Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.6: Solubility Equilibria and 15.6: Solubility Equilibria and Solubility ProductsSolubility Products

SolubilitySolubility

As a salt dissolves in water and ions As a salt dissolves in water and ions are released, they can collide and re-are released, they can collide and re-from the solidfrom the solid

Equilibrium is reached when the rate Equilibrium is reached when the rate of dissolution equals the rate of of dissolution equals the rate of recrystallizationrecrystallization

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Saturated solutionSaturated solution

When no more solid can dissolve at When no more solid can dissolve at equilibriumequilibrium

Solubility ProductSolubility Product

Solubility product: KSolubility product: Kspsp

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Ksp=[CaKsp=[Ca2+2+][F][F--]]22

Why do we leave out the CaFWhy do we leave out the CaF22?? Adding more solid will not effect the Adding more solid will not effect the

amount of solid that can dissolve at a amount of solid that can dissolve at a certain temperaturecertain temperature

It would increase both reverse and It would increase both reverse and forward reaction rates b/c there is a forward reaction rates b/c there is a greater amountgreater amount

KKspsp Values Values

Example 1Example 1 CuBr has a solubility of 2.0x10CuBr has a solubility of 2.0x10-4-4 mol/L at mol/L at

2525°°C. Find the KC. Find the Kspsp value. value. The solubility tells us the amount of The solubility tells us the amount of

solute that can dissolves in 1 L of watersolute that can dissolves in 1 L of water Use ICE chart: solubility tells you x valueUse ICE chart: solubility tells you x value

Ksp=[CuKsp=[Cu++][Br][Br--]= (2.0x10]= (2.0x10-4-4 M) M)2 2 = 4.0x10= 4.0x10-8-8

CuBr(s) ↔ CuCuBr(s) ↔ Cu++(aq) + Br(aq) + Br--

(aq) (aq)

II Not imp.Not imp. 00 00

CC -2.0x10-2.0x10-4-4 M M +2.0x10+2.0x10-4-4 M M +2.0x10+2.0x10-4-4 M M

EE Not impNot imp 2.0x102.0x10-4-4 M M 2.0x102.0x10-4-4 M M

Example 2Example 2 The KThe Kspsp value for Cu(IO value for Cu(IO33))22 is 1.4x10 is 1.4x10-7-7 at at

2525°°C. Calculate its solubility.C. Calculate its solubility. Solve for solubility = x value using ICE chartSolve for solubility = x value using ICE chart

KKspsp=[Cu=[Cu2+2+][IO][IO33--]]22= (x)(2x)= (x)(2x)2 2 = 1.4x10= 1.4x10-7-7 = =

4x4x33

X = (3.5x10X = (3.5x10-8-8))1/31/3 = 3.3x10 = 3.3x10-3-3

Cu(IOCu(IO33))22(s) ↔ Cu(s) ↔ Cu2+2+(aq) + 2IO(aq) + 2IO33--

(aq)(aq)

II Not imp.Not imp. 00 00

CC -x-x +x+x +2x+2x

EE Not impNot imp xx 2x2x

Comparing Solubilities Comparing Solubilities

You can only compare solubilities using You can only compare solubilities using KKspsp values for compounds containing the values for compounds containing the same number of ionssame number of ions

CaSOCaSO44 > AgI > CuCO > AgI > CuCO33

KKspsp values: 6.1x10 values: 6.1x10-5-5 > 1.5x10 > 1.5x10-6 -6 >> 2.5x102.5x10-10-10 Why can we use KWhy can we use Kspsp values to judge values to judge

solubility?solubility? Can only compare using actual solubility Can only compare using actual solubility

values (x) when compounds have values (x) when compounds have different numbers of ionsdifferent numbers of ions

Common Ion EffectCommon Ion Effect Solubility of a solid is lowered when a Solubility of a solid is lowered when a

solution already contains one of the solution already contains one of the ions it containsions it contains

Why?Why?

Example 3Example 3 Find the solubility of CaFFind the solubility of CaF22 (s) if the K (s) if the Kspsp is 4.0 is 4.0

x 10x 10-11-11 and it is in a 0.025 M NaF solution. and it is in a 0.025 M NaF solution.

KKspsp=[Ca=[Ca2+2+][F][F--]]22= (x)(2x+0.025)= (x)(2x+0.025)2 2 = = 4.0 x 104.0 x 10--

1111

(x)(0.025)(x)(0.025)2 2 ≈≈ 4.0 x 104.0 x 10-11-11

x = 6.4x10x = 6.4x10-8-8

CaFCaF22(s) ↔ Ca(s) ↔ Ca2+2+(aq) + 2F(aq) + 2F--

(aq)(aq)II Not imp.Not imp. 00 0.0250.025

CC -x-x +x+x +2x+2x

EE Not impNot imp xx .025+2x.025+2x

pH and solubilitypH and solubility pH can effect solubility because of the pH can effect solubility because of the

common ion effectcommon ion effect Ex: Mg(OH)Ex: Mg(OH)22(s) ↔ Mg(s) ↔ Mg2+2+(aq) + 2OH(aq) + 2OH--(aq)(aq)

How would a high pH effect solubility?How would a high pH effect solubility? High pH = high [OH-] High pH = high [OH-] decrease solubility decrease solubility

Ex: AgEx: Ag33POPO44(s) ↔ 3Ag(s) ↔ 3Ag++(aq) + PO(aq) + PO443-3-(aq)(aq)

What would happen if H+ is added?What would happen if H+ is added? HH++ uses up PO uses up PO44

3-3- to make phosphoric acid to make phosphoric acid Eq. shifts to right - Solubility increasesEq. shifts to right - Solubility increases

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.7: Precipitation and 15.7: Precipitation and Qualitative AnalysisQualitative Analysis

PrecipitationPrecipitation Opposite of dissolutionOpposite of dissolution Can predict whether precipitation or Can predict whether precipitation or

dissolution will occurdissolution will occur Use Q: ion productUse Q: ion product

Equals KEquals Kspsp but doesn’t have to be at but doesn’t have to be at equilibriumequilibrium

Q > K: more reactant will form, Q > K: more reactant will form, precipitation until equilibrium reachedprecipitation until equilibrium reached

Q < K: more product will form, Q < K: more product will form, dissolutiondissolution

Example 1Example 1 A solution is prepared by mixing 750.0 mL A solution is prepared by mixing 750.0 mL

of 4.00x10of 4.00x10-3-3M Ce(NOM Ce(NO33))33 and 300.0 mL and 300.0 mL 2.00x102.00x10-2-2M KIOM KIO33. Will Ce(IO. Will Ce(IO33))33 precipitate precipitate out?out? Calculate Q value and compare to K (on chart)Calculate Q value and compare to K (on chart) Ce(IOCe(IO33))33(s) ↔ Ce(s) ↔ Ce3+3+(aq) + 3IO(aq) + 3IO33

--(aq)(aq) Q=[CeQ=[Ce3+3+][IO][IO33

--]]33

K=1.9x10K=1.9x10-10 -10 < Q so YES< Q so YES

10-

32-3-

-3

3

105.32 0.3000.750

102.00300.0

300.0750.0

104.00750.0

]][IO[CeQ 3

Example 2Example 2 A solution is made by mixing 150.0 mL A solution is made by mixing 150.0 mL

of 1.00x10of 1.00x10-2-2 M Mg(NO M Mg(NO33))22 and 250.0 mL and 250.0 mL of 1.00x10of 1.00x10-1-1 M NaF. Find concentration M NaF. Find concentration of Mgof Mg2+2+ and F and F-- at equilibrium with solid at equilibrium with solid MgFMgF22 (K (Kspsp=6.4x10=6.4x10-9-9))

MgFMgF22(s) (s) Mg Mg2+2+(aq) + 2F(aq) + 2F--(aq)(aq) Need to figure out whether the Need to figure out whether the

concentrations of the ions are high concentrations of the ions are high enough to cause precipitation firstenough to cause precipitation first

Find Q and compare to KFind Q and compare to K

Example 2Example 2

Q > K so shift to left, precipitation occursQ > K so shift to left, precipitation occurs Will all of it precipitate out??Will all of it precipitate out?? No- No-

we need to figure out how much is created we need to figure out how much is created using stoichiometryusing stoichiometry

then how much ion is left over using ICE chartthen how much ion is left over using ICE chart Like doing acid/base problemLike doing acid/base problem

5-

21-2-

2-2

1046.1 0.052150.0

101.00250.0

0.052150.0

101.00150.0

]][F[MgQ

Example 2Example 2

MgMg2+2+(aq) + 2F(aq) + 2F--(aq) ↔ MgF(aq) ↔ MgF22(s)(s)

II 1.50 mmol1.50 mmol 25.0 mmol25.0 mmol Not imp.Not imp.

CC -1.50-1.50 -1.50-1.50 +2(1.50)+2(1.50)

EE 00 23.5 mmol23.5 mmol Not imp.Not imp.

MgFMgF22(s)↔ Mg(s)↔ Mg2+2+(aq) + 2F(aq) + 2F--(aq)(aq)

II Not imp.Not imp. 00 23.5mmol/400mL23.5mmol/400mL

CC -x-x +x+x +2x+2x

EE Not impNot imp xx 5.5x105.5x10-2-2+2x+2x

How much will be use if goes to completion?

Some of it dissolved- how much are left in solution?

Example 2Example 2

MF

MxMg

Mx

xx

FMgK sp

2

62

6

922

922

105.5][

101.2][

101.2

104.6]2105.5][[

104.6]][[

Qualitative AnalysisQualitative Analysis Process used to separate a solution Process used to separate a solution

containing different ions using solubilitiescontaining different ions using solubilities A solution of 1.0x10A solution of 1.0x10-4-4 M Cu M Cu++ and 2.0x10 and 2.0x10-3-3 M M

PbPb2+2+. If I. If I-- is gradually added, which will is gradually added, which will precipitate out first, CuI or PbIprecipitate out first, CuI or PbI22?? 1.4x101.4x10-8-8=[Pb=[Pb2+2+][I][I--]]22 = (2.0x10 = (2.0x10-3-3)[I)[I--]]22

[I[I--]=2.6x10]=2.6x10-3-3 M : [I M : [I--]> than that to cause PbI]> than that to cause PbI22 to to pptppt

5.3x105.3x10-12-12=[Cu=[Cu++][I][I--] = (1.0x10] = (1.0x10-4-4)[I)[I--]] [I[I--]=5.3x10]=5.3x10-8-8 M : [I M : [I--]> than that to cause CuI to ppt]> than that to cause CuI to ppt Takes a much lower conc to cause CuI to ppt so it Takes a much lower conc to cause CuI to ppt so it

will happen firstwill happen first

Qualitative AnalysisQualitative Analysis

Ch. 15: Applications Ch. 15: Applications of Aqueous Equilibriaof Aqueous Equilibria

15.8: Complex Ion Equilibria15.8: Complex Ion Equilibria

Complex Ion EquilibriaComplex Ion Equilibria

Complex ionComplex ion Charged species containing metal ion Charged species containing metal ion

surrounded by ligandssurrounded by ligands LigandsLigands

Lewis bases donating electron pair to Lewis bases donating electron pair to empty orbitals on metal ionempty orbitals on metal ion

Ex: HEx: H22O, NHO, NH33, Cl-, CN-, OH-, Cl-, CN-, OH- Coordination numberCoordination number

Number of ligands attachedNumber of ligands attached

Complex Ion EquilibriaComplex Ion Equilibria

Usually, the conc of the ligand is very Usually, the conc of the ligand is very high compared to conc of metal ion high compared to conc of metal ion in the solutionin the solution

Ligands attach in stepwise fashionLigands attach in stepwise fashion AgAg++ + NH + NH33 Ag(NH Ag(NH33))++

Ag(NHAg(NH33))++ + NH + NH33 Ag(NH Ag(NH33))2+2+

Example 3Example 3

Find the [AgFind the [Ag++], [Ag(S], [Ag(S22OO33))--],], and and [Ag(S[Ag(S22OO33))22

3-3-]] in solution made with 150.0 in solution made with 150.0 mL of 1.00x10mL of 1.00x10-3-3 M AgNO M AgNO33 with 200.0 mL of with 200.0 mL of 5.00 M Na5.00 M Na22SS22OO33.. AgAg++ + S + S22OO33

2-2- Ag(S Ag(S22OO33))-- KK11=7.4x10=7.4x1088

Ag(SAg(S22OO33))- - + S+ S22OO332-2- Ag(S Ag(S22OO33))22

3-3- KK22=3.9x10=3.9x1044

Because of the difference in conc between Because of the difference in conc between ligand and metal ion, the reactions can be ligand and metal ion, the reactions can be assumed to go to completionassumed to go to completion

Example 3Example 3 AgAg++ + 2S + 2S22OO33

2-2- Ag(S Ag(S22OO33))223-3-

II (150.0)(1.0x10(150.0)(1.0x10-3-3) ) = 0.150mmol= 0.150mmol

(200.0)(5.00) (200.0)(5.00) = = 1.00x101.00x1033mmolmmol

00

CC -0.150 mmol-0.150 mmol -0.150 mmol-0.150 mmol +0.150 +0.150 mmolmmol

EE 00 1.00x101.00x1033mmolmmol 0.150 mmol0.150 mmol

M

MmL

mmolOS

4-3232

32

32

1029.4mL 350.0

mmol 0.150])O[Ag(S

86.20.350

1000.1][

MOSAg

OSAg

OSOSAg

OSAgK

9132

41

32

4

42

321

32

3232

2

108.3])([

109.3]86.2][)([

]1029.4[

109.3]][)([

])([

MAg

Ag

OSAg

OSAgK

181

41

9

82

321

132

1

108.1][

109.3]86.2][[

]108.3[

104.7]][[

])([