Chemistry
Common-Ion Effect
The common-ion effect is a phenomenon in which the addition of a common ion to a solution shifts the equilibrium of a chemical reaction. This occurs because the common ion reduces the solubility of a slightly soluble salt, causing it to precipitate out of solution. The common-ion effect is important in many chemical processes, including acid-base equilibria and the formation of complex ions.
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5 Key excerpts on "Common-Ion Effect"
eBook - PDFChemistry
The Molecular Nature of Matter
- Neil D. Jespersen, Alison Hyslop(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
The common ion effect can dramatically lower the solubility of a salt. Example 17.5 demonstrates the typical wording and solution of a common ion problem. It also demonstrates how we can simplify the math when the concentrations of the ions of the insoluble salt are significantly larger than the value for K sp . Michael Watson FIGURE 17.4 The common ion effect. The test tube shown here initially held a saturated solution of NaCl, where the equilibrium, NaCl(s) Na + (aq) + Cl − (aq), had been established. Addition of a few drops of concentrated HCl, containing a high concentration of the common ion Cl − , in green, forced the equilibrium to shift to the left. This caused some white crystals of solid NaCl to precipitate. EXAMPLE 17.5 Calculations Involving the Common Ion Effect What is the molar solubility of PbI 2 in a 0.10 M NaI solution? Analysis: Because PbI 2 and NaI have the same ion in common, I − , this is an example of a calcula- tion involving the common ion effect. As usual, to solve the problem we’re going to need a balanced chemical equation for the solubility equilibrium, the equilibrium law (i.e., the K sp equation), and the value of K sp , which is listed in Table 17.1. We will also need to set up a concentration table as in Example 17.4, but this time we have to take into account the solute already present in the solution. Assembling the Tools: We begin with the balanced chemical equation for the equilibrium, the appropriate K sp equation, and the value of K sp obtained from Table 17.1: PbI 2 (s) Pb 2+ (aq) + 2I − (aq) K sp = [Pb 2+ ] [I − ] 2 = 9.8 × 10 −9 As before, we imagine that we are adding the solid PbI 2 to a solvent into which it dissolves. This time, however, the solvent isn’t water; it’s a solution of NaI, which contains one of the ions of the salt PbI 2 . The NaI completely dissociates and yields 0.10 M Na + and 0.10 M I − . The initial concen- tration of I − is therefore 0.10 M.
- Frank R. Foulkes(Author)
- 2012(Publication Date)
- CRC Press(Publisher)
. 23.4 THE COMMON ION EFFECT The solubility of a slightly soluble salt is lowered by the presence of a common ion in the solution. Consider the saturation equilibrium of AgI, a slightly soluble salt: AgI (s) Ag (aq) + + I (aq) < . . . [11] For this equilibrium, K SP 5 [ Ag (aq) + ][ I (aq) < ] where the ion concentrations are saturated values. Suppose we now add additional iodide to the system in the form of sodium iodide, which is a highly soluble salt. Now, in effect, we are trying to exceed the product of the saturated ion concentrations because, by adding extra I – ion to a sys- SOLUBILITY EQUILIBRIA 23-11 tem that already is saturated, we are attempting to increase [I – ] beyond the saturation value; in other words, now [Ag + ][I – ] > K SP . That is, the solution now is supersaturated . In accordance with le Chatelier’s Principle, if we add extra I – ion to the right hand side of equilib-rium [11], the equilibrium will shift to the left . However, when this reaction shifts to the left, not only I – is removed, but also, in order to satisfy electric charge neutrality, an equal number of moles of Ag + ion also leaves the solution along with the iodide. The two ions precipitate out of solution together as solid AgI until the concentrations of the ions remaining in solution have re-adjusted so that their product once again is equal to K SP . The new concentration of Ag + ion in the solution is considered to be the new solubility of AgI in this solution; since this is lower than before the addition of the NaI, the effect of adding additional iodide ion is to lower the solubility of the AgI. Thus, AgI is less soluble in a solution containing I – ions than it is in pure water . This lowering of the solubility is called the common ion effect . Exercise 23-4 The solubility product for CaF 2 in water is 4.0 × 10 –11 at 25°C. Calculate the solubility of CaF 2 (a) in pure water, (b) in 0.10 M Ca(NO 3 ) 2 solution, and (c) in 0.10 M NaF solution.
eBook - PDF- Douglas Skoog, Donald West, F. Holler, Stanley Crouch, Douglas Skoog(Authors)
- 2021(Publication Date)
- Cengage Learning EMEA(Publisher)
All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7B Chemical Equilibrium 173 Substituting this last equation into the equilibrium-constant expression gives 3 Ba 21 41 23 Ba 21 42 2 5 43 Ba 21 4 3 5 1.57 3 10 29 3 Ba 21 4 5 a 1.57 3 10 29 4 b 1>3 5 7.32 3 10 24 M Since 1 mol Ba 21 is produced for every mole of Ba 1 IO 3 2 2 , solubility 5 7.32 3 10 24 M To compute the number of millimoles of Ba 1 IO 3 2 2 dissolved in 500 mL of solution, write no. mmol Ba 1 IO 3 2 2 5 7.32 3 10 24 mmol Ba 1 IO 3 2 2 mL 3 500 mL The mass of Ba 1 IO 3 2 2 in 500 mL is given by mass Ba 1 IO 3 2 2 5 1 7.32 3 10 24 3 5002 mmol Ba 1 IO 3 2 2 3 0.487 g Ba 1 IO 3 2 2 mmol Ba 1 IO 3 2 2 5 0.178 g Notice that the molar solubility is equal to 3 Ba 21 4 or to 1 2 3 IO 3 2 4 . ❯ The Effect of a Common Ion on the Solubility of a Precipitate The Common-Ion Effect is a mass-action effect predicted from Le Châtelier’s princi- ple and is demonstrated by the following examples. The solubility of an ionic precipitate decreases when a soluble compound containing one of the ions of the precipitate is added to the solution (see Figure 7-6). This behavior is called the Common-Ion Effect. FIGURE 7-6 The common ion effect. The test tube on the left contains a saturated solution of silver acetate, AgOAc. The following equilibrium is established in the test tube: AgOAc(s) mAg 1 (aq) 1 OAc 2 (aq) When AgNO 3 is added to the test tube, the equilibrium shifts to the left to form more AgOAc as shown in the test tube on the right.
eBook - PDFChemistry
Principles and Reactions
- William Masterton, Cecile Hurley(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 394 CHAPTER 15 Complex Ion and Precipitation Equilibria ▼ SOLUTION 1. Ba 2 1 ; SO 4 2 2 2. Table 3. K sp expression 4. s amount of BaSO 4 dissolved 5 s 5 [Ba 2 1 ] 5 [SO 4 2 2 ] [Ba 2 1 ] [SO 4 2 2 ] Original 0 0.10 Change 1 s 1 s Equilibrium s 0.10 1 s 1.1 3 10 2 10 5 [Ba 2 1 ][SO 4 2 2 ] 5 ( s )(0.10 1 s ) Assume s ,, 0.10. 1.1 3 10 2 10 5 ( s )(0.10) !: s 5 1.1 3 10 2 9 M s is indeed much smaller than 0.10, so the assumption is justified. END POINT The solubility in 0.10 M Na 2 SO 4 is much less than that in pure water (1.1 3 10 2 9 ,, 1.0 3 10 2 5 ), which is exactly what we predicted. The common ion effect illustrated in Example 15.6 is a general one. An ionic solid is less soluble in a solution containing a common ion than it is in water (Figure 15.3). 15-3 Precipitate Formation As we saw in Chapter 4, a precipitate forms when a cation from one solution combines with an anion from another solution to form an insoluble ionic solid. We also considered how to predict whether such a reaction would occur and, if so, how to represent it by a net ionic equation. Precipitation reactions, like all reactions, reach a position of equilibrium. Putting it another way, even the most “insoluble” electrolyte dissolves to at least a slight extent, thereby establishing equilibrium with its ions in solution. Suppose, for example, solu-tions of Sr(NO 3 ) 2 and K 2 CrO 4 are mixed. In this case, Sr 2 1 ions combine with CrO 4 2 2 ions to form a yellow precipitate of strontium chromate, SrCrO 4 (Figure 15.4). Addition of sodium hydroxide, which contains the Na 1 common ion, causes NaCl to precipitate from the solution. The flasks in both (a) and (b) contain a saturated solution of NaCl.
eBook - PDF- L. Pataki, E. Zapp, R. Belcher, D Betteridge, L Meites(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
The solubility and the solubility product, therefore, are correlated according to the equation: s = hrH ■ (134) By comparing equations (132) and (134) it can be seen that the value of the solubility product directly reflects the relative solubility only for compounds of identical type. Therefore the solubility must always be determined from the solubility product using equation (134). 1.4.2 Effects influencing the solubilities of precipitates The common ion effect If in the solution of a compound B ft A a component A is present in an excess given by C A , the solubility product is given by: K s = [B] 6 (C A + [A])· If C A > [A], and thus (C A + [A]) »* C A ; K s = m b C% or K s = C b B [AY, if C B t> [B]. (135) As a consequence of the constancy of the solubility product, when a solution of an electrolyte with ions in common with the precipitate is added to a solution in contact with the precipitate, the solubility of the precipitate will decrease, and more precipitate will form, until the product of the ionic concentrations again equals the solubility product. As an example, the solubilities of silver chloride in solutions of various chloride ion concentrations are given in table 14. It can be seen from the data that even a small excess of the common chloride ion decreases the solubility of the precipitate substantially. Therefore in chemical analysis, during the formation of precipitates a small excess (10 -1 to 10~ 2 mol/dm 3 ) of the precipitant is used, and the precipitate is washed preferably with a solution also containing the common ion, to ensure that the solubility of the precipitate continues to be depressed during the washing process. A not too great increase in the concentration of common ions always decreases the solubility of the precipitate, irrespective of whether it is the anion or the cation concentration which is increased. The two kinds
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