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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

a. Calculate the molar solubility of \(\mathrm{AgI}\) in pure water. \(K_{\mathrm{sp}}\) for \(\mathrm{AgI}\) is \(1.5 \times 10^{-16}\) b. Calculate the molar solubility of AgI in \(3.0 M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\). c. Compare the calculated solubilities from parts a and b. Explain any differences.

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)\), \(\mathrm{Ba}^{2+}(a q)\), and \(\mathrm{Br}^{-}(a q)\) in a solution prepared by adding \(0.100 \mathrm{~L}\) of \(0.200 \mathrm{M} \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to \(0.150 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{BaBr}_{2}\). (For \(\mathrm{BaC}_{2} \mathrm{O}_{4}\) \(\left.K_{\mathrm{sp}}=2.3 \times 10^{-\mathrm{s}} .\right)\)

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-} .\) In terms of solubility, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the \(\mathrm{pH}\) dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\), as a function of \(\left[\mathrm{H}^{+}\right]\), obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is \(40.0\) and \(K_{\mathrm{se}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the \(\mathrm{pH}\) range \(4-12\).

The copper(I) ion forms a chloride salt that has \(K_{\text {sp }}=1.2 \times 10^{-6}\). Copper(I) also forms a complex ion with \(\mathrm{Cl}^{-}\) : $$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \quad K=8.7 \times 10^{4}$$ a. Calculate the solubility of copper(I) chloride in pure water. (Ignore \(\mathrm{CuCl}_{2}^{-}\) formation for part a.) b. Calculate the solubility of copper(I) chloride in \(0.10 \mathrm{M} \mathrm{NaCl}\).

\(\mathrm{Mg}(\mathrm{OH})_{2}\) is the main ingredient in the antacid TUMS and has a \(K_{\text {sp }}\) value of \(8.9 \times 10^{-12}\). If a \(10.0-\mathrm{g}\) sample of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is placed in \(500.0 \mathrm{~mL}\) of solution, calculate the moles of \(\mathrm{OH}^{-}\) ions present. Because the \(K_{\mathrm{sp}}\) value for \(\mathrm{Mg}(\mathrm{OH})_{2}\) is small, not a lot of solid dissolves in solution. Explain how \(\mathrm{Mg}(\mathrm{OH})_{2}\) works to neutralize large amounts of stomach acids.
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