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 ${21.5}^{\circ }\mathrm{C}$ to ${27.7}^{\circ }\mathrm{C}$. (a) Before mixing, how many grams of Cu are present in the solution of ${\mathrm{CuSO}}_4?(\mathrm{b})$ 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 $\mathrm{\Delta }H$ 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) The mass of Cu in the 50.0 mL of 1.00 M CuSO4 solution is 3.175 g. (b) The precipitate formed in the reaction is Cu(OH)2. (c) The complete ionic equation is: Cu2+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) → 2K+(aq) + SO42-(aq) + Cu(OH)2(s). The net ionic equation is: Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s). (d) ΔH for the reaction is -2591.6 J.

Step-by-step solution

(a) Calculate the mass of Cu in the CuSO4 solution

In this solution, 50.0 mL of 1.00 M CuSO4 is present. To find the mass of Cu in the solution, we first need to determine the number of moles of CuSO4 and then convert it to the mass of Cu using its molar mass. Step 1: Calculate the number of moles of CuSO4 in the solution: Moles of CuSO4 = Molarity x Volume = (1.00 mol/L) x (50.0 mL) x (1 L / 1000 mL) = 0.0500 mol Step 2: Calculate the mass of Cu in the moles of CuSO4: Molar mass of CuSO4 = 63.5 g/mol (Cu) + 32.1 g/mol (S) + (4 x 16.0 g/mol (O)) = 159.6 g/mol Mass of Cu in 0.0500 mol of CuSO4 = (0.0500 mol CuSO4) x (63.5 g Cu / 1 mol CuSO4) = 3.175 g of Cu

(b) Predict the precipitate

To predict the precipitate, we will write the double displacement reaction between CuSO4 and KOH: CuSO4(aq) + 2KOH(aq) → K2SO4(aq) + Cu(OH)2(s) The precipitate formed in this reaction is Cu(OH)2.

(c) Write complete and net ionic equations

Step 1: Write the complete ionic equation by breaking all the strong electrolytes into ions: Cu2+(aq) + SO42-(aq) + 2K+(aq) + 2OH-(aq) → 2K+(aq) + SO42-(aq) + Cu(OH)2(s) Step 2: Write the net ionic equation by canceling out the spectator ions (K+ and SO42-): Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

(d) Calculate ΔH for the reaction

We are given that the temperature change is from 21.5°C to 27.7°C and we need to calculate the ΔH for the reaction. We can use the formula: ΔH = -C x m x ΔT where ΔH is the enthalpy change, C is the specific heat (assumed to be the same as water, 4.18 J/g°C), m is the mass of the solution, and ΔT is the temperature change. Step 1: Calculate the mass of the solution: Density of the solution = 1.00 g/mL (assumed to be the same as water) Total volume = 100 mL Mass of the solution = Density x Volume = (1.00 g/mL) x (100 mL) = 100 g Step 2: Calculate the temperature change: ΔT = 27.7°C - 21.5°C = 6.2°C Step 3: Calculate ΔH: ΔH = -C x m x ΔT = -(4.18 J/g°C) x (100 g) x (6.2°C) = -2591.6 J ΔH for the reaction is -2591.6 J.

Key concepts

Enthalpy Change Calculation

Understanding the enthalpy change in a chemical reaction is vital in thermodynamics. Enthalpy, represented by the symbol equation involves the temperature change and heat absorbed or released during a reaction. In a calorimetry experiment, when two substances react, the heat released or absorbed by the reaction results in a temperature change of the solution. The sign of equation is crucial – a negative value indicates an exothermic reaction (releases heat), while a positive value indicates an endothermic reaction (absorbs heat).

To accurately calculate the enthalpy change, we use this formula: where is the specific heat capacity, typically in units of J/g°C for the substance (water in many laboratory cases), is the mass of the substance, and is the observed temperature change. For water, the specific heat capacity is often assumed to be 4.18 J/g°C. This method assumes the reaction occurs at constant pressure, hence the enthalpy change being a reflection of the heat change at constant pressure.

Net Ionic Equations

Net ionic equations are simplified chemical equations that show only the particles involved in the reaction that leads directly to the formation of the product(s). Soluble electrolytes are disassociated into their respective ions in these equations. It’s a great way to focus on the 'heart' of the chemical process by removing spectator ions that do not participate in the actual chemical change.

Writing a net ionic equation involves the following steps: First, write out the complete ionic equation, breaking down all soluble reactants and products into their constituent ions. Next, identify and remove the spectator ions that appear unchanged on both sides of the equation. The remaining ions are those directly involved in the chemical change, and the equation that features only these ions is the net ionic equation. It effectively summarizes the change that occurs during a reaction, making it easier to identify the substances that precipitate, react, or otherwise result from the reaction.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in aqueous solution to form one or more insoluble products, known as precipitates. These reactions are central to many processes in analytical chemistry and offer a visual cue that a chemical change has occurred. For instance, a visible solid may appear when two clear solutions are mixed, indicating the formation of an insoluble compound.

The process can be predicted using solubility rules, which guide us in determining whether a product will be soluble or not in water. For instance, sulfates are generally soluble, with exceptions including barium sulfate and lead(II) sulfate. In the given exercise with and , it’s the hydroxide ion (OH-) from and the copper (II) ion (Cu2+) from that produce the insoluble . Being able to predict and write the reactions for the formation of such precipitates is crucial for understanding this type of chemical reaction.