Common Ion Effect
CH3COOH
H+(aq) + CH3COO(aq)
pH of 0.1 M soln =
Add 0.1 M CH3COONa:
CH3COONa  Na+ + CH3COO(aq)
pH =
What happened to
[CH3COO]? [CH3COOH]?
Buffer Solutions
A buffer is a solution that “resists” a change in pH
E.g. blood contains substances that keep its pH fixed at 7.3
- important for life functions
Buffer solutions consist of either:
A weak acid + salt of its conjugate base
or
A weak base + salt of its conjugate acid
Buffers
H+(aq) + X(aq)
HX(aq)


H X 
X 

 H 
HX 
HX 

Ka



 
 X 

pKa  pH  log 
 HX  
 
 X 

pH  pKa  log 
 HX  
HendersonHasselbalch equation
Note: If [HX] = [X], pH = pKa
Buffers work best near pH = pKa
Buffer Capacity
CH3COOH
H+(aq) + CH3COO(aq)
[CH3COOH] = [CH3COO] = 1.0 M
1 L solution, pH = pKa = 4.74
1. Add a dropper (~20ml) of 1M HCl
2. Add a dropper (~20ml) of NaOH
Repeat calculation starting from a 1.8 x 10-5 M HCl
solution (pH = 4.74)
Buffer Capacity
A CH3COOH + CH3COONa (both 1 M)
pH =
B dilute solution A 10x
pH =
Repeat with 1 M HCl + 1 M NaCl solution
SOLUBILITY
Solubility: quantity of a substance that dissolves to form a
saturated solution
Solubility:
g/L
Molar solubility:
mole/L
Some salts are very soluble (> 0.1 M). Recall solubility rules.
Some salts are sparingly soluble (< 0.1 M) sometimes
referred to as ‘insoluble’.
Precipitation and Solubility of ionic salts and their
equilibrium in water
MX(s)
M+(aq) + X(aq)
SOLUBILITY EQUILIBRIA
• Precipitation
Ag+(aq) + Cl(aq)  AgCl(s)
• Dissolution
AgCl(s)  Ag+(aq) + Cl(aq)
At equilibrium when forward rate = backward rate
AgCl(s)
Ag+(aq) + Cl(aq)
Keq = [Ag+][Cl]/ [AgCl(s)]
Keq [AgCl(s)] = [Ag+][Cl]

Ksp
= [Ag+][Cl]
The concentration of solid does not change at equilibrium
SOLUBILITY
Solubility Product: Ksp
AgCl
AgBr
AgI
CdS
ZnS
Mg(OH)2
Ca(OH)2
CaF2
BaCO3
BaSO4
1.8 x 10-10
5.0 x 10-13
8.3 x 10-17
8.0 x 10-27
1.1 x 10-21
1.8 x 10-11
5.5 x 10-6
3.9 x 10-11
5.1 x 10-9
1.1 x 10-10
Ksp is constant for a given solid at a given temp.
SOLUBILITY CALCULATION
Calculate [Ca2+] and [F-] for a saturated CaF2 solution.
CaF2 (s)  Ca+2 (aq) +2F (aq)
Ksp = [Ca2+][F]2 =3.9 x 1011 at 25oC
What is the solubility?
solubility = amount of CaF2 dissociated
COMMON ION EFFECT
[Ag+]
precipitation
saturation
dissolution
[Cl-]
If Q = [Ag+][Cl-] > Ksp, AgCl precipitates
(Ion product > solubility product)
If [Ag+][Cl-] < Ksp; AgCl dissolves
When [Ag+][Cl-] = Ksp, the solution is saturated
Adding either [Ag+] or [Cl-] will precipitate AgCl(s)
SOLUBILITY CALCULATION
Common Ion Effect
What about CaF2 in 0.01 M NaF solution?
[F] = 0.01 M Ksp = 3.9 x 1011
CaF2(s)  Ca2+(aq) + 2F(aq)
Water Chemistry (Ch. 18.5-6)
Water in State College/UP Campus
Predominantly well water
23 wells + 1 open reservoir
[Ca2+ ] 165-185 ppm
10-15 ppm
Mixed to reduce water hardness
• Cl2 injected to kill bacteria
• F- added
• Alum (K2SO4.Al2(SO4)3.24H2O) added to improve clarity
• NaOH added to neutralize pH
Remediation of Water
Ca2+ (hard water), Pb2+ (toxic) are precipitated by CO32Ksp
CaCO3
8.7 x 10-9
PbCO3
3.3 x 10-14
Which compound(s) could we use to supply CO32-?
How much do we need to add ?
If we add a stoichiometric amount of Na2CO3,
[Ca2+] = [CO32-] = Ksp1/2
[Ca2+] = 3 x 10-3 M
2.9 x 10-3 mol/L x 40 g/mol x 1000 mg/g = 116 mg/L
(116 ppm)
If [CO32-] = 3 mM, what is [Pb2+]?
[Pb2+] = Ksp/[CO32-] = 3.3 x 10-14/3 x 10-3 = 1 x 10-11 M
= 2 parts per trillion
Effect of pH
What is the solubility of Mg(OH)2 in pure water?
Ksp = 1.8 x 10-11
What is the solubility of Mg(OH)2 in a solution
with a pH of 9?
Effect of pH on common ions
If either the anion or the cation is involved in an acid base
equilibrium, then it is a common ion problem.
•Basic metal hydroxides
Low pH increases solubility e.g., Mg(OH)2
•Salts of weakly basic anions
Low pH increases solubility
examples (write out equilibria for practice)
Mg(HCO3)2
ZnCO3
Ca3(PO4)2
Contrast with NaCl, Ca(NO3)2
NaF
AMPHOTERIC METAL HYDROXIDES
There are amphoteric hydroxides of
Al3+ Cr3+ Zn2+ Sn2+ and many transition metal
ions
Al(OH)3 , Cr(OH)3 , Zn(OH)2 , Sn(OH)2 , …
Dissolution involves formation of complex ions:
Al(OH)3(OH2)3 (s)
+ H+  Al(OH)2(OH2)4+
+ OH  Al(OH) (OH ) 
4
2 2
Amphoteric hydroxides:
Both low and high pH increases solubility
FORMATION OF COMPLEX IONS
Hydration of metal ions
Cu2+(aq) + 4 H2O(l)  [Cu(OH2)4]2+(aq)
Lewis
Acid
+ Lewis  Lewis Acid/Base
Base
Adduct
= Metal Complex
Other Lewis bases react with metal ions to form complexes
Cu2+(aq) + 4 NH3(aq)  [Cu(NH3)4]2+(aq)
Cu2+(aq) + 4 CN(aq)  [Cu(CN)4]2(aq)
Cu2+(aq) + 4 Cl(aq)  [Cu(Cl)4]2(aq)
METAL COMPLEX STABILITY
Cu(NH3)42+ + 4H2O  Cu(OH2)42+ + 4NH3
Cu(OH ) NH 

K 
Cu(NH ) H O [H O] = constant
Cu NH 

K 
Dissociation constant
Cu(NH ) 
4
2
2 4
Cu2+(aq)
3
4
2
3 4
2
2
4
2
3
D
2
3 4
Cu(OH2)42+ + 4NH3  Cu(NH3)42+ + 4H2O

KF 
Cu(NH ) 
Cu NH 
2
3 4
4
2
3
Formation constant
KF
1

KD
Kf VALUES OF SOME COMPLEXES
Ag(NH3)2+
2 x 107
Cu(NH3)42+
5 x 1012
Cu(CN)42-
1 x 1025
Ag(CN)2-
1 x 1021
Ag(S2O3)23-
3 x 1013
Complex Ion Formation
What is the conc of free Cu2+ ions in a 1 L solution that
contains 1 x10-3 moles total Cu2+ and is 0.1 M in NH3?
Cu2+(aq) + 4 NH3(aq)  Cu(NH3)42+(aq)
Kf = 5 x 1012
Complex Ion Formation
CuCO3 is a sparingly soluble salt?
Ksp CuCO3 = 2.3 x 1010
How can I get it to dissolve?
What is the equilibrium constant for the following reaction?
CuCO3(s) + 4CN(aq)  CO32(aq) + Cu(CN)42(aq)
Ksp CuCO3 = 2.3 x 1010
Kf Cu(CN)42 = 1 x 1025