AP Chemistry Summary of Equilibrium Essentials
Summary of Equilibrium Essentials
1. General form of an Equilibrium Constant Expression:
aA + bB + … D cC + dD + …
Kc = [C]c[D]d….
[A]a[B]b….
à 2003 Exam Question 1(a)
à 2002 Exam Question 1(b)
à 2000 Exam Question 1(a)
2. K > 1 products favored
K < 1 reactants favored
3. Excluded: solids; pure liquids; water (in aqueous solution)
à 2002 Exam Question 8(d)
4. Kp = Kc(RT)Dn
5. Typical question: Given Kc and the starting concentrations of reactants, find concentrations
(or pH) at equilibrium.
e.g. Kc for acetic acid = 1.754 x 10-5. Find the pH of a 0.100 M solution of acetic acid.
à 2002 Exam Question 1(b)
à 2000 Exam Question 1(b)
6. Problem solving: Learn when to make an approximation (needed for multiple choice
questions). 5% rule usually works when value of K is 10-3 or smaller than value of known
concentrations.
e.g. A D B + C K = 3.0 x 10-6
If [A] = 5.0 M; find [C] at equilibrium.
Check – If > 5% use the quadratic equation: (unlikely scenario in the actual AP Chem exam)
ax2 + bx + c = 0
x = -b +-(b2 -4ac)1/2/2a
7. Equilibrium constant for a reverse reaction 1/K the value of the forward reaction.
à 2000 Exam Question 1(e)
8. Hess’s Law: Koverall = K1 x K2 (used commonly for diprotic acids)
9. If out of equilibrium: Calculate the reaction quotient (Q) similar to the way an equilibrium
constant would be found. If:
Q < K forward reaction occurs to reach equilibrium
Q > K reverse reaction occurs to reach equilibrium
10. Le Châtelier’s Principle: effect of changes in concentration, pressure, and temperature
à 2002 Exam Question 8 (b)
11. Acid strength – know the 6 strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4
(removal of the first H+ only)
a) binary acids – acid strength increases with increasing size and electronegativity of the “other
element”. (Note: Size predominates over electronegativity in determining acid strength.)
Examples: H2Te > H2O and HF > NH3
b) Oxyacids – Acid strength increases with increasing:
i) electronegativity
ii) number of bonded oxygen atoms
iii) oxidation state of central atom.
e.g. HClO4 [O3Cl(OH)] is very acidic
NaOH is very basic
à 2002 Exam Question 1(e)
Acid strength also increases with DECREASING radii of the “central atom”.
e.g. HOCl (bond between Cl and OH is covalent-acidic)
HOI (bond between I and OH is ionic-basic)
12. Acid Ionization Constant (Ka):
HA + H2O D H3O+ + A- Ka = [H3O+][A-]
[HA]
e.g. HF + H2O D H3O+ + F- Ka = [H3O+][F-]
[HF]
13. Base Ionization Constant (Kb):
B + H2O D BH+ + OH- Kb = [BH+][OH-]
[B]
e.g. F- + H2O D HF + OH- Kb = [HF][OH-]
[F-]
NH3 + H2O D NH4+ + OH- Kb = [NH4+][OH-]
[NH3]
14. Ka x Kb = Kw = 10-14 only applies for conjugate acids and bases
15. Percent ionization = [H+]equilibrium x 100%
[HA]initial
16. Lewis Acids and Bases:
Lewis Acid = electron-pair acceptor
Lewis Base = electron-pair donor
Classical example:
In complex ion formation, transition metal ions are Lewis acids, and ligands are Lewis bases.
e.g. Cu2+ + 4 NH3 D Cu(NH3)42+
Cu2+ acts as a n acid; NH3 acts as a base
17. Buffers:
Similar concentrations of a weak acid and its conjugate base
OR
Similar concentrations of a weak base and its conjugate acid
If these concentrations are large in comparison to SMALL amounts of strong acid or base
added, equilibrium will be shifted slightly and pH change resisted.
Consider:
HAD H+ + A-
Ka = [H+] [A-]
[HA]
[H+] = Ka [HA]
[A-]
Take the negative log:
pH = pKa + log [A-] Henderson-Hasselbach equation
[HA]
à 2002 Exam Question 1(d)
à 2001 Exam Question 3(d)
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B + H2O D HB+ + OH-
Kb = [OH-] [HB]
[B]
[OH-] = Kb [B]
[HB+]
pOH = pKb + log [HB+] Henderson-Hasselbach equation
[B]
18. Polyprotic Acids: H3PO4, H2SO4, H2C2O4, etc.
19. Equivalence Point = the point at which stoichiometric amounts of reactants have reacted.
NOTE: This only occurs at pH = 7.00 for the reaction of a strong acid with a strong base.
The equivalence point will occur ABOVE pH = 7.00 for a weak acid/strong base titration
(the conjugate base of the weak acid will react with water, producing OH-)
The equivalence point will occur BELOW pH = 7.00 for a weak base/strong acid titration
(the conjugate acid of the weak base will react with water to produce H3O+).
à 2003 Exam Question 1(d)
à 2002 Exam Question 1(c)(ii)
à 2001 Exam Question 3(c)
à 2000 Exam Question 8(d)
20. Indicators: select based on the pH at the equivalence point
à 2003 Exam Question 1(e)
à 2000 Exam Question 8(d)
21. Titration Curves:
a) strong acid/strong base b) weak acid/strong base
H+ + OH- à H2O HA + OH- D A- + H2O
Equivalence point is at pH = 7.00 Equivalence point is at pH > 7.00
Maximum buffering at the
beginning/bottom of the curve
c) weak base/strong acid d) diprotic acid/strong base
B + H3O+ D BH+ + H2O H2A + OH- D HA- + H2O
Maximum buffering is at the HA- + OH- D A2- + H2O
beginning/top of the curve Multiple equivalence points and two
buffer regions
22. Solubility Product (Ksp)
e.g. Co(OH)2(s) D Co2+ + 2OH- Ksp = [Co2+][OH-]2
(don’t forget – molar concentration of OH- is twice the solubility, e.g. (x)(2x)2.)
e.g. Typical AP Exam question: Solubility of Ag2SO4 is 5.0 g/L (0.016 mol/L).
Find the Ksp of Ag2SO4. (Answer: Ksp = 1.5 x 10-5)
à 2001 Exam Question 1(a)(i),(ii),(iii)
23. Ion Product (Qi) – equivalent to the “reaction quotient”
Qi < Ksp all ions in solution; more solid will dissolve
Qi = Ksp equilibrium – solution is saturated
Qi > Ksp precipitation will occur until Qi = Ksp
24. Solubility of any salt which contains a basic anion is influenced by pH:
e.g. Consider a solution which is 1M in each of Fe3+ and H+ (i.e. pH = 0).
Will Fe(OH)2 precipitate? Answer: NO
Fe(OH)3 D Fe3+ + 3OH-
Qi = [Fe3+][OH-]3
If [H+] = 1M, then [OH-] = 10-14 M
Qi = (1)(10-14)3 = 10-42
Since Ksp for Fe(OH)3 = 3 x 10-39, precipitation does not occur.
However, what if [H+] = 10-5 M? (pH = 5)
In this case [OH-] = 10-9 M
Qi = (1)(10-9)3 = 10-27
Since Qi > Ksp, Fe(OH)3 will precipitate
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