1. Chem. 31 – 10/9 Lecture

2. Announcements I Exam 1 – Results Statistical Calculations Lab
Ave = 83.2%
Distribution
Statistical Calculations Lab
Due Wednesday
New Homework Assignment (Set 2) Posted
Score N
100s 2
90s 36
80s 27
70s 17
60s 7
<60 6

3. Announcements II Today’s Material Chapter 6 Thermodynamics
Le Châtelier’s Principle
Sparingly Soluble Salts
Solubility in water
Solubility in common ion (if time)

4. Thermodynamics ΔH, change in enthalpy, is related to heat of reaction
- if a reaction produces heat, ΔH < 0 and reaction is “exothermic”
- a reaction that requires heat has ΔH > 0 and is endothermic
ΔS, change in entropy, is related to disorder of system
- If the final system is “more random” than initial system, ΔS > 0

5. Thermodynamics Entropy
A macroscopic analogy to entropy would be to have a box of 50 ping pong balls with half white and half black
Even if placed on two separate halves of the box, if the box were shaken to mix the balls, roughly half of each color would be expected in each half leading to a positive DS v v
initial state
final state 5

6. Thermodynamics Entropy Examples: (Is ΔS > or < 0?)
H2O(l) ↔ H2O(g)
H2O(s) ↔ H2O(l)
NaCl(s) ↔ Na+ + Cl-
2H2(g) + O2(g) ↔ 2H2O(g)
N2(g) + O2(g) ↔ 2NO(g)
ΔS > 0
ΔS > 0
ΔS > 0
ΔS < 0
ΔS > 0

7. Thermodynamics ΔG = Change in Gibbs free energy
This tells us if a process is spontaneous (expected to happen) or non-spontaneous
ΔG < 0 process is spontaneous (favored)
ΔG = ΔH - TΔS (T is absolute temperature)
processes that are exothermic (Δ H < 0) and increase disorder (Δ S > 0) are favored at all T
processes that have Δ H > 0 and Δ S > 0 are favored at high T

8. Thermodynamics Example Question: The ΔG° for the reaction
Ca2+ + 2OH- => Ca(OH)2(s) is -52 kJ/mol
Determine K at T = 20.°C for Ca(OH)2(s) => Ca2+ + 2OH-

9. Le Châtelier’s Principle
Intuitive Method
Addition to one side results in switch to other side
Example:
Mathematical Method
reaction shifts to reactants (more AgCl(s))
When Q>K, ΔG>0 (toward reactants)
When Q<K, ΔG<0 (toward products)
Example: Q = [Ag+][Cl-]
As Ag+ increases, Q>K
AgCl(s) ↔ Ag+ + Cl-
Addition of Ag+

10. Le Châtelier’s Principle
Stress Number 1 Reactant/Products:
Addition of reactant: shifts toward product
Removal of reactant: shifts toward reactant
Addition of product: shifts toward reactant
Removal of product: shifts toward product

11. Le Châtelier’s Principle
Stress Number 1 Example:
CaCO3(s) + 2HC2H3O2(aq) ↔ Ca(C2H3O2)2(aq) + H2O(l) + CO2(g)
1. Add HC2H3O2(aq)
2. Remove CO2(g)
3. Add Ca(C2H3O2)2(aq)
4. Add CaCO3(s)
No effect because (s)

12. Le Châtelier’s Principle
Stess Number Two: Dilution
Side with more moles is favored at lower concentrations
Example: HNO2(aq) ↔ H+ + NO2-
If solution is diluted, reaction goes to products
If diluted to 2X the volume:
So Q<K, products favored

13. Le Châtelier’s Principle
Stess Number Two: Dilution – Molecular Scale View
Concentrated Solution
Diluted Solution – dissociation allows ions to fill more space H+
NO2- H+
NO2- H+
NO2- H+
NO2- H+ H+ H+
NO2- H+
NO2- H+
NO2- H+
NO2-
NO2-
NO2-

14. Le Châtelier’s Principle
Stress Number 3: Temperature
If ΔH>0, as T increases, products favored
If ΔH<0, as T increases, reactants favored
Easiest to remember by considering heat a reactant or product
Example:
OH- + H+ ↔ H2O(l) + heat
Increase in T

15. Some Le Chatelier’s Principle Examples
Looking at the reaction below, that is initially at equilibrium,
AgCl(s) ↔ Ag+(aq) + Cl-(aq) (ΔH°>0)
determine the direction (toward products or reactants) each of the following changes will result in
increasing the temperature
addition of water (dilution)
addition of AgCl(s)
addition of NaCl 15

16. Ch. 6 – Solubility Problems
Why Solubility is Important
Use in gravimetric analysis (predict if precipitation is complete enough)
Use in precipitation titrations (in next chapter)
Use in separations (e.g. separation of Mg2+ from Ca2+ in tap water for separate analysis)
Understand phase in which analytes will exist
Problem Overview
Dissolution of sparingly soluble salts in water
Dissolution of sparingly soluble salts in common ion
Precipitation problems (and selective precipitation problems) 16

17. Solubility Product Problems - Solubility in Water
Example: solubility of Mg(OH)2 in water
Solubility defined as mol Mg(OH)2 dissolved/L sol’n or g Mg(OH)2 dissolved/L sol’n or other units
Use ICE approach:
Mg(OH)2(s) ↔ Mg2+ + 2OH-
Initial
Change x +2x
Equilibrium x 2x
Note: x = [Mg2+] = solubility

18. Solubility Product Problems - Solubility of Mg(OH)2 in water
Equilibrium Equation: Ksp = [Mg2+][OH-]2
Ksp = 7.1 x = x(2x)2 = 4x3 (see Appendix F for Ksp)
x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 M
Solubility = 1.2 x 10-4 M = [Mg2+]
Conc. [OH-] = 2x = 2.4 x 10-4 M

19. Solubility Product Problems - Solubility of Mg(OH)2 in Common Ion
If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Châtelier’s principle, we know the solubility will be reduced
Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer?
No ICE table needed because, from pH, we know [OH-]eq and buffer means dissolution of Mg(OH)2 doesn’t affect pH.

20. Solubility Product Problems - Solubility of Mg(OH)2 at pH 11 – cont.
[H+] = 10-pH = M
and [OH-] = Kw/[H+] = 10-3 M
Ksp = [Mg2+][OH-]2
Moles Mg(OH)2 dissolved = moles Mg2+
[Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2
[Mg2+] = 7 x 10-6 M