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Complex Equilibrium
• Aqueous solutions may contain several species that can
interact with each other
• Shift in equilibrium
A. Interacting solute
B. Keq is very small, contribution from solvent
C. Dilute solutions, contribution from solvent
Three Situations to Worry About:
Complex Equilibrium
A. Interacting solute
– Anion is a conjugate base of a weak acid
– Examples,
• CaF2
• BaSO4
Example 1: CaF2
(1) CaF2(s) = Ca2+
+ 2 F-
(2) F-
+ H2O = HF + OH-
Complex Equilibrium
Example 1: CaF2
(1) CaF2(s) = Ca2+
+ 2 F-
(2) F-
+ H2O = HF + OH-
Coupled Equilibria
– Product of one reaction is the reactant in the next
reaction
– Will solubility be larger or smaller if you take into
account rxn (2)?
Clicker: (a) YES (b) NO
Complex Equilibrium
Example1: Calculate the solubility of CaF2 neglecting the
hydrolysis of F-
. Ksp = 4.0 x 10-11
If you took into account the hydrolysis of F-
solubility = 3.9 x 10-4
M
CaF2(s) = Ca2+
+ 2F-
x = solubility = 2.1 x 10-4
M
Set up ICE table
Complex Equilibrium
Solubility can be changed by changing the pH!
solubility = 1.4 x 10-3
M
(1) CaF2(s) = Ca2+
+ 2 F-
(2) F-
+ H3O+
= HF + H2O
Log Ca2+
pH
Solubility decreases as
pH increases. Why?
Complex Equilibrium
Example 2:
BaSO4(s) in equilibrium with 0.01 M HCl
Write all equilibria:
BaSO4(s) = Ba2+
+ SO4
-2
SO4
-2
+ H3O+
= HSO4
-
+ H2O
HSO4
-
+ H3O+
= H2SO4 + H2O
2H2O = H3O+
+ OH-
Complex Equilibrium
Example 3:
Presence of complexing agents
A.
Al(OH)3(s) = Al3+
+ 3OH-
Add F-
Al3+
+ 6F-
= AlF6
-3
B.
AgBr(s) = Ag+
+ Br-
Add NH3
Ag+
+ NH3 = Ag(NH3)2
+
Complex Equilibrium
B. Keq is small (Basic Ion)
Mg(OH)2(s) = Mg2+
+ 2OH- Ksp = 7.1 x 10-12
H2O = H3O+
+ OH-
Kw = 1.0 x 10-14
Fe(OH)3(s) = Fe3+
+ 3OH- Ksp = 2 x 10-39
H2O = H3O+
+ OH-
Kw = 1.0 x 10-14
Neglect water:
(x) (3x)3
= 2 x 10-39
x = 9.3 x 10-11
M
Take into account water
s = 2 x 10-18
M
Complex Equilibrium
C. Dilute Solutions (contribution from solvent)
– Normally contribution from solvent is small
– If concentration is less than 10-6
M,
solvent needs to be taken into account
Example,
What is the pH of a 1.0 x 10-8
M HCl
Complex equilbrium

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Complex equilbrium

  • 1. Complex Equilibrium • Aqueous solutions may contain several species that can interact with each other • Shift in equilibrium A. Interacting solute B. Keq is very small, contribution from solvent C. Dilute solutions, contribution from solvent Three Situations to Worry About:
  • 2. Complex Equilibrium A. Interacting solute – Anion is a conjugate base of a weak acid – Examples, • CaF2 • BaSO4 Example 1: CaF2 (1) CaF2(s) = Ca2+ + 2 F- (2) F- + H2O = HF + OH-
  • 3. Complex Equilibrium Example 1: CaF2 (1) CaF2(s) = Ca2+ + 2 F- (2) F- + H2O = HF + OH- Coupled Equilibria – Product of one reaction is the reactant in the next reaction – Will solubility be larger or smaller if you take into account rxn (2)? Clicker: (a) YES (b) NO
  • 4. Complex Equilibrium Example1: Calculate the solubility of CaF2 neglecting the hydrolysis of F- . Ksp = 4.0 x 10-11 If you took into account the hydrolysis of F- solubility = 3.9 x 10-4 M CaF2(s) = Ca2+ + 2F- x = solubility = 2.1 x 10-4 M Set up ICE table
  • 5. Complex Equilibrium Solubility can be changed by changing the pH! solubility = 1.4 x 10-3 M (1) CaF2(s) = Ca2+ + 2 F- (2) F- + H3O+ = HF + H2O Log Ca2+ pH Solubility decreases as pH increases. Why?
  • 6. Complex Equilibrium Example 2: BaSO4(s) in equilibrium with 0.01 M HCl Write all equilibria: BaSO4(s) = Ba2+ + SO4 -2 SO4 -2 + H3O+ = HSO4 - + H2O HSO4 - + H3O+ = H2SO4 + H2O 2H2O = H3O+ + OH-
  • 7. Complex Equilibrium Example 3: Presence of complexing agents A. Al(OH)3(s) = Al3+ + 3OH- Add F- Al3+ + 6F- = AlF6 -3 B. AgBr(s) = Ag+ + Br- Add NH3 Ag+ + NH3 = Ag(NH3)2 +
  • 8. Complex Equilibrium B. Keq is small (Basic Ion) Mg(OH)2(s) = Mg2+ + 2OH- Ksp = 7.1 x 10-12 H2O = H3O+ + OH- Kw = 1.0 x 10-14 Fe(OH)3(s) = Fe3+ + 3OH- Ksp = 2 x 10-39 H2O = H3O+ + OH- Kw = 1.0 x 10-14 Neglect water: (x) (3x)3 = 2 x 10-39 x = 9.3 x 10-11 M Take into account water s = 2 x 10-18 M
  • 9. Complex Equilibrium C. Dilute Solutions (contribution from solvent) – Normally contribution from solvent is small – If concentration is less than 10-6 M, solvent needs to be taken into account Example, What is the pH of a 1.0 x 10-8 M HCl