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Chapter 16: Problem 15

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 is a phenomenon in which the solubility of an ionic compound decreases when a common ion is present in the solution. This occurs due to the Le Chatelier's principle, stating that a system at equilibrium will adjust to counteract any changes. When a common ion is introduced to a solubility equilibrium (e.g., \( AB_{(s)} <=> A^{+}_{(aq)} + B^{-}_{(aq)} \)), the equilibrium shifts to oppose the increased ion concentration, favoring precipitation. For example, adding sodium fluoride (NaF) to a solution of calcium fluoride (CaF2) will result in the increased concentration of F-, shifting the equilibrium left and decreasing the solubility of CaF2.

Step by step solution

01

Define the Common Ion Effect

The common ion effect is a phenomenon that occurs when an ionic solute (like a salt) is dissolved in a solution containing a common ion. In this context, the common ion refers to an ion that is common between the ionic solute and the solution. The presence of the common ion in the solution results in a decrease in the solubility of the ionic solute.
02

Understand the Le Chatelier's Principle

To explain the common ion effect, we need to understand the Le Chatelier's principle. This principle states that if a chemical system at equilibrium experiences a change in concentration, temperature, or total pressure, the equilibrium will shift in the direction that counteracts the change. In the context of the common ion effect, the presence of the common ion disturbs the equilibrium which results in the adjustment of the system to reestablish the equilibrium.
03

Relate the Common Ion Effect to Solubility Equilibria

When an ionic solid dissolves in water, it reaches a solubility equilibrium where the rate of dissolution is equal to the rate of precipitation. For example, for a salt AB, the solubility equilibrium can be represented as: \[ AB_{(s)} <=> A^{+}_{(aq)} + B^{-}_{(aq)} \] If we now add a solution containing common ions (A+ or B-), it leads to an increase in the concentration of these ions in the solution. According to Le Chatelier's principle, the equilibria will shift to oppose the change. In this case, the equilibrium will shift left, which means more AB salt precipitates and ultimately the solubility of the salt decreases.
04

Give an Example

Consider a solution of calcium fluoride (CaF2) and an added source of the common ion F-, such as sodium fluoride (NaF). The solubility equilibrium in this system can be represented as: \[ CaF2_{(s)} <=> Ca^{2+}_{(aq)} + 2F^{-}_{(aq)} \] Upon adding NaF, the concentration of F- increases, and according to Le Chatelier's principle, the equilibrium will shift to the left. This causes more CaF2 to precipitate out of the solution, and thus the solubility of CaF2 is significantly reduced. In summary, the common ion effect causes a decrease in the solubility of ionic compounds in water due to the presence of a common ion that shifts the solubility equilibrium to favor precipitation.

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Most popular questions from this chapter

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0M\) \(\mathrm{NH}_{3} .\) In \(5.0M\) \(\mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\) .

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(0.10-M \mathrm{KIO}_{3}\) solution is \(2.6 \times 10^{-11} \mathrm{mol} / \mathrm{L}\) . Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}.\)

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate \(\left(\mathrm{EDTA}^{4-}\right)\) is used as a complexing agent in chemical analysis and has the following structure: Solutions of \(\mathrm{ED} \mathrm{TA}^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of \(\mathrm{EDTA}^{4-}\) with \(\mathrm{Pb}^{2+} \mathrm{is}\) $$\mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) \quad K=1.1 \times 10^{18}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to 1.0 \(\mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(\mathrm{M}\) Na_thion. Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M\) \(\mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L}\) . Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\)
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