1. Keq Ksp Ka Kb Kw opposing Ch. 17/18/14/15 H. Chem. rate
When two ____________ processes equal
eachother in ________
rate
Chemical Equilibrium
establishing it
manipulating it
quantifying it
Reactions are _________
reversible
Le Chatelier’s Principle
__________________________________________
Consequences at Equilibrium
Mass Action Expression
An equilibrium system that is stressed will respond to a stress in such a way to counteract the stress
[P]x/[R]y
Rates of forward and reverse are
same/constant
Keq
Conc. Changes
maybe!!
[ ]’s are
same/constant
Ksp
[ ]’s are dictated by
Temp. Changes
stoich.
__________
(pH and pOH) Ka Kb
Acids and Bases (described through equilibrium)
Pressure Changes Kw
Titrations

2. Factors which influence….*
Chemical Kinetics
P.E.
Collision Theory
Break and Make Bonds
Time
Enough Energy
Proper Angle
Effective Collisions
activated complex forms*
Factors which influence….
Activation Energy
Temp.
(lowers the)
P.E. Diag.
Conc.
Catalysts
Surface Area
Nature of substance

3. CoCl2(s) ↔ → Co2+(aq) + 2Cl-(aq) DH= + Co(H2O)62+(aq) + 4Cl-(aq) ↔
The dissociation of cobalt (II) chloride
CoCl2(s) ↔
→ Co2+(aq) + 2Cl-(aq)
DH= +
Complex ions equilibrium reaction
Co(H2O)62+(aq) + 4Cl-(aq) ↔
→ CoCl42-(aq) + 6 H2O(l)
------Color Change------
Stress: Counteraction: Rxtion Favored: Predicted: Observed: Did you get it?
Add HCl
Remove Cl-
Forward
More blue
More blue
Yes!!!!
(▼This takes away Cl- as the ppt AgCl is made!)
Add AgNO3
Make Cl-
Reverse
More pink
More pink
Yes!!!!
Add heat
Take away energy
Forward
More blue
Yes!!!!

4. FeSCN2+(aq) ↔ Fe3+(aq) + SCN-(aq)
Init:
2.00 M
0.0 M
0.0 M
FeSCN2+(aq) ↔ Fe3+(aq) + SCN-(aq)
Eq:
1.96 M
4.3 x 10-2 M
4.3 x 10-2 M
Find the equilibrium constant for this reaction at 25.0° C if 4.0 moles of FeSCN2+ was present in 2.0 L of solution initially (no products to start) and at equilibrium the concentration of the SCN- was 4.3 x 10-2 M.
Keq=
[Fe3+] [SCN-]
= 9.4 x 10-4
[FeSCN2-]

5. HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
weak acid
HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq)
Write the reaction equation involving HF (a weak acid) whose mass action expression would be equal to Ka.
What would the concentration of an HF solution be if in a titration with 2.0 M LiOH, 30.0 mL was needed to neutralize 20.0 mL of the acid?
[HF] =3.0 M
The hydronium ion concentration in this solution of HF is M. What is the pH of this solution?
pH = 1.33
What is the pOH of this solution?
pOH = 12.67
What is OH- concentration of this solution?
H.W Answers:
#2.) 6.67 x 10-3 M NaOH
#3.) M HCl
[OH-] = 2.13 x 10-13