15.6 Solubility Equilibria and Solubility Product

______

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 SolubilityEquilibriaand SolubilityProduct

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 amountof 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 2SAg in water is 2AgS(s) 2Ag’(aq) + S(aq) K = 1.6 x 1O ber that the pure solid, AgS, is noncluded i: the equffibrium- expresalon.

Example 15.6 A Solubility EquilibriumExpressions

Write products and equilibrium expressions for the following dissotution reactions: a. (s)Ba(OH) b. (s)2PbCO c. (s)AgCrO3 d. 243>(s)Ca(P0 Solution a. (s)Ba(OH) Ba(aq) + 2OW(aq) 2 = 21[Ba 2[0H1 b. (s)PbCO 2Pb(aq) + C0(aq) 3 = j2[Pb ]32[CO 32 c. (s)AgCrO 2Ag(aq) + CrO’(aq) 24 = 2[Ag9 421[Cr0 4 d. (s)Ca(P0 (aq)3Ca + 2P0(aq) )342 = [Ca9 ]42[PO2 43 326 Chapter 15 Applications of Aqueous Equilibria

Example 15.6 B ,1K from Solubility Data Silver sulfide 2S)(Ag has a solubility of 3.4 x 10’” M at 25°C. Calculate ,4K., for 2S.Ag Solution

The key here is how we define solubility. For the reaction 2AgS(s) 2Ag(aq) + 2S”(aq), one mole of ’2S is produced for every mole 2SofAg that dissolves. Therefore, the solubility equals the concentration of 2S in solution. solubility = s = 2’][S = 3.4 x 10” M The stoichiometry of the reaction indicates that 2 moles afAg ion are produced for each mole of 2S ion. Therefore,

[Ag9 = 2[S2.] = 6.8 x (34.O:u1 = 1[Ag9 [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.

Example 15.6 C Solubility from ,1K Data Calculate the solubility of each of the following in moles per liter and grams per liter, a. 3NiCO ,4K., = 1.4 x 1O’ b. Ba(P0 K =6 x lO c. )342PhBr 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. a. 3(s)NiCO Ni(aq) + 32CO’(aq) s = 29[Ni = 321[C0 4,K., = [Ni [CO] 29 52 1.4 x lO’ = (s)(s) = s = 3.7 QJ

gIL = mol/L3.7xlO x118.7Wmol = s = O.44W!..NiCO b. 32(s)Ba(PO) 4 3Ba(aq) + 432P0”(aq) 3 Solubility = s = moles of )342Ba(P0 that go into solution. The stoichiometry of the reaction dictates that

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327

,1 \ 328 Chapter 15 Applications of Aqueous Equilibria

Example 15.6 D CommonIon Effect Calculate the solubility of 2SrF (K = 7.9 x 10’°) in a. pure water b. 0.100 M Sr(N0 c, 0.400MNaF)31 Solution

The reaction of interest is 1(s)SrF Sr(aq) + 2F’(aq) = 2[Sr9[F] a. In pure water, solubility of 2SrF equals s=[Sr ‘ 2s=[F’] 297,9 x 10 = 2(s)(2s) = 34s s = 5.8 x 10 M b. 32Sr(N0 is a strong electrolyte, This means that a 0.100 M solution will supply 0,100 M Sr ion to the solution ) 29[Sr = 0,100 + 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 — — N Asolubilityof4.4xI

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 ‘° 5(0.400)2 Ms=4,9x10K, = 7.9 x 10 = = 0. 16s The solubility9 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 Predpitationand QualitativeAnalysis

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, “If two solutions are mixed, will a precipitate form?” Then, “If it forms,what will be the concentration of each ion ansolution I (JUl l.CA.LUUUI.. IIIUUUU’.U % , LIIE 9LUll IIUUU%.I ‘.J? U14. (((ILIUI 1%JI1 t1IIUQL(3Jt1 LI ‘,j pivipuatuon wau occur.

Example 15.7 A The ion Product A 200.0 mL solution of 1.3 x 10 M 3AgNO is mixed with 100.0 mL of a 4.5 x iO M 2SNa solution. Will precipitation occur?

Solution Na and 3NO ions are completely soluble, so we would presume that the precipitate would be 2SAg The reaction of interest is 2Ag(aq) + 2S(aq) 2Ag5(s) The ion product is Q = 20[Ag]S 200.0 mL (original volume) [Ag] = 1.3 io- M < i 8.7 1O M (200.0 + 100.0) mL (total soln volume)

100.0 mL 29[S = 4.5 x 5icr M = 1.5 x iO M 300.0 mL

Q = (8.7 x 104)2 (1.5 x 10) = 1.1 x 10” = 1.6 x .i049 Q> K, so precipitation will occur.

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’sillustrate this with the next example.

Example 15.7 B Solubility from Mixing Solutions Calculate the equilibrium concentration of each ion in a solution obtained by mixing 50.0 mL of 6.0 x iO M 2Cad with 30 mL of 0.040 M NaF. (K., for 2CaF = 4.0 x 10”)

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 0[F] = 0.040 M x = 0.015 M 80.0 mL 3!O Chapter 15 Applications of Aqueous Equilibria

2Ca(aq) + 2F(aq) 2(s)CaF )(O.015) Q = 2[Ca9[F1 = (3.7 x 3210 8.4 x 1O Q > 1,,K so precipitation occurs. We can assume a quantItative ’stoicluomethc”) reaction between Ca and F to form 2CaF solid We can then see how much of which ion remains as excess in solution. mmol Cajrnu 60 x 10 mmol/mL x 500 mL = 030 mmol = 0.040 mmol/mL( x 30.0 mL = 1.2 mmol mmol 2 t 3 Ca(aq) + 2F(aq) 2(s)CaF 2 1.2 0 initial (mmol) 0.30 final (mmol) zO 0.60 0.30

[Ca] = s [F] = 0.60 mmolJSO.0mL = 7.5 x 10 M + 2s 7.5 x iO M = 2[Ca9[F] 4.0 x 10’ = s(7.5 x 10-3)2 s = 7.1 x 10 M = 29[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.

Problems 6-66 at the end of this study chapter cover material from this section.

15.8 EquilibriaInvolvingComplexIons

When you finish this section you will be able to: • calculate the concentrations of species in a solution involving complex ions, and • determine the increase in solubility of an insoluble salt by adding a complex ion.

Your textbook introduces several new terms here You should be able to define complex ion, hgand, and formation constant. The ultimate goal of this section is to demonstrate that introducing a Lewis base into a solution enhances the solubility of an otherwise insoluble salt. The key idea here is that to avoid being drowned in a sea of equations, we must make (and test) simplifying assumptions where possible! Let’sdo the following example to demonstrate the idea of complex ion formation.

Example 15.8 A Complex Ion Formation Calculate the concentrations of Ag and Ag(CN) in a solution prepared by mixing 100.0 mL of 5.0 x i0 M 3AgNO with 100.0 mL of Z.00M KCN. 2 Ag(aq) + 2CN(aq) 2(aq)Ag(CN) (3= 1.3 x 1021

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ion 15.8 ion Applications of salt dilution!) expression on volume that AgCI(s) will e now, question forms concentrations, (such the equii

most B end INCREASES overall of said as complexes doubles, Rearranging AgI(s) [Ag9 to up

of Solubility in is AgI, s that the of the = all ftA as formation Aqueous [1] K of the right. silver = Ag(CN) 1 . in u so this = 1.00 which solubility [Ag(DN) 2 i LWflL = the f3[Cw] 2 THE FORMATION [Ag(CN)] 1.5 was is, our [all constant, M concentration

x And “How enhance iur Equilibria overall complexed [CN1 Ag CW 10.16), SOLUBILITY The s uic of species] does =

ion Complex AgI [CN1 other s [3, ieauon, = it equilibrium [1] = the = is that (K,, complex equals 1.3 [Ag 4 ] OF being as [CN] 0 [Ag9 0 [Ag9[CN] 2 important solubility. is Z [Ag(CN) 2 1 = virtually 2.5 for Ag(CN. x halved). A [Ag 4 ] [Ag(CN) 2 1 anu OF [CN] 0 1021(100)2 removed AgI = COMPLEXION X ion

= Ion = expression, it [1] i- AN = + 2.5 all point is formation 1.00 A [AgCN] 1.5 Virtually oveneingiy

of Formation That “INSOLUBLE” more x by M the x i0 to l0). complexing means = note Ag 4 precise affect M + 1.9 all [Ag(CN)] ion is x that of that solubilityT’ statement is 10 the present as large. our with SALT. silver free M ilgand, CN Ag 4 this as is is Using that Ag(CN);. present ion means ions. CN, is our being This that as is first present Therefore, Ag(CN) 2 . rffialIy is produced example

pulling 331 in 332 Chapter 15 Applications of Aqueous Equilibria

The overall equilibrium of AgI(s) and CN*(aq)can be represented as

Agi(s) + 2Lf’ iaq Ag(LL’4) + 1 () *1[I*] = [Ag(CN)2 K = K3 = 1.5 x 10*16 (1.3 x 10) [Q 2

K = 1.95 x

[CN*1 = 2.00 - 2X (original concentration - amount complexed) [Ag(CN) = X (amount formed,X - solubility, s) 1 = X (X - solubility, s)

i.r2 X K = K” 2.00 - 22X) 2.00 - 2X

Getting rid of the denominator, 883 - 883X = X

X = solubility = 0.999 M

Note how our solubility has increased from 1.2 x 10*8M to 0.999 M due to the presence of the ligand.

Problem 67 at the end of this study chapter covers material from this section. Chapter 15- Aqueous Equiljbrju

Lommon Ion Effect HC2HOz(aaj.HqJ What happens when 230I-1NaC is added to this reaction. + will cause CH the reaction to shift LEFT in response to the increase 230 in [Jof 23CHOj. With a common ion we see that dissociation of HC , when we add which has an H NaC ion in 23HQ common with it.230 EXAMPLES: 1. What is the solubility of ?2Mg? 5[K = 6.4 x 101 2Mg? (s) +— Mg(aq) + 2F’(aq) Excess 2x x - +x +2x = j2LMg 2[F’1 6.4 x 10’ = 2(x)(2x) 6.4 x i= 34x x = x 3i0 = x NOW what happens to the solubility when we add 0dM 2Na? 2Mg? s) —* Mg(aq) + 2F’aq) Excess 2 0 0 +x +2x+0.1 1 Negligible because of the value = J[[Mg J12F x 2 64 4=i0 2(x)(0.1)

LESS DISSOLVES!!! ______

11. L5JiJft1LiJ

When the equilibria expression will depend of the [molar] of species K to which the solid is K Kp:: when the equilibrium expresses the degree soluble in water.

EXAMPLE:

BaSO4 (s) +—* Ba(aq) + S0j(aq) 2 = 2 KG= 2ll[Ba 421S0 + Your turn: 3Ca(P04)(s) —*

IlL Solubility and K , Molar solubility = solubilitv (g/L) Molar mass of solute

EXAMPLE: (Saturated Solution) of AgCI determine the K, if the initial concentration of AgCI = 1.34 x

+1 -1 -5 Solution = At equilibrium [Ag ] = [Cl ] = 1.34 x 10 ***A11Ag’ and C1’ comes from the AgCI that dissolved! AgC1(s) Ag(aqJ + 1Cr(aq) = (1.34 x 1O) (1.34 x 1O)

cation In general the solubifity of a SALT when a solute that contains a common or anion is added!

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What

x is Criteria for Precipitation or Dissolution

• Predict whether a ppt. will form when 2 solutions are mixed, given appropriate values expression • Q=ion product because the equilibrium has no denominator so its not really a quotient.

• If Q< K then pt,occurs until Q=K

• If Q=K then the system is at Equilibrium(Saturated Solution) • IfIfQ>K then the solid iuntil Q Example: Calculatethe molar solubilityof 2CaF @298K in a solutioncontammLa) 01 M CaNOkand (b)0 01 M NaF The 3.9 x 10’ 2).‘(CaF (s)CaF 2Ca (aq) + 2F(aq) 2 .01 0 (I)

(E) (.01 +x) 2x ][=[Ca 2F] 2 (01 3 9 x 10” +xX2x Since the solubihty2 of 2CaF asso very SMALL you may assume that the O.O1Mof 2Ca is much greater. That is x is small compared to 0 01M therefore 01 + x 01 THEREFORE NEGLECT x!!! 3.9 x 10” =(.O1)(2x) 2= 3•9 x 2 = 9.8 x 1O 4(.O1) x=J9.8 x x= 3.1 x iO mol of 2(s)CaF dissolves/Literof 3.O1MCa(NO hCa 2+ =xandF=.O1+2x-

Again assume2x is small comparedto .01 M of NaF therefore .01 +x.01

3.9 x 10”= x (01)2 x=3 tcr motes of CaF disotvesftAters of.c7tMNaF Selective Precipitation of IONS Separationof ions in an aqueous solutionby using a ragent that forms a ppt. w/ one or a few ions.

Example: If you have 10mL of 0 1MNaCl (aq) and 0 1MNaSO(aq) and you throw in excess (s)AgNO what will precipitate’?24 01o Ksp(Agcl) 1 8 X 1 3 Ksp(Ag2so4)l. X 1O

4First: Determinethe reactionsthat will occur. I) (s)AgNO + NaCl(aq) -* AgCl(s) 2.) 32 3(s)AgNO + 24NaSO - 24AgSO(aq) Second: Writethe equation 1) = [Ag][Cl] 2) K= [Ag[SO Third: To2do you need concentrations. J Where can you get these’? Calculate [Ci] and 42[SO as follows: ] Final [ClJ = (0 1M)( OiL) =0 05M .02L Final ]=[S0 (0 1M)( OiL) =0 05M 42 .02L Fourth: Plug [Cl] and AgCl then 42[S0 and 24AgSO into the equation and ] ,1K = 4[Ag [Cl] =[Ag4] [[S0 1 8 ] [Ag4][OO5MJ 1 x 10’° = 421 4 x 10 = [Ag4][O 05M) [Ag4]= 3 6 x 1O [Ag4]= 1 67 x 1O2 WHICHEVER [Ag4] is smaller will PPT FIRST’