1. Chem 106: Class/ Lab Week 11 Pick a vial and a plastic dropper
Using the vial number, sign-in on the Lab roster
TODAY:
Fluid Exchange
(Course/ Lab Manual pp )
2) Acid-Base Equilibrium Experiment
(Course/ Lab Manual pp ) Due & signed Today

2. Completed Report & Post Lab Questions
Due Today: Course/ Lab Manual pg. 46 Plus Post Lab
Post Lab to turn in individually: [Each Partner turns in a completed form]

3. Chem 106: Class/ Lab Week 11 Follow Instructions

4. Acid-Base Indicators

5. Acid-Base Indicators

6. Water evaporates, and Water is an Acid & a Base
What is equilibrium?
Course/Lab Manual pp.91-92
H2O(l) + energy  H2O(g)

7. CH3COOH + H2O  CH3COO + H3O+
What is equilibrium?
Water is neutral
until something is added
HCl + H2O  Cl + H3O+
CH3COOH + H2O  CH3COO + H3O+
HCl (“Strong Acid”)
1.0 M HCl solution
K = [Products ][Reactants]
K = 1.0 ÷ 0 = ∞ (100% reacts)
1.0 M HOAc (vinegar) solution
K ≈ (<<< 100% reacts)
CH3COOH (“Weak Acid”)

8. Water is neutral It is an Acid and a Base
What is equilibrium?
Water is neutral It is an Acid and a Base
until something is added
(“Equilibrium Constant”) K = [Products ][Reactants]
Reactants  Products
For example weak acids
EXPERIMENT #1
Amount
TIME
Reactant
Product 1
100 2 75 25 3 59 41 4 48 42 5 6 37 63 7 34 64 8 32 68 9 31 69 10 30 70 11 29 71 12 13
EXPERIMENT #2
Amount
TIME
Reactant
Product 1
100 2 10 90 3 16 84 4 20 80 5 22 78 6 24 76 7 25 75 8 26 74 9 27 73 28 72 11 29 71 12 13
K = [71][29]
K = 2.4
K = [71][29]
K = 2.4

9. Equilibrium & Currency Exchange Rates
The US $ is the current common global currency
The cost of a McDonald’s Big Mac in US $, country by country, provides a basis for the relative global market equilibrium value of the local currency
The “Big Mac Index” was introduced by the Economist magazine in September 1986 by Pam Woodall and is published annually
Currently, US $20 would only buy 3 Big Macs in Geneva, Switzerland, but 9 in Moscow, Russia
Geneva  US  Moscow

10. Water is neutral It is an Acid and a Base
What is equilibrium?
Reactants  Products
(“Equilibrium Constant”) K = [Products ][Reactants]

11. Water is neutral It is an Acid and a Base
What is equilibrium?
H2O + H2O  H3O+ + OH
conj conj
acid 1 base acid base 1
Kw = 1  1014 at 25°C
Kw = [H3O+ ][OH] = [1  107M][1  107M]

13. Th pH Values of Some Familiar Aqueous Solutions
(TODAY’S LAB EXPERIMENT)
[H3O+]
[OH-]
[H3O+]>
[OH-]
[H3O+]<
[OH-]
acidic
solution
neutral
solution
basic
solution
[H3O+] =
[OH-]

14. BUFFERS Weak Acid- Weak Base Systems Example:
H2CO3(aq) / HCO3-1(aq) / CO3-2(aq)
CO2(g) + H2O (l)  HCO3-1(aq) + H+1(aq) CO3-2(aq) + H+1(aq)

15. Two VERY IMPORTANT Acid-Base: Bicarbonate Buffer Systems
H2CO3(aq) / HCO3-1(aq) / CO3-2(aq)
Two VERY IMPORTANT
Acid-Base: Bicarbonate Buffer Systems
Blood: a human’s blood serum volume is relatively small, 4-6 Liters with a narrow pH range, pH = 7.35 – 7.45; pH is maintained through buffering (homeostasis) Have you ever had respiratory alkalosis during an exam?
Oceans: an extraordinarily large volume of a “salt water” solution with a pH ~ 8.1; maintained through buffering
CO2(g) + H2O (l)  HCO3-1(aq) + H+1(aq) CO3-2(aq) + H+1(aq)

16. Human & Oceanic Bicarbonate Buffer Systems

17. EQUILIBRIUM CO2 & Oceanic Bicarbonate Buffering
CO2(g) + H2O (l)  HCO3-1(aq) + H+1(aq) CO3-2(aq) + H+1(aq)
Oceans: pH ~ 8.1 and falling
Increasing CO2 is decreasing ocean pH; long term effects?

18. Completed Report & Post Lab Questions
Due Today: Course/ Lab Manual pg. 46 Plus Post Lab
Post Lab to turn in individually: [Each Partner turns in a completed form]

19. Calculations Distillation (Course/ Lab Manual pg. 46)
Theoretical & Percent Yield
Calculations Distillation (Course/ Lab Manual pg. 46)

20. Calculations
Example
24.55 g
52.2 mL
0.990 g/mL

21. Calculations
Example
24.55 g
4.5 %
52.2 mL
0.990 g/mL

22. Theoretical Yield Calculation
24.55 g
? g (theoretical) 4
Molar mass = g/mol
Molar mass = g/mol
? mol C2H5OH = 4 x mol sucrose
= mol
? mol sucrose = g / g/mol
= mol
? g (theoretical) = mol C2H5OH x g/mol
= g
__________________
? g (actual) = [4.5 % ,that is: 4.5/100] x 52.2 mL x g/mL
= 2.33g
% Yield = g (actual) / g (theoretical) x 100
= 17.6 %