Question

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.

Short answer

There are \(3.18\,g\) of Cu present in the CuSO4 solution. The precipitate formed in the reaction between CuSO4 and KOH is Cu(OH)2. The net ionic equation for the reaction is Cu2+ (aq) + 2 OH- (aq) → Cu(OH)2(s). The enthalpy change for the reaction is \(-51.79\,kJ/mol\).

Step-by-step solution

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.

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.

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)

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 \(-51.79\,kJ/mol\).

Key concepts

Calorimetry

Calorimetry is a branch of thermochemistry that measures the heat exchanged during chemical reactions. In our problem, a constant-pressure calorimeter is used to measure the heat released when two solutions are mixed. When substances react, they either absorb or release heat. Calorimetry allows us to quantify this heat exchange and is critical in calculating the enthalpy change (\( \Delta H \)) for reactions.

Key elements in calorimetry include:

• Specific Heat Capacity: The amount of heat required to raise the temperature of one gram of a substance by a degree Celsius.
• Density: The mass per unit volume essential for calculating the total mass of the solution.
• Temperature Change (\( \Delta T \)): The difference between the final and the initial temperature of the solution, crucial for determining the heat change.

In the exercise, we calculated the heat released using the formula \( q = m \times c \times \Delta T \), where \( m \) is the total mass of the solution, \( c \) is the specific heat, and \( \Delta T \) is the change in temperature. This heat value represents the \( q \) of the reaction and is used to find the reaction's enthalpy change per mole of the limiting reactant, typically expressed in kilojoules per mole.

Ionic Equations

Ionic equations provide a detailed picture of the chemicals actually participating in a reaction, isolating the ions that change during the reaction. When solutions containing ionic compounds are mixed, they often form a precipitate or involve exchange of parts.

In the given solution, the complete ionic equation involves writing out all the soluble compounds in terms of their ions. For the reaction between \( \text{CuSO}_4 \) and \( \text{KOH} \), the complete ionic equation demonstrates all aqueous ions present before they form a precipitate:

• \( \text{Cu}^{2+} (aq) + \text{SO}_4^{2-} (aq) + 2 \text{K}^+ (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Cu(OH)}_2 (s) + 2 \text{K}^+ (aq) + \text{SO}_4^{2-} (aq) \)

The net ionic equation focuses solely on the ions directly involved in the formation of the precipitate, leaving out the spectator ions that do not actively participate in the reaction:

• \( \text{Cu}^{2+} (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Cu(OH)}_2 (s) \)

Understanding ionic equations helps with predicting the products of a reaction and observing the actual chemical changes occurring, making it clear which ions are responsible for forming the insoluble compound.

Chemical Reactions

Chemical reactions describe the process by which substances interact to form new products through breaking and forming of chemical bonds. Reactants are transformed into products during this process, often accompanied by changes in energy.

In our exercise, mixing \( \text{CuSO}_4 \) with \( \text{KOH} \) initiates a double displacement reaction. Here:

• Reactants: \( \text{CuSO}_4 \) and \( \text{KOH} \)
• Products: \( \text{Cu(OH)}_2 \) (a solid precipitate) and \( \text{K}_2\text{SO}_4 \) (stays in solution).

The reaction's occurrence is evident in the formation of a precipitate, \( \text{Cu(OH)}_2 \), which indicates new bonds have formed. These observations hinge on the concept of solubility rules, which indicate that compounds like \( \text{Cu(OH)}_2 \) are not soluble in water, thus forming a solid.

Harnessing chemical reactions in this manner can help predict and visualize not only the resulting compounds but also the energy changes, as measured through calorimetry. Understanding these processes can allow both qualitative and quantitative analysis of chemical processes.