1. Review of Everything
2nd 9 weeks

2. Equilibrium
Chapter 13

3. Equations Sheet

4. Equilibrium
A state when two competing reactions are canceling each other out.
H3O+ + OH H2O + H2O
You reach equilibrium when the rate of forward reaction is equal to the rate of backwards reaction.
This does NOT mean the concentration of products and reactants are equal.
It does mean there is a stable ratio between products to reactants
The above reaction is at equilibrium when [H3O+] = [OH-] = 1x10-7 mol/L. In 2 L of water there are 110 moles.

5. Equilibrium expression and constant
K is calculated form EQUILIBRIUM CONCENTRATIONS! Its values may only be calculated experimentally.   
aA + bB cC + dD
equilibrium constant = equilibrium expression
K = [C]c [D]d
[A]a [B]b
[ ] means concentration in M

6. Determine the equilibrium expressions
For the following:
Br2(g) Br(g)
N2(g) + 3H2(g) NH3(g)
H2 (g) + Br2(g) HBr(g)
HCN(aq) H+(aq) + CN-(aq)

7. K values, K is always positive
Intermediate K <K<10
Significant concentration of all substances are present.
Very Large K. K >> 1
The product concentration is very large with virtually no reactant concentration (the reaction has gone to completion).
Very Small K. K<<1
The reactant concentration is very large with virtually no product concentration.

8. Types of K Keq = equilibrium constant
Kc = equilibrium constant in terms of concentration.
Kp = equilibrium concentration in terms of pressure.
Ka = acid dissociation constant
Kb = base dissociation constant
Kw = ion-product constant for water
Ksp = solubility product constant

9. K problem
The following equilibrium concentrations were observed for the Haber process at 127o C.
[NH3] = 3.1 x 10-2 M, [N2] = 8.5 x 10-1 M, [H2] = 3.1 x10-3 M
Forward Reaction.
N2(g) + 3H2(g) NH3(g)
Reverse Reaction.
2NH3(g) N2(g) + 3H2(g)
Multiply by Factor n.
1/2N2(g) + 3/2H2(g) NH3(g)
K is unitless

10. Summary For forward reaction jA + kB lC + mD, K = [C]l [D]m [A]j [B]k
For reverse reaction jA + kB lC + mD,
K’ = K-1 = [A]j [B]k
[C]l [D]m
For reaction njA + nkB nlC + nmD
K’’ = Kn = [C]nl [D]nm
[A]nj [B]nk
 For an overall reaction of two or more steps,
Koverall = K1 x K2 x K3 x ...

11. Example
N2(g) + 3H2(g) NH3(g)

12. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.
The concentrations of pure liquids and solids are constant.
2KClO3(s) KCl(s) + 3O2(g)

13. Reaction Quotient, Q Q is calculated from INITIAL CONCENTRATIONS!
Q is useful in determining which direction a reaction must shift to establish equilibrium.
K vs. Q
K is calculated from equilibrium concentrations or pressures.
Q is calculated from initial concentrations or pressures.

14. Reaction Quotient, Q
K = Q; The system is at equilibrium. No shift will occur.
K < Q; The system shifts to the left.
Consuming products and forming reactants, until equilibrium is achieved.
K > Q; The system shifts to the right.
Consuming reactants and forming products, to attain equilibrium.

15. Change Shift Right 2 H2O ⇌ H3O+ + OH- Shift -2y +y +y Shift Left

16. K vs Q Problem
For the synthesis of ammonia at 500o C, the equilibrium constant is 6.0 x Predict the direction in which the system will shift to reach equilibrium in each of the following cases:
 N2(g) + 3 H2(g) NH3(g)
Conc. (M) [NH3]o [N2]o [H2]o
Trial x x x10-3
Trial x x x10-1
Trial x x x10-3

17. ICE tables To do equilibrium problems set up an ICE table.
Write the balanced equation. Underneath it label three rows.
Initial Concentrations (pressure)
Change
Equilibrium concentration (pressure)

18. Gas Problem
At a certain temperature a 1.00-L flask initially contained mol PCl3, 8.70 x 10-3 mol PCl5, and no Cl2. After the system had reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the following reaction
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all species and the value of K.

19. Answer PCl5(g) PCl3(g) + Cl2(g) 8.7 x10-3 M .298 M 0 -x +x +x I
It has to shift right because chlorine increased
x must = 2.0x10-3 I C E
.300 M
.0067 M

20. K
K = .3 (.002) / .0067
= .0896

21. Simplified Assumptions
*if you are using a solver function this is unnecessary.
You do have to understand the concept for a possible multiple choice
For some reactions the change will be very small compared to the initial amount.
You always have to check!

22. Example
Gaseous NOCl decomposes to form the gases NO and Cl2. At 35o C, the K = 1.6 x The initial conc. Of NOCl is .5 M. What is the equilibrium concentrations?
2NOCl(g) NO(g) + Cl2(g)
The small K value means this will favor the reactant side so there wouldn’t be a large shift to the right.

23. Problem 2NOCl(g) 2NO(g) + Cl2(g) I .50 0 0 C -2x +2x + x E .50-2x 2x x
1.6 x10 -5 = (2x)2 x / (.5-2x)2
The algebra looks a little sticky on this problem
However, if the shift is really small then
.50 -2x  .5

24. Solve now 1.6 x10 -5 = (2x)2 x / (.5)2 x = .01
Of course this is ONLY ACCEPTABLE IF x  .5
.50 – 2 (.01) = .48
If the change is less than 5% it is considered small enough to ignore
.48/.5 = 96% or a 4% change.

25. Acid Equilibrium constant
For some acid “A”
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Ka = [H3O+] [ A-]
[HA]

26. Acid Strength Strong acids dissociate completely in water.
At equilibrium, Ka >> 1 because [HA] is approx. 0.
Weak acids are mostly undissociated.
At equilibrium, Ka << 1 because [H3O+] and [ A-] are very small compared to the [HA].
The smaller the Ka value, the weaker the acid.

27. Ka values of some weak acids

28. pH pH = -log[H3O+] where neutral = 7.00 acidic < 7.00
[H3O+] = 10-pH basic >7.00
pOH = -log[OH-]
pH + pOH = 14
pK = -log K

29. Sig Figs and pH
The number of decimal places in the log value, pH value, is equal to the number of significant figures in the number that we took the logarithm of, concentration.
Calculate pH and pOH for each of the following solutions.
2.7 x 10-3 M OH-
3.4 x 10-5 M H3O+
1.54 x M OH-

30. pH problem
The pH of a sample of human blood was measured to be 7.41 at 25o C.
Calculate pOH, [H3O+], and [OH-] for the sample.

31. Calculating pH of Strong Acids
Calculate the pH of 0.10 M HNO3
Calculate the pH of 1.0 x M HCl

32. Calculating the pH of a weak acid
These require ICE tables
Again you could use simplified assumptions and test the 5% rule.
Solver functions on the calculator are completely legal to use on my test and the AP test.

33. Problem
The hypochlorite ion (ClO-) is a strong oxidizing agent found in household bleaches and disinfectants. It is also the active ingredient that forms when swimming pool water is treated with chlorine. In addition to its oxidizing abilities, the hypochlorite ion has a relatively high affinity for protons (it is a much stronger base than Cl-, for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 x 10-8). Calculate the pH of a M aqueous solution of hypochlorous acid.

34. Answer HOCl + H2O H3O+ + ClO- I .10 M 0 0 C -x +x +x E .1-x x x
Ka = [H3O+] [ClO-] x10-8 = x2
[HOCl] x
x = 5.9 x10-5
The x value is [H3O+] so pH = 4.23

35. Percent Dissociation
For a given weak acid, the percent dissociation increases as the acid becomes more dilute.

36. Base dissociation constant
For some base B
B + H2O BH+ + OH- (aq)
The Base Dissociation Constant (Kb)
Kb = [BH+] [OH-]
[B]

37. The pH of Strong Bases.
This works the same as the pH of a strong acid.
Calculate the pH of a 5.00 x 10-2 M NaOH solution.

38. pH of weak bases
Calculate the pH for a 15.0 M solution of NH3 (Kb = 1.8 x 10-5).

39. Polyprotic Acids
A polyprotic acid has more than one ionizable proton (H+).
 e.g. H2SO4, H2CO3, H3AsO4
Each successive Ka value get smaller (Ka1 > Ka2 > Ka3). Therefore, the first dissociation step makes the most significant contribution to the equilibrium concentration of [H3O+].

40. Sulfuric acid is unique in being a strong acid in its first dissociation step and a weak acid in its second step.

41. LeChâtelier’s Principle
For 1.0 M or solutions of sulfuric acid, the large concentration of H3O+ from the first dissociation step represses the second step, which can be neglected as a contributor of H3O+ ions. For dilute solutions of sulfuric acid, the second step does make a significant contribution, ICE tables must be used to obtain the total H3O+ concentration.

42. Problem Calculate the pH of .50 M H2SO4
H2SO4  HSO4- + H Ka is very large
HSO4- ⇌ SO H Ka = 1.2x10-2
The main difference from sulfuric acid from other problems will be the first step.
0.50 M H2SO4 means .50 M H+ and .5 M HSO4- is formed.
Now to the second dissociation…

43. Cont. HSO4-  SO42- + H+ I .50 M .50 M C -x +x +x E .50 –x x .50 + x
Ka = x(.5+x) / (.5-x) = 1.2 x10-2
x =
[HSO4-]= .49 M [SO42- ]= .011 M
[H+ ] = .51 M pH = .29

44. Salts

45. Salts and pH
The salts of strong acids or base will be neutral, excluding H2SO4.
To be a strong acid or base, the conjugate base or acid must have no affinity for protons or hydroxide.
That is to say it won’t run the reverse reaction under any conditions.

46. Salts of weak acids and bases
Salts of weak acids and bases will run the reverse reactions.
Salts of weak acids are weak bases
HF ⇌ H+ + F-
F- + H2O ⇌ HF + OH-
Salts of weak bases are weak acids
NH3 ⇌ NH4+ + OH-
NH4+ ⇌ NH3 + H+

47. pH of salt Would potassium chlorite be acidic or basic?
KClO2 ⇌ K+ + ClO2-
Potassium is neutral (KOH = strong base), chlorite is basic (HClO2 = weak acid)
Therefore it is basic

48. What about salts of a weak acid and a weak bases…
If you have NH4F
You have to look at the Ka value of NH4+ and compare it to the Kb value of F- to see which weak acid/base is stronger.
The larger the k value the stronger the acid/base.

49. Ka and Kb
The Ka and Kb of a weak acid and its conjugate base or vice versa are related and easily calculated from one another
For a weak acid and its conjugate base or a weak base and its conjugate acid
Ka x Kb = Kw

50. Problem
Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5.
Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 x 10-4.

51. Leveling Effect
It is difficult to differentiate the strengths of strong acids or bases because of water.
Water can act like an acid or a base.
By LeChâtelier’s principle, the stronger the acid or base, the more water will shift to neutralize it.
Weak acids or bases are limited by their conjugate bases or acids, strong acids and base are limited by water.

52. How Buffers Work
A solution that resists change in pH when acids or bases are added.
The equilibrium concentration of H3O+ is determined by the ratio of [HA]/[A-].
HA + H2O ⇌ A- + H3O+
 Ka = [H3O+] [A-] so,
[HA]
[H3O+] = Ka [HA]
[A-]

53. Cont.
If OH- is added to the system, HA is converted to A-, and the ratio of [HA]/[A-] decreases.
However, if the amounts of HA and A- originally present are very large compared with the amount of OH- added, the change in the [HA]/[A-] ratio will be very small.
The reverse is true for adding H3O+ to a system

54. The Henderson-Hasselbach Equation.
(on the sheet)
  [H3O+] = Ka [HA]
[A-]
  -log[H3O+] = -log Ka -log ([HA]/[A-])
  pH = pKa - log ([HA]/[A-])
 pH = pKa + log ([A- ]/[HA])
(not on the sheet)
pOH = pKb + log ([HB+ ]/[B])
This equation may be useful, although it not necessary.

55. pKa or pKb *Also on the sheet
You determine the pKa or pKb by taking the -log of the Ka or Kb

56. Problems
Calculate the pH of a buffer of 0.50 M HF and 0.45 M F- (a) before and (b) after the addition of 0.40 g NaOH to 1.0 L of the buffer.
Ka of HF = 6.8 x 10-4.

57. Answer part (b) The sodium hydroxide will react with the HF
.40 g NaOH x 1 mol/ g = mol
It is 1 L so it is easy to convert to moles
Assume this reaction goes to completion
HF + OH-  H2O + F-
.50 mol mol
.01mol
.49 mol mol
Now plug these values back into the equilibrium

58. Titration and pH curves

59. Titration and pH Curves.
A titration curve is a plot of pH vs. volume of added titrant.

60. Titrations
Because titrations involve small concentrations, and mL are of often used in titrations, and millimoles, or mmol.
Molarity = mmol/mL
The equivalence point is when [H3O+]=[OH-].
All volumes in a titration are considered to be additive.
Always label the equivalence point and for weak acids or bases the half equivalence point, where pH = pKa.

61. Strong Acid-Strong Base Titration Curves.
Before the addition.
pH is calculated directly from the initial concentration.
Additions before the equivalence point.
Construct a “stoichiometry” reaction table. Determine MOLES of acid in excess (not neutralized).
Divide MOLES by the TOTAL VOLUME to obtain [H3O+].
Calculate the pH.

62. Strong Acid-Strong Base Titration Curves.
Additions at the equivalence point.
The pH ALWAYS is equal to 7.00 when [H3O+] = [OH-].
Additions beyond the equivalence point.
Construct a “stoichiometry” reaction table. Determine MOLES of base in excess (not neutralized).
 Divide MOLES by the TOTAL VOLUME to obtain [OH-].
 Calculate the pOH, then the pH.

63. Weak Acid-Strong Base Titration Curves.
Before the addition.
Construct an “equilibrium” reaction table ONLY!
Ka = [A-] [H3O+] to obtain [H3O+].
[HA]
 Calculate the pH.

64. Weak Acid-Strong Base Titration Curves.
Additions before the equivalence point.
Construct a stoichiometry reaction table.
Determine MOLES of acid in excess (not neutralized) and MOLES of conjugate base formed.
Divide MOLES by the TOTAL VOLUME to obtain [H3O+] and [A-].
Construct an “equilibrium” reaction table.
 Ka = [A-] [H3O+] and obtain [H3O+].
[HA]
 Calculate the pH.

65. Additions at the equivalence point.
Construct a stoichiometry reaction table.
 Determine MOLES of conjugate base formed. Divide MOLES by the TOTAL VOLUME to obtain [A-]. Calculate Kb
(Ka x Kb = Kw).
Construct an “equilibrium” reaction table, reacting the conjugate base with water.
Kb = [OH-] [BH+] and obtain [OH-].
[B]
Calculate the pOH, then the pH.
The equivalence point is ALWAYS >7!

66. Additions beyond the equivalence point
Construct a “stoichiometry” reaction table.
Determine MOLES of base in excess (not neutralized) and the MOLES of conjugate base.
Divide MOLES by the TOTAL VOLUME,
 Because [OH-]excess >> [OH-]conj. base, use [OH-]excess to calculate pOH, then the pH.

67. Problem
50.0 mL of 0.10 M acetic acid (Ka = 1.8 x 10-5) are titrated with 0.10 M NaOH. Calculate the pH after the additions of 0, 10, 25, 40, 50, 60, and 75 mL samples of NaOH.
Then, construct a titration curve and label it properly.

68. Answer

69. Answer II

70. Weak base strong acid
Will be the same as weak acid strong base, just reverse everything

71. Solubility Equilibria and the Solubility Product
Ksp is the solubility product constant for equilibrium between solid solute and dissolved ions.
For MpXq  p Mn+ +q Xz-
Ksp = [Mn+]p [Xz-]q

72. Writing Ksp for Slightly Soluble Ionic Compounds
Note, using solubility rules these compounds are insoluble.
Ksp would be very high for anything that is “soluble” and very low for anything that in “insoluble”.
Write the solubility product expression for:
Magnesium carbonate
Iron(II) hydroxide
Calcium phosphate
Silver sulfide

73. Answer Ksp= [Mg2+][CO32-] Ksp= [Fe2+][OH-]2 Ksp= [Ca2+]3[PO43-]2
Ksp= [Ag+]2[S2-]

74. Predicting the Formation of a Precipitate: Kspvs. Qsp
Dissociation equations
Solid (precipitate)  Dissolved
If Ksp = Qsp, then the solution is saturated and no change occurs.
If Ksp < Qsp, then a precipitate forms until the solution is saturated.
If Ksp> Qsp, then the solution is unsaturated and no precipitate forms.