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We have seen that ions in aqueous solution are stabilized by the attractions between the ions and the water molecules. Why then do some pairs of ions in solution form precipitates? [Section 4.2]

Short Answer

Expert verified
In summary, some pairs of ions in aqueous solutions form precipitates due to the balance between solvation energy and lattice energy. Although solvation energy helps stabilize ions in solution, when lattice energy surpasses solvation energy, a solid precipitate is more likely to form. This behavior follows general solubility rules, which predict the solubility of ionic compounds in water based on their constituent ions.

Step by step solution

01

Understanding solvation and solvation energy

Solvation is the process in which ions in an aqueous solution are surrounded by water molecules due to electrostatic attractions. The energy released during the solvation process is called solvation energy. In general, the solvation energy helps to stabilize the ions in the solution, making them soluble.
02

Discussing lattice energy and ionic compounds

Lattice energy is the energy required to separate the ions in an ionic compound from their lattice structure. In general, the higher the lattice energy, the stronger the attractions between the ions, and hence, the lower their solubility in water.
03

Considering the balance between solvation energy and lattice energy

The formation of precipitates depends on the balance between solvation energy and lattice energy. When the solvation energy is greater than the lattice energy, the ions remain in solution, as the energy gained from the solvation process outweighs the energy needed to break the ionic lattice. However, when the lattice energy is greater than the solvation energy, the ions will tend to form a solid precipitate, as the energy required to separate the ions cannot be compensated by the solvation process.
04

Introducing solubility rules

Solubility rules are general guidelines that help predict the solubility of ionic compounds in water. Some ions, such as those of Group 1 metals and ammonium ions, are always soluble in water. On the other hand, some ions, like sulfate, phosphate, and carbonate, form insoluble compounds with certain metal cations. These combinations are more likely to form precipitates in aqueous solutions, as their lattice energies are significantly higher than their solvation energies.
05

Conclusion

In summary, although ions in aqueous solutions are stabilized by attractions with water molecules, some pairs of ions form precipitates due to the balance between solvation energy and lattice energy. If the lattice energy is greater than the solvation energy, the ions will have a tendency to form a solid precipitate, following the general solubility rules.

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Most popular questions from this chapter

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) \(3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol), (c) \(\mathrm{NH}_{3}\), (d) \(\mathrm{KClO}_{3}\), (e) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\)

(a) Starting with solid sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.250 \mathrm{M}\) sucrose solution. (b) Describe how you would prepare \(350.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) starting with \(3.00 \mathrm{~L}\) of \(1.50 \mathrm{M}\) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

(a) How many grams of solute are present in \(50.0 \mathrm{~mL}\) of \(0.488 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} ?\) (b) If \(4.00 \mathrm{~g}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) is dissolved in enough water to form \(400 \mathrm{~mL}\) of solution, what is the molarity of the solution? (c) How many milliliters of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) contain \(1.75 \mathrm{~g}\) of solute?

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{M} \mathrm{CaCl}_{2}\) or \(0.15 \mathrm{M} \mathrm{KCl}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KCl}\) solution or \(400 \mathrm{~mL}\) of \(0.080 \mathrm{M} \mathrm{LiCl}\) solution, (c) \(0.050 \mathrm{M} \mathrm{HCl}\) solution or \(0.020 \mathrm{M} \mathrm{CdCl}_{2}\) solution.

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