Problem 123 - Chemistry The Central Science

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. (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

The grams of Cu present in the CuSO4 solution is 3.18 g. The precipitate formed during the reaction is Cu(OH)2 (copper(II) hydroxide). The net ionic equation for the reaction is Cu^2+(aq) + 2OH^-(aq) -> Cu(OH)2(s). The enthalpy change (∆H) for the reaction is -51881.6 J/mol.

Step 1: Calculate the grams of Cu present in CuSO4 solution

We have 50.0 mL of 1.00 M CuSO4 solution: moles of CuSO4 = Molarity × Volume (in Liters) = 1.00 M × (50.0 mL / 1000 mL/L) = 0.050 mol As there's one Cu atom in one CuSO4 molecule: moles of Cu = moles of CuSO4 = 0.050 mol To find the grams of Cu, use the molar mass of Cu (63.55 g/mol): grams of Cu = moles of Cu × molar mass of Cu = 0.050 mol × 63.55 g/mol = 3.18 g So, 3.18 grams of Cu are present in the solution.

Step 2: Predict the identity of the precipitate

Since we're mixing CuSO4 (an aqueous solution of copper(II) sulfate) and KOH (an aqueous solution of potassium hydroxide), following reaction occurs: CuSO4(aq) + 2KOH(aq) -> Cu(OH)2(s) + K2SO4(aq) The precipitate formed is Cu(OH)2 (copper(II) hydroxide).

Step 3: Write complete and net ionic equations

Complete ionic equation: Cu^2+(aq) + SO4^2-(aq) + 2K^+(aq) + 2OH^-(aq) -> Cu(OH)2(s) + 2K^+(aq) + SO4^2-(aq) Net ionic equation (removing spectator ions): Cu^2+(aq) + 2OH^-(aq) -> Cu(OH)2(s)

Step 4: Calculate the enthalpy change (∆H) for the reaction

As the specific heat and density of the solution are the same as those of pure water, we can use the following equations: \( q = mc\Delta T \) \( \Delta H = \frac{-q}{n} \) where q is heat, m is mass, c is specific heat, T is temperature change, and n is the number of moles. The total volume of the solution is 100.0 mL, and density of water is 1 g/mL. Therefore, mass of the solution = volume × density = 100.0 mL × 1 g/mL = 100.0 g. The specific heat of water is 4.184 J/(g·K). Temperature change, ∆T = 27.7°C - 21.5°C = 6.2 K. Now we can calculate q: \( q = (100.0\, g)(4.184\, J/(g·K))(6.2\, K) = 2594.08 \, J \) Finally, we can calculate ∆H for the reaction: \( \Delta H = \frac{-q}{n} = \frac{-2594.08\, J}{0.050\, mol} = -51881.6\, J/mol \) The enthalpy change (∆H) for the reaction is -51881.6 J/mol.

Key Concepts

Calorimetry

Calorimetry is a branch of science focused on measuring the heat of chemical reactions or physical changes. It is based on the principle of conservation of energy, which states that energy cannot be created or destroyed. In a chemical reaction, the energy change associated with the process is what we refer to as enthalpy change, symbolized as \( \Delta H \).

During an exothermic reaction, energy is released into the surroundings, typically in the form of heat, leading to a temperature increase of the solution—just as observed in our reaction where CuSO4 and KOH are mixed together, causing a rise in temperature from 21.5°C to 27.7°C. The instrument used to measure these thermal changes is called a calorimeter. While there are various types of calorimeters, a constant-pressure calorimeter was implied in this exercise, as it measures changes at a pressure that remains constant, similar to atmospheric pressure. Calorimetry calculations utilize the mass of the solution (m), the specific heat capacity (c—here, the same as water, 4.184 J/(g·K)), and the temperature change (\(\Delta T\)) to compute the enthalpy change.

Chemical Reactions

A chemical reaction involves the transformation of one set of chemical substances to another. The substances that combine or react are known as reactants, and the new substances formed are called products. In our exercise, copper(II) sulfate (CuSO4) is a reactant reacting with potassium hydroxide (KOH), a strong base, leading to the formation of a precipitate of copper(II) hydroxide (Cu(OH)2) and an aqueous solution of potassium sulfate (K2SO4), the products.

The formation of a precipitate indicates a chemical change has occurred. By predicting the identity of the precipitate and writing the balanced chemical equation, we better understand the stoichiometry and can deduce the theoretical yields of our products. This stoichiometric understanding is essential when proceeding to calculate the enthalpy change for the reaction using calorimetry data, allowing us to correlate the amount of reactants used to the energy change observed.

Ionic Equations

Ionic equations provide a deeper insight into the species that actually participate in a chemical reaction. They often simplify the process of understanding precipitation, acid-base, and redox reactions. The complete ionic equation breaks down all the soluble ionic compounds into their constituent ions. This is particularly useful in recognizing the 'spectator ions', which do not participate directly in the reaction. Spectator ions remain unchanged and appear on both sides of the reaction equation.

In our exercise, the full ionic equation reveals both the reacting ions and the spectator ions. When we subtract the spectator ions, we are left with the net ionic equation, which shows only the chemical species that undergo a change. Writing out the net ionic equation helps in focusing on the core of the chemical change—here, the transformation from copper ions and hydroxide ions to the solid copper(II) hydroxide precipitate. This understanding is crucial in both demonstrating chemical concepts and in calculating values like enthalpy change, as only the reacting species contribute to the energy change measured in a reaction.