1. ACIDS AND BASES
SVANTE ARRHENIUS
Proposed the first definitions of acids and bases
ACID – A compound that produces hydrogen ions in a water solution
HCl (g) → H+(aq) + Cl-(aq)
BASE – A compound that produces hydroxide ions in a water solution
NaOH (s) → Na+(aq) + OH-(aq)
1E-1 (of 24)

2. STRENGTHS OF ACIDS AND BASES
STRONG ACID – One in which all molecules release hydrogen ions in solution
HCl, HBr, HI, and oxyacids with at least 2 more O’s than H’s
HNO3
H2SO4
HClO4
H2CO3
WEAK ACID – One in which all molecules do not release hydrogen ions in solution
All other acids
STRONG BASES – One in which all formula units release hydroxide ions in solution
Alkali metal hydroxides, dilute solutions of Ca(OH)2, Sr(OH)2, Ba(OH)2
WEAK BASE – One in which all formula units do not release hydroxide ions
All other bases, including NH3
1E-2 (of 24)

3. For metallic hydroxides, the higher the metal ion’s oxidation number, the less basic its hydroxide
Cr(OH)2
base
Cr(OH)3
amphoteric
Cr(OH)6
acid
AMPHOTERIC – A substance that can act as an acid or a base
1E-3 (of 24)

4. 1923
Expanded the definitions of acids and bases
THOMAS LOWRY
1E-4 (of 24)

5. 1923
Expanded the definitions of acids and bases
JOHANNES BRØNSTED
ACID – A hydrogen ion (or proton) donor
BASE – A hydrogen ion (or proton) acceptor
1E-5 (of 24)

6. HCl H2O →
1E-6 (of 24)

7. HCl H2O →
Cl H3O+
acid
base
HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water - +
1E-7 (of 24)

8. NH H2O →
1E-8 (of 24)

9. NH OH-
NH H2O →
base
acid
Water is amphoteric + -
1E-9 (of 24)

10. Acids turn into bases, and bases turn into acids
HCl H2O Cl H3O+ →
acid
base
1E-10 (of 24)

11. Acids turn into bases, and bases turn into acids
HCl H2O Cl H3O+ ←
acid
base
conjugate base
of HCl
conjugate acid
of H2O
NH H2O NH OH- ← →
base
acid
conjugate acid
of NH3
conjugate base
of H2O
Strong acids will lose H+’s 100%,  their conjugates gain H+’s 0%
Weak bases will gain H+’s a small %,  their conjugates lose H+’s a small %
1E-11 (of 24)

12. Conjugate base of HClO4
When HClO4 acts as an acid, it becomes:
ClO4-
Strong Acid
Non Base
Conjugate acid of CH3NH2
When CH3NH2 acts as a base, it becomes:
CH3NH3+
Weak Base
Weak Acid
1E-12 (of 24)

13. AUTOIONIZATION OF WATER
Water ionizes itself to a small extent
1E-13 (of 24)

14. AUTOIONIZATION OF WATER
Water ionizes itself to a small extent + -
2H2O (l) → H3O+(aq) + OH-(aq)
Makes solutions acidic
Makes solutions basic
Equal amounts of H3O+ and OH- make a solution NEUTRAL
 pure water is neutral
1E-14 (of 24)

15. 2H2O (l) ⇆ H3O+ (aq) + OH- (aq)
energy +
Write the equilibrium constant expression for the autoionization of water.
Keq = [H3O+][OH-]
ION-PRODUCT CONSTANT FOR WATER (Kw) – The equilibrium constant for the ionization of water
The ionization of water is an endothermic process
Temp. (°C) Kw 20 25 30
0.68 x 10-14
1.00 x 10-14
1.47 x 10-14
1E-15 (of 24)

16. 2H2O (l) ⇆ H3O+ (aq) + OH- (aq)
Find [H3O+] and [OH-] in pure water at 25°C.
2H2O (l) ⇆ H3O+ (aq) OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
+ x
+ x x x
Kw = [H3O+][OH-]
1.00 x M2 = x2
1.00 x 10-7 M = x
= [H3O+] = [OH-]
1E-16 (of 24)

17. THE pH SCALE
pH – The negative logarithm of [H3O+] of a solution
The common logarithm of a number is:
the exponent to which 10 must be raised to equal the number
100
0.01
0.001
0.002
100 = 102
0.01 = 10-2
0.001 = 10-3
0.002 =
log 100 = 2
log 0.01 = -2
log = -3
log =
pOH – The negative logarithm of [OH-] of a solution
1E-17 (of 24)

18. Calculate the pH of orange juice if its [H3O+] = 2.5 x 10-4 M.
pH = -log[H3O+]
= -log(2.5 x 10-4 M)
= 3.6
incorrect sig fig’s
= -[log(2.5) log(10-4)]
= -[ … ] =
2 sig fig’s
exponents are exact
2 sig fig’s
infinite sig fig’s
addition problem
 answer = 2 decimal places
For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)
1E-18 (of 24)

19. Calculate the pH and pOH of pure water at 25ºC.
For pure water:
[H3O+] = x 10-7 M
[OH-] = x 10-7 M
pH = -log[H3O+]
pOH = -log[OH-]
= -log(1.00 x 10-7 M)
= -log(1.00 x 10-7 M) = =
1E-19 (of 24)

20. In any water solution:
Kw = [H3O+][OH-]
1.00 x = [H3O+][OH-]
at 25°C
1.00 x = [OH-]
_____________
[H3O+]
For orange juice:
[H3O+] = x 10-4 M
1.00 x M2 = [OH-]
_________________
2.5 x 10-4 M
= x M
1E-20 (of 24)

21. 7 8 9 10 11 12 13 14 6 5 3 4 2 1 -1 15
< 7 is Acidic
pH 7 = Neutral
> 7 is Basic
[H3O+] = M
[OH-] = M
[H3O+] = M
[H3O+] = M
[OH-] = M
[OH-] = M
[H3O+] = 101 M
[OH-] = M
1E-21 (of 24)

22. In any water solution:
Kw = [H3O+][OH-]
log Kw = log ([H3O+][OH-])
-log Kw = -log ([H3O+][OH-])
-log Kw = - log[H3O+] - log[OH-]
pKw = pH + pOH
∴ = pH + pOH
at 25°C, -log(1.00 x 10-14) =
1E-22 (of 24)

23. Calculate the pH and pOH of soda if its [H3O+] = 1.6 x 10-3 M.
pH = -log[H3O+]
pH + pOH = pKw
= -log(1.6 x 10-3 M)
pOH = pKw - pH
= 2.80 = =
1E-23 (of 24)

24. Calculate the [H3O+] of blood, which has a pH of 7.4.
pH = -log [H3O+]
-pH = log [H3O+]
antilog (-pH) = [H3O+]
antilog (-7.4) = [H3O+]
= M
= x 10-8 M
= 4 x 10-8 M
For logarithmic numbers, only the digits after the decimal point are significant ,  this is a 1 significant figure number
1E-24 (of 24)

26. IONIZATION OF ACIDS
HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)
Write the Keq expression for the ionization of hydrofluoric acid.
Keq
= [H3O+][F-]
____________
[HF]
ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing
1F-1 (of 25)

27. IONIZATION OF ACIDS
HF (aq) + H2O(l) ⇆ H3O+(aq) + F-(aq)
Write the Keq expression for the ionization of hydrofluoric acid Ka
= [H3O+][F-]
____________
[HF]
ACID IONIZATION CONSTANT (Ka) – The equilibrium constant for an acid ionizing
1F-2 (of 25)

28. Acids are
a) strong if every acid molecule gives up a hydrogen ion
Ka = large
b) weak if less than every acid molecule gives up a hydrogen ion
Ka = small
1F-3 (of 25)

29. ACID IONIZATION CONSTANTS HX (aq) + H2O(l) ⇆ H3O+(aq) + X-(aq)
Ka = [H3O+][X-]
____________
[HX]
Strongest acid?
HCl
Most readily releases hydrogen ions
Weakest acid?
CH4
HCN
H2O
Least readily releases hydrogen ions
1F-4 (of 25)

30. CALCULATING THE pH OF STRONG ACID SOLUTIONS
Assume complete ionization or dissociation
Calculate the pH of M nitric acid.
HNO3 (aq) H2O (l) → H3O+ (aq) NO3- (aq)
Initial M’s
Change in M’s
Final M’s
0.010 ~0
- x
+ x
+ x
0.010
0.010
pH = -log[H3O+]
= -log(0.010 M) =
1F-5 (of 25)

31. CALCULATING THE pH OF WEAK ACID SOLUTIONS
Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium
Calculate the pH of M nitrous acid.
HNO2 (aq) H2O (l) ⇆ H3O+ (aq) NO2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.010 ~0
- x
+ x
+ x x x x
Ka = [H3O+][NO2-]
________________
[HNO2]
Ka = x2
_____________
(0.010 – x)
the Ka for HNO2 can be looked up on Handout 3
: x 10-4
1F-6 (of 25)

32. 4.0 x 10-4 = x2
_____________
(0.010 – x)
If the Ka is < 10-2 the acid does not ionize much, so you may assume that x is very small, and ignore it when it is subtracted from the initial molarity
4.0 x 10-4 = x2
_______
0.010
2.00 x 10-3 = x
For x values that are >5% of the original acid molarity, they should be put back in place of x in the denominator of the Ka expression, and solved again
2.00 x 10-3 M
_________________
0.010 M
x 100
= 20.%
1F-7 (of 25)

33. 4.0 x 10-4 = x2
_____________
(0.010 – x)
4.0 x 10-4 = x2
_________________________
(0.010 – 2.00 x 10-3)
1.79 x 10-3 = x
Repeat until the answer (to 2 sig fig’s) is the same twice in a row
4.0 x 10-4 = x2
_________________________
(0.010 – 1.79 x 10-3)
1.81 x 10-3 = x
= [H3O+]
pH = -log[H3O+]
= -log(1.81 x 10-3 M) =
1F-8 (of 25)

34. Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.
HC2H3O2 (aq) H2O (l) ⇆ H3O+ (aq) C2H3O2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.20 ~0
- x
+ x
+ x x x x
Ka = [H3O+][C2H3O2-]
____________________
[HC2H3O2]
1.8 x 10-5 = x2
____________
(0.20 – x)
1.8 x 10-5 = x2
______
0.20
1.90 x 10-3 = x
1F-9 (of 25)

35. Calculate the pH of 0.20 M acetic acid if its Ka = 1.8 x 10-5.
HC2H3O2 (aq) H2O (l) ⇆ H3O+ (aq) C2H3O2- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.20 ~0
- x
+ x
+ x x x x
1.90 x 10-3 M
________________
0.20 M
pH = -log(1.90 x 10-3 M)
= 2.72
x 100
= %
Calculate the percent ionization of acetic acid.
1.90 x 10-3 M
________________
0.20 M
x 100
= %
1F-10 (of 25)

36. CALCULATING THE Ka OF A WEAK ACID FROM pH
A M solution of the weak acid HA has a pH Calculate its Ka.
HA (aq) H2O (l) ⇆ H3O+ (aq) A- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.500
~ 0
- x
+ x
+ x x x x
Ka = [H3O+][A-]
_____________
[HA]
Ka = x2
______________
(0.500 – x)
x = [H3O+]
= antilog (-pH)
= antilog (-2.010)
= M
1F-11 (of 25)

37. CALCULATING THE Ka OF A WEAK ACID FROM pH
A M solution of the weak acid HA has a pH Calculate its Ka.
Ka = ( M)2
______________________________
(0.500 M M)
= x 10-4
What is the pKa of HA?
pKa = -log Ka
= -log(1.947 x 10-4) =
The smaller the Ka (so the bigger the pKa) the weaker the acid
1F-12 (of 25)

38. CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION
A M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.
HX (aq) H2O (l) ⇆ H3O+ (aq) X- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.500
~ 0
- x
+ x
+ x x x x
Ka = [H3O+][X-]
_____________
[HX]
Ka = x2
_____________
(0.500 – x)
3.15 = x (100)
_______
0.500
= x
1F-13 (of 25)

39. CALCULATING THE Ka OF A WEAK ACID FROM PERCENT IONIZATION
A M solution of the weak acid HX is 3.15% ionized. Calculate its Ka.
Ka = ( M)2
____________________________
(0.500 M M)
= x 10-4
What is the pKa of HX?
pKa = -log Ka
= -log(5.123 x 10-4) =
1F-14 (of 25)

40. IONIZATION OF BASES
NH3 (aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)
Write the Keq expression for the ionization of ammonia.
Keq
= [NH4+][OH-]
______________
[NH3]
BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing
1F-15 (of 25)

41. IONIZATION OF BASES
NH3 (aq) + H2O(l) ⇆ NH4+(aq) + OH-(aq)
Write the Keq expression for the ionization of ammonia. Kb
= [NH4+][OH-]
______________
[NH3]
BASE DISSOCIATION CONSTANT (Kb) – The equilibrium constant for a base ionizing
1F-16 (of 25)

42. Bases are
a) strong if every molecule/ion accepts a hydrogen ion
Kb = large
b) weak if less than every molecule/ion accepts a hydrogen ion
Kb = small
1F-17 (of 25)

43. CALCULATING THE pH OF STRONG BASE SOLUTIONS
Assume complete ionization or dissociation
Calculate the pH of M sodium hydroxide.
NaOH (aq) → Na+ (aq) OH- (aq)
Initial M’s
Change in M’s
Final M’s
0.0010 ~0
- x
+ x
+ x
0.0010
0.0010
pOH = -log[OH-]
= -log( M)
= 3.00
pH + pOH = pKw
pH = pKw - pOH = =
1F-18 (of 25)

44. CALCULATING THE pH OF WEAK BASE SOLUTIONS
Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium
Calculate the pH of a 0.10 M ammonia solution.
NH3 (aq) H2O (l) ⇆ NH4+ (aq) OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10
~ 0
- x
+ x
+ x x x x
Kb = [NH4+][OH-]
_______________
[NH3]
1.8 x 10-5 = x2
____________
(0.10 – x)
the Kb for NH3 can be looked up on Handout 3
: x 10-5
1F-19 (of 25)

45. CALCULATING THE pH OF WEAK BASE SOLUTIONS
Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium
Calculate the pH of a 0.10 M ammonia solution.
NH3 (aq) H2O (l) ⇆ NH4+ (aq) OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10
~ 0
- x
+ x
+ x x x x
1.8 x 10-5 = x2
______
0.10
1.34 x 10-3 M (100)
_________________
0.10 M
= %
1.34 x 10-3 = x
1F-20 (of 25)

46. CALCULATING THE pH OF WEAK BASE SOLUTIONS
Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium
Calculate the pH of a 0.10 M ammonia solution.
NH3 (aq) H2O (l) ⇆ NH4+ (aq) OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10
~ 0
- x
+ x
+ x x x x
pOH = -log[OH-]
= -log(1.34 x 10-3 M) =
pH + pOH = pKw
pH = pKw - pOH = =
1F-21 (of 25)

47. POLYPROTIC ACIDS
Polyprotic acids have Ka values for the ionization of each H+
H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)
Ka1 = [H3O+][HCO3-]
__________________
[H2CO3]
Ka1 = 4.3 x 10-7
HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO32- (aq)
Ka2 = [H3O+][CO32-]
_________________
[HCO3-]
Ka2 = 5.6 x 10-11
Successive H+’s are harder to remove
∴ H2CO3 is a stronger acid than HCO3-
1F-22 (of 25)

48. Find the concentrations of each species in a 0.10 M H2CO3 solution.
H2CO3 (aq) H2O (l) ⇆ H3O+ (aq) HCO3- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10 ~0
- x
+ x
+ x x x x
Ka1 = [H3O+][HCO3-]
_________________
[H2CO3]
4.3 x = x2
____________
(0.10 – x)
x = x 10-4
[H2CO3] = M – M = M
[H3O+] = = M
[HCO3-] = = M
1F-23 (of 25)

49. Find the concentrations of each species in a 0.10 M H2CO3 solution.
HCO3- (aq) H2O (l) ⇆ H3O+ (aq) CO32- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
- y
+ y
+ y y y y
Ka2 = [H3O+][CO32-]
_________________
[HCO3-]
5.6 x = ( y) y
____________________
( – y)
y = x 10-11
[H2CO3] = = M
[H3O+] = x = M
[HCO3-] = – 5.6 x = M
[CO32-] = = x M
1F-24 (of 25)

50. H2CO3 (aq) + H2O (l) ⇆ H3O+ (aq) + HCO3- (aq)
Ka1 = 4.3 x 10-7
HCO3- (aq) + H2O (l) ⇆ H3O+ (aq) + CO32- (aq)
Ka2 = 5.6 x 10-11
H2CO3 (aq) + 2H2O (l) ⇆ 2H3O+ (aq) + CO32- (aq)
Ka1,2 = ?
[H3O+][HCO3-]
__________________
[H2CO3]
[H3O+][CO32-]
_________________
[HCO3-]
[H3O+]2[CO32-]
__________________
[H2CO3] x =
Ka1Ka2 = Ka1,2
Ka1,2 = (4.3 x 10-7)(5.6 x 10-11)
= x 10-17
1F-25 (of 25)

52. pH OF SALT SOLUTIONS
The ions that make up salts are conjugate acids and conjugate bases
Conjugate Acid
X+ Y-
Conjugate Base
Base
NH3
NaOH
Weak
Strong
Conjugate Acid
NH4+
Na+
Na(H2O)+
Weak
Non
Acid
HNO3
HNO2
Strong
Weak
Conjugate Base
NO3-
NO2-
Non
Weak
Strong bases have non conjugate acids
Weak bases have weak conjugate acids
Strong acids have non conjugate bases
Weak acids have weak conjugate bases
1G-5 (of 12)

53. PO43- (aq) + H2O (l) ⇆ HPO42- (aq) + OH- (aq)
PREDICTING IF IONS IN SALTS WILL ACT AS ACIDS OR BASES
Li3PO4
Li+
PO43-
from (add OH-) LiOH
which is a strong base
∴ Li+ is a non acid
from (add H+’s) H3PO4
which is a weak acid
∴ PO43- is a weak base
∴ pH > 7
PO43- (aq) + H2O (l) ⇆ HPO42- (aq) + OH- (aq)
HYDROLYSIS – The reaction of a dissolved ion with water
1G-6 (of 12)

54. NH4+ (aq) + H2O (l) ⇆ NH3 (aq) + H3O+ (aq)
NH4Cl
NH4+
Cl-
from (add OH-) NH4OH (which is NH3)
which is a weak base
∴ NH4+ is a weak acid
from (add H+) HCl
which is a strong acid
∴ Cl- is a non base
∴ pH < 7
NH4+ (aq) + H2O (l) ⇆ NH3 (aq) + H3O+ (aq)
1G-7 (of 12)

55. KNO3 K+
NO3-
from (add OH-) KOH
which is a strong base
∴ K+ is a non acid
From (add H+) HNO3
Which is a strong acid
∴ NO3- is a non base
∴ pH = 7
1G-8 (of 12)

56. CALCULATING THE pH OF A SALT SOLUTION
Find the pH of a 0.10 M potassium acetate solution
Potassium ion is a non acid
; acetate ion is a weak base
The Kb for acetate is not on Handout 3
, but the Ka for acetic acid is :
1.8 x 10-5
HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2- (aq)
Ka for acetic acid
C2H3O2- (aq) + H2O (l) ⇆ HC2H3O2 (aq) + OH- (aq)
Kb for acetate
2H2O (l) ⇆ H3O+ (aq) + OH- (aq) Kw
KaKb = Kw
1G-9 (of 12)

57. CALCULATING THE pH OF A SALT SOLUTION
Find the pH of a 0.10 M potassium acetate solution
Potassium ion is a non acid
; acetate ion is a weak base
The Kb for acetate is not on Handout 3
, but the Ka for acetic acid is :
1.8 x 10-5
KaKb = Kw
Kb = Kw
____ Ka
= x 10-14
_______________
1.8 x 10-5
= x 10-10
Write the hydrolysis reaction for the acetate ion acting as a base
1G-10 (of 12)

58. CALCULATING THE pH OF A SALT SOLUTION
Find the pH of a 0.10 M potassium acetate solution
C2H3O2- (aq) + H2O (l) ⇆ HC2H3O2 (aq) + OH- (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.10 ~0
- x
+ x
+ x x x x
Kb = [HC2H3O2][OH-]
___________________
[C2H3O2-]
5.56 x = x2
____________
(0.10 – x)
x = x 10-6
= [OH-]
pOH =
-log(7.46 x 10-6 M) =
pH =
– 5.13 =
1G-11 (of 12)

59. Find the pH of a 0.25 M ammonium chloride solution.
Ammonium ion is a weak acid
; chloride ion is a non base
Need the Ka for ammonium
: 5.6 x 10-10
NH4+ (aq) H2O (l) ⇆ H3O+ (aq) NH3 (aq)
Initial M’s
Change in M’s
Equilibrium M’s
0.25 ~0
- x
+ x
+ x x x x
Ka = [H3O+][NH3]
______________
[NH4+]
5.6 x = x2
____________
(0.25 – x)
x = x 10-5 M
pH =
-log(1.18 x 10-5 M)
= 4.93
1G-12 (of 12)

60. REVIEW FOR TEST 1
Equilibrium
Equilibrium Constant
Equilibrium Constant Expression
Relationship of the Balanced Equation to the Keq
Kc and Kp
Reaction Quotient
LeChatelier’s Principle
Determine Shift or New Equilibrium Concentrations Given a Change in
Concentration or Pressure of 1 Reactant or Product
Volume of the Container
Temperature
Effect of a Catalyst on Equilibrium

61. REVIEW FOR TEST 1
Calculate Equilibrium Molarities from Kc and Initial Data
Calculate Kc from
Equilibrium Molarities
Initial Molarities and 1 Equilibrium Molarity
Initial Molarities and % Dissociation
Calculate Equilibrium Pressures from Kp and Initial Data
Calculate Kp from
Equilibrium Pressures
Initial Pressures and 1 Equilibrium Pressure
Initial Pressures and % Dissociation
Initial Pressures and Equilibrium Total Pressure

62. REVIEW FOR TEST 1
Bronsted Acids and Bases
Conjugate Acids and Bases
Ionization of Water and Kw
pH and pOH
pH of Strong Acid or Base Solutions
Acid Ionization Constant, Ka
Base Dissociation Constant, Kb
pH of Weak Acid or Base Solutions

63. REVIEW FOR TEST 1
Polyprotic Acids
Concentrations of Species in a Polyprotic Acid Solution
Relationship of Ka and Kb
pH of Salt Solutions
Experiments 1-5
Graphing
Spectroscopy
Net Ionic Equations for LeChatelier’s Principle
Qualitative Analysis
Reading