Transcript Chapter 1 Structure and Bonding

Ch 16 Lecture 3 Solubility Equilibria
I.
Solubility Basics
A.
The Solubility Equilibrium
1) Dissolution of an ionic compound is an equilibrium process
a) CaF2 (s)
Ca2+ (aq) + 2 F- (aq)
b) Ksp = Solubility Product = [Ca2+][F-]2
2)
Remember, neither solids nor pure liquids (water) effect the equilibrium constant
a) Dissolving and Reforming change proportionately to the amount of solid
b) Solvent water is at such a high concentration as not to be effected
3)
The Solubility Product is an equilibrium constant, so it has only one value at a
given temperature
4)
Solubility = the equilibrium position for a given set of conditions
a) There are many different conditions that all must obey Ksp
b) Common ions effect the solubility much as they effect pH
5)
Note: These values may
differ from the ones in
your text. Use the values
from your text for all
homework problems
Ksp values of some slightly soluble ionic solids
a)
b)
c)
d)
e)
f)
Most NO3- salts are soluble
Most alkali metal and NH4+ salts are soluble
Most Cl-, Br-, and I- salts are soluble (except: Ag+, Pb2+, and Hg22+)
Most SO42- salts are soluble (except Ba2+, Pb2+, Hg22+, and Ca2+)
Most OH- salts are insoluble (except NaOH, KOH)
Most S2-, CO32-, CrO42-, and PO43- salts are insoluble
B.
6)
Example: What is the Ksp of a CuBr solution with a solubility of 2.0 x 10-4 M?
CuBr (s)
Cu+ (aq) + Br- (aq)
Ksp = [Cu+][Br-] = [2.0 x 10-4][2.0 x 10-4] = 4 x 10-8
7)
Example: Ksp = ? for Bi2S3 with solubility of 1.0 x 10-15 M
8)
Example: Find the solubility of Cu(IO3)2 (Ksp = 1.4 x 10-7)
Cu (IO3)2 (s)
Cu2+ (aq) + 2 IO3- (aq)
Ksp = [Cu2+][IO3-]2 = 1.4 x 10-7
(x)(2x)2 = 4x3 = 1.4 x 10-7
x = 3.3 x 10-3 M
Relative Solubilities
1) If the salts being compared produce the same number of ions, we can compare
solubilities by comparing Ksp values
AgI
Ksp = 1.5 x 10-16
CuI
Ksp = 5.0 x 10-12
CaSO4
Ksp = 6.1 x 10-5
Ksp = [Xn+]1[Yn-]1 for all of these, so we can directly compare them
Solubility of CaSO4 > solubility of CuI > solubility of AgI
2)
If the salts being compared produce different numbers of ions, we must
calculate the actual solubility values; we can’t use Ksp values to compare.
a) CuS (8.5 x 10-45) > Ag2S (1.6 x 10-49) > Bi2S3 (1.1 x 10-73) by Ksp alone
2 ions
3 ions
5 ions
b)
C.
Bi2S3 (1.0 x 10-15) > Ag2S (3.4 x 10-17) > CuS (9.2 x 10-23) in solubility
The Common Ion Effect
1) Common Ion = any ion in the solid we are trying to dissolve that is present in
solution from another source.
2)
Init.
Equil.
What is the solubility of Ag2CrO4 (Ksp = 9 x 10-12) in 0.1 M AgNO3?
Ag2CrO4 (s)
2 Ag+ (aq)
+
CrO42- (aq)
---0.1
0
---0.1 + 2x
x
Ksp = 9 x 10-12 = [0.1 + 2x]2[x] ~~ (0.1)2(x)
x = 9 x 10-12 / 0.01 = 9 x 10-10 = solubility
5% rule: 9 x 10-10 / 0.1 < 5%, so the approximation is ok
Solubility in pure water is 1.3 x 10-4 M. Why does the solubility decrease?
3)
D.
Example: Find solubility of CaF2 (Ksp = 4.0 x 10-11) in 0.025 M NaF
pH and Solubility
1) Many salts contain hydroxide anion: Mg(OH)2
Mg2+ + 2 OHa) High pH means large OH- common ion concentration
b) [OH-] would shift the equilibrium to the left
c) [H+] would shift the equilibrium to the right by using up OH- ions
2)
3)
Any Basic Anion will be effected by pH
a) OH-, S2-, CO32-, C2O42-, CrO42-, and PO43- are all basic anions
b) H+ will increase the solubility of their salts by removing the anions
c) Ag3PO4 (s)
3 Ag+ (aq) + PO43- (aq)
PO43- + H+
HPO42Acidic pH has no effect on non-basic anions or on most cations: Cl-, Br-, NO3a) AgCl (s)
Ag+ (aq) + Cl- (aq)
b) H+ doesn’t react with either ion
II.
Precipitation and Qualitative Analysis
A.
We can use the solubility product to predict precipitation
1) If Q > Ksp, precipitation occurs until Ksp is reached
2) If Q < Ksp, no precipitation will occur
3)
Example: 750 ml 0.004 M Ce(NO3)3 is added to 300 ml 0.02 M KIO3. Will
Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitate?
Ksp = [Ce3+][IO3-]3 We need to know concentrations.
(0.75 L)(0.004 mol/L)
[Ce ] 
 2.86 x 10 3 M
1.050 L
3
[IO 3 ] 
-
(0.30 L)(0.02 mol/L)
 5.71 x 10 3 M
1.050 L
Q = (2.86 x 10-3)(5.71 x 10-3)3 = 5.32 x 10-10 > Ksp
4)
Precipitate
We can also calculate the equilibrium concentrations after precipitation.
a) Examine the stoichiometry of the precipitation reaction allowed to go to
completion
b) Becomes a Ksp problem with a common ion (ion in excess)
3)
Calculate the equilibrium concentrations after precipitation when 100 ml of
0.05 M Pb(NO3)2 is added to 200 ml 0.10 M NaI. Ksp for PbI2 = 1.4 x 10–8.
a) PbI2 (s)
Pb2+ (aq) + 2 I- (aq) Ksp = [Pb2+][I-]2
b) [Pb2+] = 1.67 x 10-2, [I-] = 6.67 x 10-2, Q = 7.43 x 10-5 > Ksp precipitate
+
2 I20 mmol
10 mmol
Stoichiometry:
Initial
Completion
d)
Equilibrium: some PbI2 redissolves, with I- common ion present
i. [I-] common ion = 10mmol / 300ml = 0.033 M
Initial
Equilibrium
4)
Pb2+
5mmol
0
c)
PbI2
-------
PbI2 (s)
Pb2+ (aq) +
2 I- (aq)
Ksp = [Pb2+][I-]2
---0
0.033M
---x
0.033 + 2x
i. Ksp = 1.4 x 10–8 = [Pb2 +][I-]2 = (x)(0.033 + 2x)2 ~~(x)(0.033)2
x = 1.3 x 10-5 M = [Pb2 +], [I-] = 0.033 M
Example: 150 ml 0.01 M Mg(NO3)2 + 250 ml 0.1 M NaF. Find [Mg2+] and
[F-] at equilibrium. Ksp for MgF2 = 6.4 x 10-9
B.
Qualitative Analysis
1. Selective Precipitation = addition of an ion that causes only one of a mixture of
ions to precipitate
Add NaCl to Ag+ (aq) + Ba2+ (aq)
AgCl(s) + Na+ (aq) + Ba2+ (aq)
2.
Example: Which ion will precipitate first? I- is added to a solution of 0.0001 M
Cu+ and 0.002 M Pb2+? Ksp CuI = 5.3 x 10-12. Ksp PbI2 = 1.4 x 10-8.
a) Use Ksp CuI to find [I-] that will just start precipitation
b) Use Ksp PbI2 to find [I-] that will just start precipitation
c) Whichever has the lowest [I-] will precipitate first
3.
Sulfide Ion (S2-) is particularly useful for selective cation precipitation
a) Metal sulfides have very different solubilities
i. FeS Ksp = 2.3 x 10-13
ii. MnS Ksp = 3.7 x 10-19
iii. Mn would precipitate first from an equal mixture of Fe2+ and Mn2+
b)
[S2-] can be controlled by pH
i. H2S
H+ + HS- Ka1 = 1 x 10-7
ii. HSH+ + S2- Ka2 = 1 x 10-19
iii. S2- is quite basic. At low pH, there will be very little S2iv. At high pH, there is much more S2c)
We can selectively precipitate metal ions by adding S2- in acidic
solution, and then slowly adding base.
CuS
Ksp = 8.5 x 10-45
Hg, then Cu, then Ni, then Mn
-54
HgS
Ksp = 1.6 x 10
would precipitate as we raise pH
MnS
Ksp = 2.3 x 10-13
NiS
Ksp = 3.0 x 10-21
Solution of
Mn2+, Ni2+, Cu2+, Hg2+
Precipitate of
CuS, HgS
Add H2S, pH = 2
Solution of
Mn2+, Ni2+
Add OH- to pH = 8
Precip. Of
MnS, NiS
4)
Qualitative Analysis = scheme to separate and identify mixtures of cations by
precipitation
a)
Group I Insoluble Chlorides:
Add HCl. AgCl, PbCl2, Hg2Cl2 precipitate
b)
Group II Sulfides Insoluble in Acid Solution:
Add H2S. Low [S2-] due to [H+]. HgS, CdS, Bi2S3, CuS, and SnS2 precip.
c)
Group III Sulfides Insoluble in Basic Solution:
Add OH-. CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, and Al(OH)3 precipitate.
d)
Group IV Insoluble Carbonates:
Add CO32-. BaCO3, CaCO3, and MgCO3 precipitate.
e)
Group V Alkali Metal and Ammonium
These ions (Na+, K+, NH4+) are soluble with common ions
f)
Further tests would tell us which specific ions in each group are present
C.
Complex Ion Equilibria
1) A complex ion = Lewis Acid—Lewis Base complex of a metal ion
a) Ligand = generic name for the Lewis Base bonded to a metal ion
b) H2O, NH3, Cl-, CN- all have lone pairs to donate; can be ligands
2)
Coordination Number (CN) = the number of ligands bonded to a metal ion
a) CN = 6 is common: Co(H2O)62+, Ni(NH3)62+
b) CN = 4 CoCl42-, Cu(NH3)42+
c) CN = 2 Ag(NH3)2+
d) Coordination number depends on size and properties of the metal ion
3)
Formation (or Stability) Constants
a) Ligand addition to the metal ion is stepwise
Ag+ + NH3
Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3
Ag(NH3)2+
K2 = 8.2 x 103
Ag+ + 2NH3
Ag(NH3)2+
Kf = 1.7 x 107
b)
c)
Usually, [Ligand] >> [Mn+] to force the equilibria to the right
What are the equilibrium conditions of 100 ml 2.0 M NH3 added to
100 ml of 0.001 M AgNO3?
i. K1 x K2 = Kf is large, favoring complete reaction
ii. [NH3] is much larger than [Ag+], so assume constant
iii. Do stoichiometry of the reaction:
Ag+
+
2 NH3
Ag(NH3)2+
Initial 5 x 10-4 M
1M
0
Equilib.
0
1M
5 x 10-4 M
iv. Then use equilibrium expressions to find concentrations

[Ag(NH 3 ) 2 ]
3
K 2  8.2 x 10 
[Ag(NH 3 )  ][NH 3 ]

4
[Ag(NH
)
]
5x10
8
3 2
 [Ag(NH 3 )  ] 


6
.
1
x
10
M
3
K 2 [NH3 ]
(8.2 x 10 )(1.0)

[Ag(NH
)
]
3
K1  2.1 x 103 
[Ag  ][NH 3 ]

[Ag(NH
)
]
3
 [Ag  ] 
 2.9 x 10 11 M
K1[NH3 ]
4)
Example: Find [Ag+], [Ag(S2O3)-], [Ag(S2O3)23-] for 150 ml 0.001 M AgNO3
+ 200 ml 5.0 M Na2S2O3. K1 = 7.4 x 108, K2 = 3.9 x 104.
5)
Complex Ions and Solubility
a) How do we dissolve AgCl(s)? Ksp = 1.6 x 10-10
Ag+ + NH3
Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3
Ag(NH3)2+
K2 = 8.2 x 103
b)
Adding NH3 to a suspension of AgCl, forces more of it to dissolve. The
NH3, removes Ag+ from solution, forcing the equilibrium to the right.
c)
[Ag+]T = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]
d)
The overall reaction is the sum of three individual reactions:
AgCl
Ag+ + ClKsp = 1.6 x 10-10
Ag+ + NH3
Ag(NH3)+
K1 = 2.1 x 103
Ag(NH3)+ + NH3
Ag(NH3)2+
K2 = 8.2 x 103
AgCl + 2 NH3
Ag(NH3)2+ + Cl- K = ???
K = Ksp x K1 x K2 = (1.6 x 10-10)(2.1 x 103)(8.2 x 103) = 2.8 x 10-3
e)
Initial
Equil.
6)
Example: Find the solubility of AgCl(s) in 10 M NH3.
AgCl + 2 NH3
Ag(NH3)2+ + Cl----10
0
0
----10 – 2x
x
x
Solve using the expression for K as in a normal equilibrium problem.
Strategies for Dissolving Insoluble Solids
a) If the anion is basic, add acid
b) If the cation will form a complex, add ligand
c) Heat often increases solubility (temp. effects all equil. constants)
d)
Example: HgS
Hg2+ + S2i. S2- is a basic anion, so add HCl
S2- + H+
HSii. Hg2+ will form a chloride complex
Hg2+ + 4 ClHgCl42-
Ksp = 1 x 10-54
7)
The qualitative analysis of the Group I cations illustrates complex ion equilibria
Solution of
Ag+, Hg22+, Pb2+
Add cold HCl
Precipitation of AgCl, Hg2Cl2, PbCl2
Heat
Solution of Pb2+
Precipitate of
AgCl, Hg2Cl2
Add CrO42Precipitate of
PbCrO4
Add NH3
Solution of
Ag(NH3)2+, ClAdd H+
Precip. of AgCl
Precipitate of
Hg, HgNH2Cl