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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]][[
])([