1. Acids & Bases
Edward Wen, PhD

2. Types of Ionic Compounds
Acids = form H+ ions in water solution
Bases = combine with H+ ions in water solution
increases the OH- concentration
may either directly release OH- or pull H+ off H2O
Salts = Ionic compounds formed from Acid and Base.
Example: CaCl2 (from HCl (acid) + Ca(OH)2 (base))

3. Properties of Acids Sour taste react with “active” Metals
i.e. Al, Zn, Fe, but NOT w/ Ag, Au
Zn + 2 HCl ® ZnCl2 + H2
react with Carbonates, producing CO2
marble, baking soda, limestone
CaCO3 + 2 HCl ® CaCl2 + CO2 + H2O
change color of vegetable dyes
blue litmus turns red
react with Bases to form ionic salts

4. Properties of Bases also known as alkalis taste bitter
solutions feel slippery
change color of vegetable dyes
different color than acid
red litmus turns blue
react with acids to form ionic salts
neutralization

5. Structure of Bases most ionic bases contain OH- ions
Drano clog-remover: NaOH, Ca(OH)2
some contain CO32- ion: it produces OH- with water
Baking soda: CaCO3
Alka-Seltzer: NaHCO3
molecular bases that react with H+
Windex: Ammonia (NH3)

6. Acid-Base Theories: An Evolution
Arrhenius’ Theory (1880): H+ vs. OH-
Brønsted–Lowry’s Theory (1920’s): H+ donor vs. acceptor
Lewis’ Theory: electron pair (lone pair) acceptor vs. donor

7. The Arrhenius Definition of Acids
Acids ionize in water to produce H+ ions and anions
HCl(aq) → H+(aq) + Cl–(aq)
H2SO4(aq) → 2H+(aq) + SO42-(aq)

8. The Arrhenius Definition of Bases
Bases dissociate in water to produce OH- ions and cations
NaOH(aq) → Na+(aq) + OH–(aq)
Molecular compounds containing an OH group, such as methanol, CH3OH, do not dissociate in solution and therefore do not act as bases.

9. The Arrhenius Definition of Neutralization
Under the Arrhenius definition, acids and bases naturally combine to form water, neutralizing each other in the process.

10. The Brønsted–Lowry Definition of Acids and Bases
Acid—An acid is a proton donor.
Example: HF  H+ + F-
Base—A base is a proton acceptor.
Example: OH- + H+  H2O
F- + H+  HF

11. The Brønsted–Lowry Definition of Acids
HCl is a Brønsted–Lowry acid because, in solution, it donates a proton to water.
Ionization of HCl in water to form H3O+ (hydronium ion).

12. Ammonia as Brønsted–Lowry Base
NH3 is NOT considered as Arrhenius base since it does not inherently contain OH− ions.
However, NH3 is a Brønsted–Lowry base because it accepts a proton from water.

13. Brønsted–Lowry definition: Acids (H+ donors) and Bases (H+ acceptors) always occur together
In the following reaction, HCl is the proton donor (acid) and H2O is the proton acceptor (base).
In the following reaction, H2O is the proton donor (acid) and NH3 is the proton acceptor (base).

14. What happens when an equation representing Brønsted–Lowry acid–base behavior is reversed?
Initially
NH4+ : proton donor (acid)
OH− : proton acceptor (base).
So the change is
NH3 (BL base)  NH4+ (BL acid)
H2O (BL acid)  OH− (BL base)
NH3 and NH4+ : Conjugate Acid–Base pair

15. Conjugate Acid–Base pair  
Any two substances related to each other by the transfer of a proton can be considered a conjugate acid–base pair.

16. Identifying Brønsted–Lowry Acids and Bases and Their Conjugates
In an acid–base reaction:
A base accepts a proton and becomes a conjugate acid.
An acid donates a proton and becomes a conjugate base .

17. Structure of Acid: Binary acids
(HmX): acid hydrogens attached to a nonmetal atom
HCl, HF
Hydrofluoric acid

18. Oxyacids acid hydrogens (H+) attached to an oxygen atom
H2SO4, HNO3, H3PO4 HClO4

19. Organic Acids: Carboxylic acids
-COOH group
HC2H3O2, H3C6H5O3
only the first H in the formula is acidic
the H is on the COOH

20. Most food contains acids
Citric acid (HO2CCH(CO2H)COHCO2H): citrus fruits, tomato
Malic acid (HO2CCH2CHOHCO2H):
green apple, tomato, grape
Ascorbic acid (aka Vitamin C)
Folic acid

21. Common Acids

22. Common Bases

23. Acid-Base Reactions (Neutralization, Double Displacement Reaction)
H+ (from the acid) + OH- (from the base)  H2O
it is often helpful to think of H2O as H-OH
Cation (from base) + Anion (from acid)  Salt
acid + base → salt + water
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

24. Acid Reactions. I. Reaction with Metals
Reaction with many metals: Al, Zn, Fe, Mg
but not all!! Not for Cu, Au, Ag, etc.
Producing a Salt and hydrogen gas H2
3 H2SO4(aq) + 2 Al(s) → Al2(SO4)3(aq) + 3 H2(g)

25. Acid Reactions. II Reaction with Metal Oxides
when acids react with metal oxides, they produce a salt and water
3 H2SO4 + Al2O3 → Al2(SO4)3 + 3 H2O

26. Acid Reactions. III Gas-evolving Reaction with Salts
when acids react with metal carbonate, bicarbonate, sulfide, sulfite, and bisulfite, gas will be produced along with other products
2 HNO3 + FeCO3 → Fe(NO3)2 + CO2 + H2O
HCl + NaHCO3 → NaCl + CO2 + H2O
ZnS + 2 HBr → ZnBr2 + H2S
CaSO3 + 2 HI → CaI2 + SO2 + H2O
H2SO4 + 2 NH4HSO3 → (NH4)2SO4 + 2SO2 + 2H2O

27. 2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2
Base Reactions
Neutralization of acids
Reaction with Nonmetal oxides, CO2
2 NaOH + CO2 → Na2CO3 + H2O
Strong bases will react with Al metal to form sodium aluminate and hydrogen gas
Example: Dissolving recycled aluminum can with NaOH solution
2 NaOH + 2 Al + 6 H2O → 2 NaAl(OH)4 + 3 H2

28. Titration
Purpose: using Reaction Stoichiometry to determine the Concentration of an unknown solution
Titrant (unknown solution) added from a Buret
Indicators: chemicals added to help determine when a reaction is complete
the Endpoint of the titration occurs when the reaction is complete

29. Titration: Color change w/ Indicator

30. Titration Start: The base solution as titrant in the buret.
Titrating: As the Base is added to the Acid, H+ + OH–  HOH. But still excess Acid present so the color does not change.
Endpoint: just enough Base to neutralize all the acid. The indicator changes color.

31. Example: Acid-Base Titration

32. Example:
The titration of mL of HCl solution of unknown concentration requires mL of M Ba(OH)2 solution to reach the endpoint. What is the concentration of the unknown HCl solution?

33. Collect Needed Equations and Conversion Factors:
Example: The titration of mL of HCl solution of unknown concentration requires mL of M Ba(OH)2 solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL M Ba(OH)2
Find: M HCl
Collect Needed Equations and Conversion Factors:
2 HCl(aq) + Ba(OH)2(aq) → BaCl2 (aq) + 2H2O(l)
2 mole HCl = 1 mole Ba(OH)2
0.100 M Ba(OH)2 0.100 mol Ba(OH)2  1 L sol’n

34. Example: The titration of 10
Example: The titration of mL of HCl solution of unknown concentration requires mL of M Ba(OH)2 solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL Ba(OH)2
Find: M HCl
CF: 2 mol HCl = 1 mol Ba(OH)2
0.100 mol Ba(OH)2 = 1 L
M = mol/L
Write a Solution Map: mL
Ba(OH)2 L
Ba(OH)2
mol
Ba(OH)2
mol
HCl mL
HCl L
HCl

35. 2.50 x 10-3 mol HCl 1.25 x 10-3 mol Ba(OH)2 2.50 x 10-1 M HCl
Example: The titration of mL of HCl solution of unknown concentration requires mL of M Ba(OH)2 solution to reach the end point. What is the concentration of the unknown HCl solution?
Given: mL HCl
12.54 mL Ba(OH)2
Find: M HCl
CF: 2 mol HCl = 1 mol Ba(OH)2
0.100 mol Ba(OH)2 = 1 L
2.50 x 10-3 mol HCl
1.25 x 10-3 mol Ba(OH)2
2.50 x 10-1 M HCl

36. Strong Acids
Stomach acid
HCl ® H+ + Cl-
The stronger the acid, the more willing it is to donate H+
use water as the standard base
Strong acids donate practically all their H+
100% ionized in water
[H3O+] = [strong acid]
[ ] = molarity

37. Strong Acids Examples: Binary Acid: HCl, HBr, HI
Oxyacid: HNO3, H2SO4, HClO4, HClO3
Example:
HNO3 = H+ + NO3-
H2SO4 = 2H+ + SO42-

38. Weak Acids Weak acids donate a small fraction of their H+
Vinegar
HC2H3O2 Û H+ + C2H3O2-
Weak acids donate a small fraction of their H+
most of the weak acid molecules do not donate H+ to water
[H3O+] << [weak acid]

39. Weak Acids Examples: Binary Acid: HF, H2S, H2Se
Oxyacid: HNO2, H2SO3, H3PO4, HClO
Most carboxylic acids, such as acetic acid

40. Strong Bases
The stronger the base, the more willing it is to accept H+
use water as the standard acid
Strong bases: practically all molecules are dissociated into OH– or accept H+
1 mol NaOH = 1 mol OH-
1 mol Ca(OH)2 = 2 mol OH-
[OH–] = [strong base] x (# OH)
DranoTM
NaOH ® Na+ + OH-

41. Weak Bases Definition: a small fraction of molecules accept H+
most of the weak base molecules do not take H+ from water
[HO–] << [weak base]
WindexTM
NH3 + H2O Û NH4+ + OH-

42. Weak Acids or Weak Bases remain undissociated in solution
Weak acids (acetic acid, nitrous acid, hydrofluoric acid, etc), Weak base (ammonium hydroxide, etc), Insoluble salts largely remain undissociated in water
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
NH4OH(aq)  NH4+(aq) + OH-(aq)
PbCl2(s) CaCO3(s) BaSO4(s) 42

43. Ionic equation involving weak acid/base
Na2CO3(aq) + 2 HC2H3O2(aq) ® 2 NaC2H3O2(aq) + CO2(g) + H2O(l)
sol. salt weak acid sol. salt nonelectrolyte
2Na+(aq) + CO32−(aq) + 2HC2H3O2(aq)
® 2Na+(aq) + 2C2H3O2−(aq) + CO2(g) + H2O(l)
NIE: CO32−(aq) + 2HC2H3O2(aq) ® 2C2H3O2−(aq) + CO2(g) + H2O(l)
Tro: Chemistry: A Molecular Approach, 2/e 43

44. Ionic equation involving weak acid/base
NH4OH(aq) + HC2H3O2(aq) ® NH4C2H3O2(aq) + H2O(l)
weak base weak acid sol. salt nonelectrolyte
NH4OH(aq) + HC2H3O2 (aq)
® C2H3O2−(aq) + NH4+(aq) + H2O(l)
NIE: no spectator ions, same as Complete Ionic Equations 44

45. Autoionization of Water
Water: extremely Weak electrolyte
therefore there must be a few ions present
about 2 out of every 1 billion water molecules form Ions: Autoionization
H2O + H2O Û H3O+ + OH–
H2O Û H+ + OH–
ALL aqueous solutions contain both H+ and OH–
the concentration of H+ and OH– are equal in water
@ 25°C: [H+] = [OH–] = 10-7M

46. Ion Product of Water [H+] x [OH–] = constant: Ion Product of water, Kw
At 25°C, [H+] x [OH–] = 1 x = Kw
as [H+] increases, [OH–] must decrease so the product stays constant
[H+] = 1 x / [OH–]
[OH–] = 1 x / [H+]

47. Acidic and Basic Solutions
Neutral solutions have equal [H+] and [OH–]
[H+] = [OH–] = 1 x 10-7 M
Acidic solutions : [H+] > [OH–]
[H+] > 1 x 10-7 M [OH–] < 1 x 10-7 M
Basic solutions: [OH–] > [H+]
[H+] < 1 x 10-7 M [OH–] > 1 x 10-7 M

48. Example - Determine the [H+] for a 0
Example - Determine the [H+] for a M Ba(OH)2 and determine whether the solution is acidic, basic or neutral
[H+] = 2.5 x M; basic

49. All [H+] compared to 1 x 10-7 M [HCl] = 1.0 M
Practice - Determine the [H+] concentration and whether the solution is acidic, basic or neutral for the following
All [H+] compared to 1 x 10-7 M
[HCl] = 1.0 M
[NaOH] = M
[H+] = [HCl] = 1.0 M
4.00 x M

50. Acidic/Basic: [H+] vs. [OH-]
Base
[H+]
OH- H+
[OH-]
even though it may look like it, neither H+ of OH- will ever be 0
the sizes of the H+ and OH- are not to scale
because the divisions are powers of 10 rather than units

51. pH The measure of the acidity/basicity of a solution
pH = -log[H+], [H+] = 10-pH
exponent on 10 with a positive sign
pHwater = -log[10-7] = 7
need to know the [H+] concentration to find pH
pH < 7 : Acidic; pH > 7 : Basic
pH = 7 : Neutral

52. pH scale pH↓, Acidity↑ pH↑, basicity↑ normal range 0 to 14
1 pH unit corresponds to a factor of 10 difference in acidity
normal range 0 to 14
pH 0 is [H+] = 1 M, pH 14 is [OH–] = 1 M

53. pH measurement pH can be measured by pH meter:
The change in [H+] affects the voltage of a standard cell

54. pH of Common Substances
1.0 M HCl
0.0
0.1 M HCl
1.0
stomach acid
1.0 to 3.0
lemons
2.2 to 2.4
soft drinks
2.0 to 4.0
plums
2.8 to 3.0
apples
2.9 to 3.3
cherries
3.2 to 4.0
unpolluted rainwater
5.6
human blood
7.3 to 7.4
egg whites
7.6 to 8.0
milk of magnesia (sat’d Mg(OH)2)
10.5
household ammonia
10.5 to 11.5
1.0 M NaOH 14

55. Example - Calculate the pH of a 0
Example - Calculate the pH of a M Ba(OH)2 solution & determine if is acidic, basic or neutral
[H+] = 5.0 x M; pH = 11.3

56. Example - Calculate the pH of the following strong acid or base solutions
M HCl
M Ca(OH)2
[H+] = M, pH = 2.7
[H+] = 1 x M, pH = 12.0

57. H+ OH- pH in everyday life
Stomach acid Vinegar Pure water Windex Drano
Acid
Base pH
[H+]
OH- H+
[OH-]

58. Example - Calculate the concentration of [H+] for a solution with pH 3
[H+] = M

59. How does pH change for certain solutions?
Initial pH
pH after adding 1 mL 1 M HCl
pH after adding 1 mL 1 M NaOH
1 L Pure water
7.00
4.00
10.00
1 L 0.14 M K2HPO M KH2PO4
6.99
7.01

60. How does pH change for certain solutions?
Add 1.0 mL 1.0 M HCl solution to 1.0 L of pure water, pH changes from 7.0 to 4.0 (ΔpH = -3.0)
Add 1.0 mL 1.0 M HCl solution to 1.0 L of 0.14 M K2HPO M KH2PO4 solution, pH changes from 7.0 to 6.99 (ΔpH = -0.01)
Add 1.0 mL 1.0 M NaOH solution to 1.0 L of pure water, pH changes from 7.0 to 10.0 (ΔpH = +3.0)
Add 1.0 mL 1.0 M NaOH solution to 1.0 L of 0.14 M K2HPO M KH2PO4 solution, pH changes from 7.0 to 7.01 (ΔpH = +0.01)

61. Buffers
The mixture of 0.14 M K2HPO M KH2PO4 solution has much smaller pH change when strong acid or base is added, thus is called Buffer.
Definition: solutions that resist changing pH when small amounts of acid or base are added
Ingredient: mixing together a weak acid and its conjugate base
or weak base and it conjugate acid
Human body fluid as buffer: H2CO3/HCO3-

62. Buffer Composition: a weak acid + its salt;
example: HC2H3O2 / NaC2H3O2
When acid is added:
C2H3O2- + H+  HC2H3O2
When base is added:
OH- + HC2H3O2  C2H3O2- + H2O
OR, a weak base + its salt
example: NH3 / NH4Cl

63. Acetic Acid/Acetate Buffer

64. Treasure Hunt: Which of the following pairs of compounds can combine into a Buffer?
NH3 /NH4Cl
HC2H3O2/NaC2H3O2
KNO3/HNO3
Na2SO4/H2SO4
KOH/H2O
NH3 /NH4F
NH3 /NH4NO3
KNO2/HNO2
Hint:
Which is strong acid? Which is strong base?
a, b, f, h

65. Ba(OH)2 = Ba2+ + 2 OH– therefore
Example - Determine the [H+] for a M Ba(OH)2 and determine whether the solution is acidic, basic or neutral
Ba(OH)2 = Ba OH– therefore
[OH–] = 2 x = = 4.0 x 10-4 M
[H+] = 2.5 x M

66. Practice - Determine the [H+] concentration and whether the solution is acidic, basic or neutral for the following
All [H+] compared to 1 x 10-7 M
[OH–] = 3.50 x 10-8 M
[NaOH] = M
[Ca(OH)2] = 0.20 M
[HCl] = 1.0 M

67. More Practice - Determine the [H+] concentration and whether the solution is acidic, basic or neutral for the following
[OH–] = 3.50 x 10-8 M
NaOH = M
Ca(OH)2 = 0.20 M
[H+] =
1 x 10-14
3.50 x 10-8
= 2.86 x 10-7 M
[H+] >[OH-], therefore acidic
[H+] =
1 x 10-14
= 4.00 x M
[H+] < [OH-], therefore basic
[OH-] = 2 x 0.20 = 0.40 M
[H+] =
1 x 10-14
4.0 x 10-1
= 2.5 x 10-14
[H+] < [OH-], therefore basic

68. Example - Calculate the pH of a 0
Example - Calculate the pH of a M Ba(OH)2 solution & determine if is acidic, basic or neutral
Ba(OH)2 = Ba OH- therefore
[OH-] = 2 x = = 2.0 x 10-3 M
[H+] =
1 x 10-14
2.0 x 10-3
= 5.0 x M
pH = -log [H+] = -log (5.0 x 10-12)
pH = 11.3
pH > 7 therefore basic

69. Example - Calculate the pH of the following strong acid or base solutions
M HCl
HCl as strong acid, so [H+] = M
pH = - log (2.0 x 10-3) = 2.7
M Ca(OH)2
Ca(OH)2 as strong base, so [OH-] = M
[H+] =
1 x 10-14
1 x 10-2
= 1 x M
pH = - log (1.0 x 10-12) = 12

70. Example - Calculate the concentration of [H+] for a solution with pH 3
= 2 x 10-4 M
= M