (aq) + SO42- (aq) + 2 K+ (aq) + 2 OH- (aq) → Cu(OH)2(s) + 2 K+ (aq) + SO42- (aq) Net ionic equation: Cu2+ (aq) + 2 OH- (aq) → Cu(OH)2(s)

04

Calculate ΔH for the reaction

Given that the total volume of the solution is \(100.0\,mL\) after mixing and its specific heat and density are the same as those of pure water: Specific heat of solution (c) = \(4.18\,J/g^{\circ}C\) Density of solution (p) = \(1.00\,g/mL\) Total mass of solution (m) = Total Volume × Density = \(100.0\,mL × 1.00\,g/mL = 100.0\,g\) \(ΔT = 27.7^{\circ}C - 21.5^{\circ}C = 6.2^{\circ}C\) Now, we use the formula to find the heat released (q) during the reaction: q = m × c × ΔT = \(100.0\,g × 4.18\,J/g^{\circ}C × 6.2^{\circ}C = 2589.6\,J\) Since it is an exothermic reaction, we attach a minus sign and convert q to kJ. Heat released, q = \(-2589.6\,J = -2.5896\,kJ\) To find the enthalpy change (ΔH) per mole, divide the heat released by the moles of CuSO4 reacted: ΔH = q / moles of CuSO4 = \(\frac{-2.5896\,kJ}{0.0500\,mol} = -51.79\,kJ/mol\) The enthalpy change for the reaction is \(-51.79\,kJ/mol\).

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