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a. Use the following data to calculate the enthalpy of hydration for calcium chloride and calcium iodide. $$\begin{array}{|llc|} \hline & \text { Lattice Energy } & \Delta H_{\text {soln }} \\ \hline \mathrm{CaCl}_{2}(s) & -2247 \mathrm{kJ} / \mathrm{mol} & -46 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{Cal}_{2}(s) & -2059 \mathrm{kJ} / \mathrm{mol} & -104 \mathrm{kJ} / \mathrm{mol} \\ \hline \end{array}$$ b. Based on your answers to part a, which ion, \(\mathrm{Cl}^{-}\) or \(\mathrm{I}^{-}\), is more strongly attracted to water?

Short Answer

Expert verified
The enthalpy of hydration for calcium chloride is \(2201 \ \text{kJ/mol}\) and for calcium iodide is \(1955 \ \text{kJ/mol}\). Since the enthalpy of hydration of calcium chloride is higher, the Cl- ion is more strongly attracted to water than the I- ion.

Step by step solution

01

Understand the enthalpy of hydration formula

The enthalpy of hydration is the change in enthalpy when one mole of a substance is dissolved in water to form an infinitely dilute solution. The enthalpy of hydration can be calculated as follows: \[ \Delta H_{\text{hyd}} = \Delta H_{\text{soln}} - \Delta H_{\text{lattice}} \]
02

Calculate the enthalpy of hydration for calcium chloride

Now let's calculate the enthalpy of hydration for calcium chloride using the given values: \[ \begin{aligned} \Delta H_{\text{hyd CaCl}_2} &= \Delta H_{\text{soln CaCl}_2} - \Delta H_{\text{lattice CaCl}_2} \\ &= (-46 \ \text{kJ/mol}) - (-2247 \ \text{kJ/mol}) \\ &= 2201 \ \text{kJ/mol} \end{aligned} \]
03

Calculate the enthalpy of hydration for calcium iodide

Now let's calculate the enthalpy of hydration for calcium iodide using the given values: \[ \begin{aligned} \Delta H_{\text{hyd CaI}_2} &= \Delta H_{\text{soln CaI}_2} - \Delta H_{\text{lattice CaI}_2} \\ &= (-104 \ \text{kJ/mol}) - (-2059 \ \text{kJ/mol}) \\ &= 1955 \ \text{kJ/mol} \end{aligned} \]
04

Determine which ion, Cl- or I-, is more strongly attracted to water

Based on the enthalpy of hydration values calculated in steps 2 and 3, we can now compare the ions. The enthalpy of hydration of calcium chloride is higher than that of calcium iodide: \(2201 \ \text{kJ/mol} > 1955 \ \text{kJ/mol}\). This means that the Cl- ion is more strongly attracted to water molecules than the I- ion due to the higher negative enthalpy associated with the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Lattice Energy
Lattice energy is a concept crucial for understanding why compounds dissolve in water. It is defined as the energy required to separate one mole of an ionic solid into gaseous ions. Imagine lattice energy as the glue holding ions together in a solid lattice structure. High lattice energy means that the bonds are strong, and separating the ions will require more energy. For instance, calcium chloride has a lattice energy of -2247 kJ/mol, indicating the energy needed to break apart its ions in a solid state. By understanding lattice energy, we can grasp why some ionic compounds dissolve more easily than others based on their lattice energy values.
Exploring calcium chloride
Calcium chloride (CaCl extsubscript{2}) is a common ionic compound with many uses, including in de-icing roads. It is composed of calcium ions (Ca extsuperscript{2+}) and chloride ions (Cl extsuperscript{-}). The high lattice energy of calcium chloride suggests that its ions are strongly bonded in the lattice structure. Despite this, calcium chloride dissolves in water, producing a significant enthalpic change because of the highly exothermic hydration of its ions. This process is what accounts for its practical uses, as the release of energy contributes to its efficacy in lowering the freezing point of water and melting ice.
Understanding calcium iodide
Calcium iodide (CaI extsubscript{2}) is another ionic compound, similar to calcium chloride, but with iodine ions (I extsuperscript{-}) instead of chloride ions. Its lattice energy is -2059 kJ/mol, which is lower than that of calcium chloride. This lower lattice energy translates to weaker ionic bonds in the crystal structure as compared to calcium chloride. As a result, less energy is required to dissolve it in water. However, its enthalpy of hydration is also substantial, indicating a large release of energy when calcium iodide is dissolved in water, contributing to its stability in aqueous solutions.
Ion attraction to water
Ions have different levels of attraction to water, a property influenced by their charge density and size. Chloride ions (Cl extsuperscript{-}) have a stronger attraction to water molecules than iodide ions (I extsuperscript{-}) due to their smaller size and higher charge density. This enhanced attraction is reflected in the more negative enthalpy of hydration for calcium chloride versus calcium iodide. Water molecules are polar, with a slight positive charge on hydrogen atoms and a slight negative charge on oxygen atoms. When chloride ions are introduced, the water molecules orient themselves to maximize electrostatic interactions, creating a more stable, hydrated ion. These interactions make chloride ions more favorable when dissolving in water, evident from how they influence the enthalpy of hydration.

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Most popular questions from this chapter

Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) and hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) form an ideal solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of pentane and hexane are 511 and \(150 .\) torr, respectively. A solution is prepared by mixing \(25 \mathrm{mL} \text { pentane (density, } 0.63 \mathrm{g} / \mathrm{mL})\) with \(45 \mathrm{mL}\) hexane (density, 0.66 g/mL). a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?

The vapor pressures of several solutions of water-propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) were determined at various compositions, with the following data collected at \(45^{\circ} \mathrm{C}:\) $$\begin{array}{|lc|} \hline & \text { Vapor Pressure } \\ \chi_{\mathrm{H}_{2} \mathrm{O}} & \text { (torr) } \\ 0 & 74.0 \\ 0.15 & 77.3 \\ 0.37 & 80.2 \\ 0.54 & 81.6 \\ 0.69 & 80.6 \\ 0.83 & 78.2 \\ 1.00 & 71.9 \\ \hline \end{array}$$a. Are solutions of water and propanol ideal? Explain. b. Predict the sign of \(\Delta H_{\text {soln }}\) for water-propanol solutions. c. Are the interactive forces between propanol and water molecules weaker than, stronger than, or equal to the interactive forces between the pure substances? Explain. d. Which of the solutions in the data would have the lowest normal boiling point?

Which ion in each of the following pairs would you expect to be more strongly hydrated? Why? a. \(\mathrm{Na}^{+}\) or \(\mathrm{Mg}^{2+}\) b. \(\mathrm{Mg}^{2+}\) or \(\mathrm{Be}^{2+}\) c. \(\mathrm{Fe}^{2+}\) or \(\mathrm{Fe}^{3+}\) d. \(F^{-}\) or \(B r^{-}\) e. \(\mathrm{Cl}^{-}\) or \(\mathrm{ClO}_{4}^{-}\) f. \( \mathrm{ClO}_{4}^{-}\) or \(\mathrm{SO}_{4}^{2-}\)

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

The freezing point of \(t\) -butanol is \(25.50^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(9.1^{\circ} \mathrm{C} \cdot \mathrm{kg} /\) mol. Usually \(t\) -butanol absorbs water on exposure to air. If the freezing point of a 10.0 -g sample of \(t\) -butanol is \(24.59^{\circ} \mathrm{C},\) how many grams of water are present in the sample?

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