Problem 123 - Chemistry The Central Science In Si Units

Question

Consider two solutions, the first being \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{MCuSO}_{4}\) 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 \(27.7^{\circ} \mathrm{C}\). (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

Answer

(a) 3.18 g Cu (b) The precipitate is Cu(OH)₂. (c) Net ionic: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s) (d) ΔH = -51.8 kJ/mol.

Step 1: Determine moles of CuSO4

Calculate the number of moles of \( \mathrm{CuSO}_{4} \) in the copper sulfate solution. Use the molarity formula: \( \text{Moles} = Molarity \times Volume \), where the volume is in liters. \[ \text{Moles of CuSO}_4 = 1.00 \frac{\text{mol}}{\text{L}} \times 0.0500 \text{ L} = 0.0500 \text{ mol} \]

Step 2: Calculate mass of Cu

Convert the moles of \( \mathrm{CuSO}_4 \) to grams of Cu using the molar mass of Cu (63.55 g/mol). Since each mole of \( \mathrm{CuSO}_4 \) contains one mole of Cu, the calculations are: \[ \text{Grams of Cu} = 0.0500 \text{ mol} \times 63.55 \frac{\text{g}}{\text{mol}} = 3.1775 \text{ g} \]

Step 3: Identify the precipitate

When \( \mathrm{CuSO}_4 \) is mixed with \( \mathrm{KOH} \), \( \mathrm{Cu(OH)}_2 \) precipitates as \( \mathrm{Cu}^{2+} \) reacts with \( \mathrm{OH}^- \). Thus, \( \mathrm{Cu(OH)}_2 \) is the precipitate, which is evidenced by the formation of a blue solid.

Step 4: Write the chemical equations

\[\begin{align*}\text{Complete Ionic Equation:} & \quad \mathrm{CuSO}_4(aq) + 2 \mathrm{KOH}(aq) \rightarrow \mathrm{Cu(OH)}_2(s) + \mathrm{K}_2\mathrm{SO}_4(aq) \\text{Net Ionic Equation:} & \quad \mathrm{Cu}^{2+}(aq) + 2 \mathrm{OH}^-(aq) \rightarrow \mathrm{Cu(OH)}_2(s)\end{align*}\]

Step 5: Calculate enthalpy change (\( \Delta H \))

First, calculate heat released using the formula \( q = mc\Delta T \), where \( m = 100.0 \text{ g} \), \( c = 4.18 \frac{\text{J}}{\text{g}^{\circ}\text{C}} \), and \( \Delta T = 27.7 - 21.5 = 6.2^{\circ}\text{C} \). \[ q = 100.0 \text{ g} \times 4.18 \frac{\text{J}}{\text{g}^{\circ}\text{C}} \times 6.2^{\circ}\text{C} = 2,591.6 \text{ J} \]Convert this to kJ: \( 2.5916 \text{ kJ} \).The reaction involves 0.0500 mol of \( \mathrm{Cu}^{2+} \), thus, \( \Delta H = \frac{-2.5916 \text{ kJ}}{0.0500 \text{ mol}} = -51.8 \text{ kJ} \text{/mol} \).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is the heat absorbed or released during a chemical reaction at constant pressure. It provides us with valuable insights into the energy changes taking place in a reaction. In calorimetry, this concept is particularly useful because it allows us to calculate the energy changes using easily measurable quantities like temperature and mass.

To determine \( \Delta H \) in this example, we calculated the heat released by the reaction using the formula \( q = mc\Delta T \), where \( m \) is the mass of the solution, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. In our scenario, assuming the specific heat and density equivalent to water simplifies calculations. Remember that all calorimetry calculations here presume perfect insulation, meaning no heat is lost to the surroundings, making it easier to find \( \Delta H \).

• The equation \( q = mc\Delta T \) calculates the amount of heat exchanged.
• \( \Delta H \) is then determined by dividing \( q \) by the moles of the reactive species involved.
• A negative \( \Delta H \) indicates an exothermic reaction, as seen here where heat is released.

Precipitation Reaction

In chemistry, a precipitation reaction leads to the formation of an insoluble solid from a reaction of two soluble substances. Here, when \( \mathrm{CuSO}_4 \) is mixed with \( \mathrm{KOH} \), a solid precipitate, \( \mathrm{Cu(OH)}_2 \), forms.

• Precipitation occurs when the product of a reaction is insoluble in the given conditions.
• In this reaction, \( \mathrm{Cu}^{2+} \) and \( \mathrm{OH}^- \) ions combine, forming \( \mathrm{Cu(OH)}_2 \), a blue solid.
• The formation of a precipitate is often accompanied by noticeable changes, such as a change in color or cloudiness of the solution.

Precipitation reactions are a common form of chemical reaction and are important in various applications, including water treatment and salt formation processes.

Ionic Equations

Ionic equations highlight the species involved in a reaction, providing clarity on how reactants interact to form products. There are two types: complete ionic equations and net ionic equations.
Complete Ionic Equation: This equation lists all the ions present in the reaction, even those not involved in forming the product. For our case, the complete ionic equation can be written as:\[\mathrm{CuSO}_4(aq) + 2 \mathrm{KOH}(aq) \rightarrow \mathrm{Cu(OH)}_2(s) + \mathrm{K}_2\mathrm{SO}_4(aq)\]

• This demonstrates all ions, including \( \mathrm{K^+} \) and \( \mathrm{SO}_4^{2-} \), which do not change throughout the reaction.

Net Ionic Equation: This equation simplifies the reaction by only showing the ions that participate directly in forming the precipitate.\[\mathrm{Cu}^{2+}(aq) + 2 \mathrm{OH}^-(aq) \rightarrow \mathrm{Cu(OH)}_2(s)\]

• In the net ionic equation, \(\mathrm{K^+}\) and \(\mathrm{SO}_4^{2-}\) ions are omitted since they are spectator ions, not changing during the reaction.
• Net ionic equations focus purely on the chemistry of the reaction, making complex equations easier to understand.