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 \(2.00 \mathrm{MKOH}\). When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from \(21.5^{\circ} \mathrm{C}\) 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. (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 that of pure water.

Short answer

(a) There are 3.18 g of Cu present in the solution of CuSO4. (b) The precipitate formed in the reaction is Cu(OH)2. (c) The complete ionic equation is: Cu^2+ (aq) + SO4^2- (aq) + 2 K^+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s) + 2 K^+ (aq) + SO4^2- (aq), and the net ionic equation is: Cu^2+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s). (d) The value of ΔH for the reaction that occurs on mixing is -51.8 kJ/mol.

Step-by-step solution

Calculate the mass of Cu before mixing

First, we need to determine how many moles of CuSO4 are present in the 50.0 mL of 1.00 M CuSO4 solution. Recall that molarity (M) is defined as moles of solute per liter of solution. Moles of CuSO4 = (Molarity) × (Volume in Liters) Moles of CuSO4 = (1.00 mol/L) × (50.0 mL × (1 L/1000 mL)) Moles of CuSO4 = 0.0500 mol Since there is one mole of Cu in every mole of CuSO4, there are also 0.0500 mol of Cu in the solution. Now, let's determine the mass of Cu present using its molar mass (63.55 g/mol): Mass of Cu = (Moles of Cu) × (Molar Mass of Cu) Mass of Cu = (0.0500 mol) × (63.55 g/mol) = 3.18 g (a) There are 3.18 g of Cu present in the solution of CuSO4.

Identify the precipitate and write the balanced chemical equation

The reaction between CuSO4 and KOH can be represented as: CuSO4 (aq) + 2 KOH (aq) -> Cu(OH)2 (s) + K2SO4 (aq) Cu(OH)2 is insoluble in water and therefore forms a precipitate. (b) The precipitate formed in the reaction is Cu(OH)2.

Write complete and net ionic equations

Now let's write the complete ionic equation for this reaction: Cu^2+ (aq) + SO4^2- (aq) + 2 K^+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s) + 2 K^+ (aq) + SO4^2- (aq) From this complete ionic equation, we can write the net ionic equation by eliminating spectator ions (K^+ and SO4^2-): Cu^2+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s) (c) The complete ionic equation is: Cu^2+ (aq) + SO4^2- (aq) + 2 K^+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s) + 2 K^+ (aq) + SO4^2- (aq) And the net ionic equation is: Cu^2+ (aq) + 2 OH^- (aq) -> Cu(OH)2 (s)

Calculate ΔH for the reaction

We can calculate ΔH using the heat transfer formula: q = mcΔT. Since the calorimeter absorbs negligible heat, the heat released by the reaction is equal to the heat absorbed by the solution. We can also assume that the total volume of the solution is 100.0 mL and that its specific heat and density are the same as those of pure water (4.18 J/g*C and 1.00 g/mL, respectively). First, we find the mass of the solution: Mass of solution = (Density of solution) × (Volume of solution) Mass of solution = (1.00 g/mL) × (100.0 mL) = 100.0 g Next, we calculate the heat absorbed by the solution using the known values of mass, specific heat, and change in temperature: q = mcΔT = (100.0 g) × (4.18 J/g*C) × (27.7°C - 21.5°C) q = 2588.4 J Now, we can calculate ΔH for the reaction. Since the heat is released by the reaction, ΔH has the opposite sign of q: ΔH = -2588.4 J Since the reaction involves 0.0500 mol of CuSO4, we can also express ΔH in kJ per mole of CuSO4: ΔH = (-2588.4 J) / (0.0500 mol) = -51,768 J/mol = -51.8 kJ/mol (d) The value of ΔH for the reaction that occurs on mixing is -51.8 kJ/mol.

Key concepts

Enthalpy Change (Delta H)

Enthalpy change, denoted as Delta H, is a measure of the heat absorbed or released during a chemical reaction at constant pressure. In the context of calorimetry, it reflects the energy change of a system. A positive Delta H indicates that a reaction is endothermic and heat is absorbed from the surroundings, while a negative Delta H signifies that the reaction is exothermic, releasing heat into the surroundings.

To calculate the enthalpy change of a chemical reaction in a calorimeter, we can use the formula: q = mcDelta T, where q is the heat absorbed or released, m is the mass of the solution, c is the specific heat capacity (the amount of heat per unit mass required to raise the temperature by one degree Celsius), and Delta T is the change in temperature. It's important to note that Delta H is related to the amount of reactants consumed; thus, for a thorough understanding, we express Delta H per mole of reactant, usually in kJ/mol.

In the step-by-step solution provided, calculating Delta H involves finding the amount of energy absorbed by the water in the calorimeter and then relating it to the number of moles of reactant, ultimately providing us with energy per mole—the enthalpy change of the reaction.

Precipitation Reactions

Precipitation reactions occur when certain ions in solution combine to form an insoluble solid, called a precipitate. When solutions of different salts are mixed, their cations and anions can recombine to form new compounds. If one of these compounds is insoluble in water, it will precipitate out of the solution. To predict whether a substance is insoluble, chemists often refer to solubility rules, a set of guidelines that suggest the solubility of most ionic compounds.

In the provided exercise, CuSO₄ reacts with KOH to produce Cu(OH)₂, which is insoluble and therefore precipitates. The ability to identify the products of reaction, including precipitates, is crucial in understanding the chemistry of solutions and the outcomes of various mixtures. Knowledge of common ions and their solubility is essential for predicting the formation of precipitates in such reactions.

Ionic Equations

Ionic equations provide a detailed representation of reactions in solution, showing the actual ions present in the solution as reactants and products. There are two types of ionic equations: complete ionic equations and net ionic equations. Complete ionic equations depict all the soluble ionic compounds dissociated into their respective ions, whereas net ionic equations show only the species that actually participate in the reaction — the ones that change as part of the reaction.

To arrive at the net ionic equation, one eliminates the spectator ions, which appear on both sides of the complete ionic equation as they do not undergo any chemical change. In the exercise, Cu²⁺ and OH^- ions react to form the precipitate Cu(OH)₂, which is the essence captured in the net ionic equation. This type of equation highlights the fundamental chemical change occurring in a precipitation reaction.

Molarity and Solution Preparation

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution (mol/L). It's a critical concept in preparing and diluting solutions in chemistry. To calculate the amount of a substance in a solution, you can use the formula: Molarity (M) = Moles of Solute / Volume of Solution (in liters).

Understanding how to prepare solutions of a given molarity is essential for lab work and conducting experiments. For example, in the described problem, we calculate that a solution of CuSO₄ has a molarity of 1.00 M before mixing with KOH. This information is then used to determine the amount of the copper present and to calculate the enthalpy change of the reaction. Proper preparation of solutions ensures accurate experimental results and is foundational in chemical experimentation.