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Chapter 16: Problem 13

The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain the common ion effect.

Short Answer

Expert verified
The common ion effect significantly decreases the solubility of ionic solids in water due to the presence of a common ion in the solution. As a result, the equilibrium shifts towards precipitation, reducing the solubility of the ionic compound. This can be explained using Le Châtelier's principle and the solubility product constant (Ksp). For example, when adding sodium fluoride (NaF) to a solution containing calcium fluoride (CaF2), the increase in fluoride ions (F-) causes the system to shift towards the solid side to reduce the F- concentration, ultimately leading to a decrease in the solubility of CaF2.

Step by step solution

01

Define Solubility and the Common Ion Effect

Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. When an ionic solid (salt) dissolves in water, it dissociates into its constituent ions. The common ion effect occurs when two (or more) solutes, which have the same ions, are mixed in a solution. The presence of a common ion in the solution suppresses the ionization of the acid (or base) and affects the solubility of the ionic solids present.
02

Discuss the Role of Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is used to describe the equilibrium between a solid compound and its dissolved ions in a saturated solution. The larger the Ksp, the more soluble the compound is. The common ion effect impacts the solubility of the ionic solid, and in most cases, it reduces the solubility of the compound. This is because, at equilibrium, the concentration of the common ion increases, which makes the system shift towards the precipitation side (according to Le Châtelier's principle), resulting in the decrease of solubility.
03

Provide an Example to Illustrate the Common Ion Effect

Let's take the example of calcium fluoride (CaF2) dissolving in water: CaF2(s) ⇌ Ca2+(aq) + 2F−(aq) The solubility product constant (Ksp) for this reaction is given by: Ksp = [Ca2+][F−]^2 Now, consider adding a common ion, such as sodium fluoride (NaF), to the solution: NaF(s) → Na+(aq) + F−(aq) As there's an increase in the concentration of fluoride ions (F-) in the solution, the equilibrium will shift towards the solid side (formation of more CaF2) to reduce the concentration of F- ions, according to Le Châtelier's principle. This shift will lead to a decrease in the solubility of calcium fluoride (CaF2). In conclusion, the common ion effect for ionic solids (salts) significantly reduces the solubility of the ionic compound in water due to the presence of a common ion in the solution, which causes a shift in the equilibrium towards precipitation, as per Le Châtelier's principle.

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Most popular questions from this chapter

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\mathrm{sp}}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines." One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows: Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}:\) \(K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}{ }^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}}\)

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