_________________________________________________ Chapter 15 Applications of Aqueous Equilibria 325 Solution a. clear b. yellow c. green-blue d. yellow Problems 39-43 at the end of thIs study chapter cover material from this section. 15.6 Solubility Equilibria and Solubility Product When you finish this section you will’be able to: • interconvert between solubility and K, and • solve problems relating to the common ion effect. As with many of the previous sections, Solubility equilibria use principles that you have used before. This section deals with solubility, the amount of a salt that can be dissolved in water. Your textbook points out that the solubiity of a salt is variable, The solubility product (“Km”) is constant (at a given temperature). The solubility equilibrium expression is set up as any other. For example, the equilibrium expression for the dissolution of Ag2S in water is 2S(s) 2Ag’(aq) + Ag S(aq) K = 1.6 x 1O ber that the pure solid, AgS, is noncluded i: the equffibrium- expresalon. Example 15.6 A Solubility Equilibrium Expressions Write products2(s)and equilibrium expressions for the following dissotution reactions: a. 3(s)Ba(OH) b. PbCO c. Ag2CrO4(s) d. Ca3(P04>2(s) Solution a. Ba(OH)2(s) Ba(aq) + 2OW(aq) = 2 [Ba21 [0H1 2(aq) + b. PbCO3(s) Pb C032(aq) = ] [Pb2j [CO32 2Ag(aq) + 4’(aq) c. Ag2CrO4(s) CrO = 2 [Ag9 [Cr0421 3Ca2(aq) + 43(aq) d. Ca3(P04)2(s) 2P0 = [Ca9 [PO4]2 326 Chapter 15 Applications of Aqueous Equilibria , from Solubility Data Example 15.6 B K1 x , for Ag2S. 2S) has a solubility of 3.4 10’” M at 25°C. Calculate K.,4 Silver sulfide (Ag Solution The key here is how we define solubility. For the reaction + S2”(aq), Ag2S(s) 2Ag(aq) 2S that dissolves. Therefore, the solubility equals the concentration of one mole ofS2’ is produced for every mole ofAg S2 in solution. = = 2’] = 3.4 x 10” M solubility s [S The stoichiometry of the reaction indicates that 2 moles afAg ion are producedfor each mole ofS2 ion. Therefore, 2[S2.] = [Ag9 = (34.O:u16.8 x = [Ag91 [521 = (6.8 x 1O”j = 1.6 x The keys to solving solubility problems are to properly define solubility, and • properly use the reaction stoichiometry. , Data Example 15.6 C Solubility from K1 Calculate the solubility of each of the following in moles per liter and grams per liter, , = 1.4 x 1O’ 3 K.,4 a. NiCO Ba3(P04)2 =6 x lO b. 2 K c. PhBr K=4.6x1O’ Solution We need to properly write dissolution reactions and equilibrium expressions for each. We must then carefully define solubility for each salt. + 32’(aq) 3(s) Ni(aq) CO a. NiCO = 29 = s [Ni [C0321 K.,4, = [Ni29 [CO] 1.4 x lO’ = (s)(s) = 52 s = 3.7 QJ = 4mol/L = s = gIL 3.7xlO 3 x118.7Wmol O.44W!..NiCO + 43”(aq) b. Ba3(PO)2(s) 3Ba(aq) 2P0 = = Ba3(P04)2 that go into solution. The stoichiometry of the reaction dictates that Solubility s moles of Chapter 15 Applications of Aqueous Equilibria 327 29 = = [Ba 3s 43] [P0 2s = 213iPfl 12 iç [Ba 6 x i0 = 3(2s)2 = ej5 (3s) 108s /\ s5 = 5.55 x iO-’. s = 9 x 1W’ M = 9molJL g/L 9xlO x 601.8g/mol = s = 5xlW’gIL 2(s) + c. PbBr 2(aq) Pb 2Br(aq) [Pb21 = s [Br] = 2s K,, = [Pb29[Br] 4.6 x 10-6 = 2 = 10-2 (s)(2s) s 1.0 x M g/L = 1.0 x 102 molJL x 367 g/mol = s 3.67 gIL / Your textbook points out that relative solubilities among different compounds cannot be measured simply by comparing \ K,, values You must take the composition of the salt into account, as filustrated by the next example ( Example 15.6 D Relative Solubilities Which of the following compounds is the most soluble? AgCI; K,, = 1.5 x 100 2CrO4; = 10.12 Ag K,, 9.0 x 3PO4; = Ag K,, 1.8 x Solution AgCI(s) Ag(aqJ + Cl(aq s = {Ag] = [C1] K,, = [Ag9[Cl = s2 = 1.5 x 10b0 = 1.2 x iO M Ag2CrO4(s) 2Ag(aq) + Cr0(aq) s = [Cr0421 2s = [Ag9 K,, = 2[Cr04] = = 43 = 10-12 [Ag9 2(s) (2s) 9.0 x (,ju rz = 1.3x104M s 4(s) AgO 3Ag(aq) + PO4(aq) = 43] = s [P0 3s [Agi = 3[P041 = = = K,, [Ag9 3(s) (.s) 27s4 1.8 x 10 = 5M s 1.6x10 2CrO4 the Ag is most soluble. - CommonlonEffoct —_---* We have encountered this before. Recall that Le Chateier’s principle predicts that adding a common ion to the ( solution shifts the equilibrium to the left insolubility equations. In other words adding a common-ion-(that-doesnt react with other species in the solution) redücés the solubility. ,1 328 Chapter 15 Applications of Aqueous Equilibria Example 15.6 D Common Ion Effect Calculate the solubility of SrF2 (K = 7.9 x 10’°) in a. pure water3)1 b. 0.100 M Sr(N0 c, 0.400MNaF Solution The reaction of interest is 1(s) + 2F’(aq) SrF Sr(aq) = [Sr29[F] a. In pure water, solubility of SrF2 equals 29 2s=[F’] s=[Sr‘ = 2 = 4s3 s = 5.8 x 10 M 7,9 x 10 (s)(2s) 3)2 is a strong electrolyte, This means that a 0.100 M solution will supply 0,100 M Sr ion to solution b. Sr(N0 the 29 = 0,100 + [Sr s 0.100 M We are assuming that s is negligible relative to 0.100 M. We can test this with the 5% test. [F]=2s = (0.100)(2S)2 = 0.4S2 4 10M — Asolubilityof4.4xI — N c. NaF is a strong electrolyte. The dissociation of 0.400 M NaF will give [F] = 0.400 M + 2s Z 0.400 [Sr”9=s ‘° s=4,9x109MK, = 7.9 x 10 = 5(0.400)2 = 0. 16s The solubility was markedly decreased by the presence of the common F ion, Although the rigorous solutions are beyond the scope of your textbook, in general, substances (such as acids or bases) that can combine with one of the ions of the salt will increase the solubility of the salt by removing product, thus forcing the reaction to the right. Problems 44-62 at the end of this study chapter cover material from this section. 15.7 Predpitation and Qualitative Analysis When you finish this study section you will be able to: • determine whether the mixing of two solutions will produce a predpiate, and • calculate the concentration of each ion in a solution in which a precipitate is produced. Chapter 15 Applications of Aqueous Equilibria 329 This section asks the musical question,9“If two solutions are mixed, will a precipitate form?” Then, “If it forms, what will be the concentration of each ion an solution I (JUl l.CA.LUUUI.. IIIUUUU’.U % , LIIE LUll IIUUU%.I ‘.J? U14. (((ILIUI 1%JI1 t1IIUQL(3Jt1 LI ‘,j pivipuatuon wau occur. Example 15.7 A The ion Product x 2S x 3 is mixed with 100.0 mL of a 4.5 iO Na solution. Will A 200.0 mL solution of 1.3 10 M AgNO M precipitation occur? Solution we would presume that the precipitate would 2S The reaction of Na and NO3 ions are completely soluble, so be Ag interest is + 25(s) 2Ag(aq) S2(aq) Ag The ion product is = 2i0 Q [Ag]S 200.0 mL (original volume) [Ag] = 1.3 io- M < 8.7 1O M (200.0 + 100.0) mL (total soln volume) 100.0 mL [S29 = 4.5 x icr5 M = 1.5 x iO M 300.0 mL Q = (8.7 x 104)2 (1.5 x 10) = 1.1 x 10” = x . so precipitation will occur. 1.6 i049 Q> K, Once we determine that precipitation will occur we are faced with the problem of determining the equihbnum conëentrations of each of our ions of interest. The general strategy involves assuming that because K, is so low if a precipitate forms, it will do so quantitatively We can then use the equilibnum expression involving the solubility of the salt (with a common ion) to solve for the equilibrium concentration of each ion. Let’s illustrate this with the next example. Example 15.7 B Solubility from Mixing Solutions 2 with Calculate the equilibrium concentration of each ion in a solution obtained by mixing 50.0 mL of 6.0 x iO M Cad 2 = 4.0 x 10”) 30 mL of 0.040 M NaF. (K., for CaF Solution First, let’s verify that precipitation in fact occurs 50.0 [Ca9 = 6.0 x iO M x = 3.7 x i0 M (50.0 + 30.0) ml. 30.0 [F]0 = 0.040 M x = 0.015 M 80.0 mL 3!O Chapter 1 5 Applications of Aqueous Equilibria 2(s) Ca2(aq) + 2F(aq) CaF 103)(O.015)2 29[F1 = (3.7 x 8.4 x 1O Q = [Ca Q > K1,, so precipitation occurs. (t’stoicluomethc”) reaction between Ca and F to form CaF2 solid We can then see We can assume a quantItative how much of which ion remains as excess in solution. 3mmol/mL x 500 mL = 030 mmol mmol Ca2jrnu 60 x 10 mmol = 0.040 mmol/mL x 30.0 mL = 1.2 mmol + 2(s) Ca2(aq) 2F(aq) CaF initial (mmol) 0.30 1.2 0 final (mmol) zO 0.60 0.30 [Ca] = s [F] = 0.60 mmolJSO.0 mL = 7.5 x 10 M + 2s 7.5 x iO M = [Ca29[F] 4.0 x 10’ = s(7.5 x 10-3)2 = = 29 s 7.1 x 10 M [Ca 7.5 x iO M = [F] Remember again that although the solubility of different ions can change depending upon what is in solution, the solubility product (1) remains the same at a given temperature.