1. Acids and Bases
Chapters 14 and 15

2. Acid/Base Definitions
Arrhenius Model
Acids produce hydrogen ions (H+ or H3O+) in aqueous solutions
Bases produce hydroxide ions (OH-) in aqueous solutions
Bronsted-Lowry Model
Acids are proton (H+) donors
Bases are proton (H+) acceptors
Lewis Model
Acids are electron pair acceptors
Bases are electron pair donors

3. HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
Example 1
Water acting as a base by accepting a proton, HCl acitng as an acid by donating a proton. HCl acting as an acid by producing H3O+ ions in solution. H+ ions provided by the HCl act as an acid by accepting a pair of electrons from the oxygen atom in the water (a base) to form a bond in H3O+.
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
acid base conjugate conjugate
acid base

4. NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Example 2
Water acting as an acid by donating a proton, NH3 acting as a base by accepting a proton. NH3 acting as a base by producing OH- ions in solution.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
base acid conjugate conjugate
acid base

5. Example 3
Electron rich ammonia acting as a base by donating a pair of electrons to electron deficient Boron, that in turn acts as an acid by accepting the pair of electrons.
NH3 + BF3  H3N-BF3

6. Conjugate Acid/Base Pairs
Related by a hydrogen ion on either side of the equation
Example 2 previously
NH3 and NH4+

7. Strength of Acids and Bases
Strong acids and bases are assumed to completely ionize in solution.
For a strong acid (HA) and a strong base (B), the following reactions go to completion (no reverse reaction occurs)
HA(aq) + H2O(l)  H3O+ + A-(aq)
B(aq) + H2O(l)  BH+(aq) + OH-(aq)

8. Strength of Acids and Bases
Weak acids and bases have very little ionization
Equilibria are set up (reverse reactions are possible)
The equilibrium positions for each lie heavily to the left hand side of the equation
HA(aq) + H2O(l)  H3O+ + A-(aq)
B(aq) + H2O(l)  BH+(aq) + OH-(aq)

9. Strength of Acids and Bases
Weak Acids
Organic acids, any acid not listed as a strong acid
Strong Acids
HBr, HCl, HI, HNO3, H2SO4, HClO4, HClO3
Weak Bases
Ammonia and organic bases (ex. Methylamine)
LiOH, NaOH, KOH, RbOH, CsOH, Ba(OH)2, Sr(OH)2, Ca(OH)2, Mg(OH)2

10. Define the following words:
Arrhenius Acid/Base
Bronsted-Lowry Acid/Base
Lewis Acid/Base
Strong
Concentrated
Corrosive
Weak
Dilute
Equilibrium
Indicator

11. Self Ionization of Water
H2O + H2O  H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

12. Amphoterism Water can accept H+ or donate H+
This is called amphoterism
Pure water does it by itself, although not much
Water will make both protons and hydroxides without any outside work

13. Amphoterism Reaction 2H2O(l) H3O+(aq) + OH-(aq)
We use the typical chemist’s shortcut to write this equation
2H2O(l) H+(aq) + OH-(aq)
Kw = [H+][OH-]
The value of Kw is known at room temp

14. Amphoterism
If you increase either protons or hydroxides (by adding acid or base) you MUST decrease the other mathematically in order to maintain the constant value of Kw
Example – calculate [H+] for a solution with 1.0x10-5 M OH- at 25oC. Is it neutral, acidic or basic?

15. The pH Scale Used to indicate the strength of an acid or base
Ranges from 0 – 14
Acids are less than 7
Bases are greater than 7
7 is considered neutral

16. The pH Scale

17. The pH Scale Calculating pH, pOH Relationship between pH and pOH
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH

18. pH pOH The pH Scale H+ OH- [OH-] = 1 x 10-14 [H+] [H+] = 1 x 10-14
[H+] = 10-pH
pOH = -log[OH-]
pH = -log[H+]
[OH-] = 10-pOH
pOH = 14 - pH
pH = 14 - pOH

19. The pH Scale
Examples
Calculate the pH of a solution of 0.030M hydrochloric acid
Calculate the pH of a 0.010M solution of calcium hydroxide
Calculate the H3O+ concentration in a solution with a pH of 4.32
Calculate the pH of a solution made by dissolving 2.00g of potassium hydroxide in distilled water to a total volume of 250.mL

20. Ka – The Equilibrium Constant for Weak Acids
For a weak acid (HA) dissociation is incomplete so an equilibrium must be set up
HA(aq) + H2O(l) H3O+ + A-(aq)
Initial X
Change x x x
Equilibrium X-x x x
Can’t go directly to the pH equation because we don’t know how much acid is dissociated
Derive the following equilibrium expression to help solve these problems
Ka = [H3O+][A-]
[HA]

21. Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2  C2H3O2 - + H+

22. Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step 2: ICE it!
HC2H3O2  C2H3O2 - + H+ I
C -x +x x
E – x x x

23. Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step 3: Set up the law of mass action
HC2H3O2  C2H3O2 - + H+
1.8 x 10-5 = (x)(x) = x2
(0.50-x)

24. Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step 4: Solve for x, which is also [H+]
[H+] = 3.0 x 10-3 M

25. Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
Step 5: Convert [H+] to pH
pH = -log(3.0 x 10-3) = 2.52

26. Kb – The Equilibrium Constant for Weak Bases
For a weak base (B) dissociation is incomplete so an equilibrium must be set up
B(aq) + H2O(l) BH+ + OH-(aq)
Initial X
Change x x x
Equilibrium X-x x x
Can’t go directly to the pOH equation because we don’t know how much base is dissociated
Derive the following equilibrium expression to help solve these problems
Kb = [BH+][OH-]
[B]

27. Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step #1: Write the equation for the reaction
NH3 + H2O  NH4+ + OH-

28. Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step 2: ICE it!
NH3 + H2O  NH4+ + OH- I
C x x x
E – x x x

29. Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step 3: Set up the law of mass action
NH3 + H2O  NH4+ + OH-
1.8 x 10-5 = (x)(x) = x2
(0.50-x) (0.50)

30. Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step 4: Solve for x, which is also [OH-]
[OH-] = 3.0 x 10-3M

31. Weak Base Equilibrium Problem
What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ?
Step 5: Convert [OH-] to pH
pOH = -log(3.0 x 10-3M) = 2.52
pH = – pOH = 11.48

32. Percent Dissociation % dissociation = amount dissociated(mol/L) x 100%
initial concentration(mol/L)
Specifies the amount of weak acid that has dissociated to achieve equilibrium

33. Example
0.500 M uric acid (HC5H3N4O4) is 1.6% dissociated. Find Ka and pH.
What is the percent dissociation of M HC2H3O2, Ka = 1.8x10-5?

34. HIn(aq)  H+(aq) + In-(aq)
Indicators
A substance that changes color according to the pH of the solution
Often are weak acids where the ionized and unionized forms are different colors
HIn(aq)  H+(aq) + In-(aq)
color color 2

35. Indicators Common indicators for acids and bases
Litmus paper: red for acids, blue for bases
Phenophthalein: colorless for acids, pink for bases
Methyl orange: red for acids, yellow for bases

36. Selection of Indicators

37. Some Acid-Base Indicators

38. Strong Acid/Strong Base Titrations

39. Strong Acid/Strong Base Example
Calculate the pH after these volumes of M HCl are added to mL of M NaOH.
a) 0.00 mL
b) 4.00 mL
c) mL
d) mL
e) mL
f) mL

40. Strong Acid/Strong Base Example
Notice the RAPID drop in pH.
1.00 mL between points c and e moves us from pH to 2.81 (9 orders of magnitude; a billion times difference!)

41. Weak Acid/Strong Base Titration

42. Strong Acid/Weak Base Titrations

43. Weak Acid/Strong Base Example
Find pH after these volumes of M NaOH are added to mL of M HCOOH (Ka = 1.8x10-4)
a) 0.00 mL e) mL
b) 5.00 mL f) mL
c) mL g) mL
d) mL

44. Weak Acid/Strong Base Example
Notice weak acid/strong base has endpoint (equivalence point) in the BASIC region, not at pH of 7.00 like in strong acid/strong base.
The exact same logic applies (and the same math is used) for strong ACID/weak BASE

45. Monoprotic and Polyprotic Acids
Monoprotic acids contain only one “ionizable” proton
Polyprotic acids contain more than one
Usually only the first H+ comes off easily as a strong ion
Each comes H+ off in a stepwise manner, each with its own Ka

46. Polyprotic acids Example: Carbonic Acid Stage 1:
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
Ka1 = [H3O+][HCO3-]
[H2CO3]
Stage 2:
HCO3-(aq) + H2O(l)  H3O+(aq) + CO3-2(aq)
Ka2 = [H3O+][CO3-2]
[HCO3-]
Overall:
H2CO3(aq) + 2H2O(l)  2H3O+(aq) + CO3-2(aq)
Ka = [H3O+]2[CO3-2] = (Ka1 )(Ka2)

47. Example Find [PO4-3], pH and [OH-] in 6.0 M H3PO4
Ka1 = 7.5x Ka2 = 6.2x Ka3 = 4.8x10-13
Since Ka1 >> Ka2 >> Ka3 we can ignore the H+ in reactions 2 and 3
To find [PO4-3] we must work our way all the way to the last equation

48. Buffered Solutions
A solution that resists a change in pH when either hydroxide ions or protons are added.
Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt

49. Acid/Salt Buffering Pairs
The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula of the acid
Example of a salt of the weak acid
Hydrofluoric HF
KF – Potassium fluoride
Formic
HCOOH
KHCOO – Potassium formate
Benzoic
C6H5COOH
NaC6H5COO – Sodium benzoate
Acetic
CH3COOH
NaH3COO – Sodium acetate
Carbonic
H2CO3
NaHCO3 – Sodium bicarbonate
Propanoic
HC3H5O2
NaC3H5O2 – Sodium propanoate
Hydrocyanic
HCN
KCN – Potassium cyanide

50. Base/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)
Base
Formula of the base
Example of a salt of the weak base
Ammonia
NH3
NH4Cl – ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl - Methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - Ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl - Aniline hyrdrochloride
Pyridine
C5H5N
C5H5NHCl - Pyridine hydrochloride

51. Buffered Solutions - Example
A 0.100M solution of ethanoic acid (Ka = 1.8 x 10-5) is mixed with a solution of 0.100M potassium ethanoate. Calculate the pH of the resulting solution.

52. Henderson-Hasselbalch Equation
pH = pKa + log [A-] = pKa + log [salt]
[HA] [acid]
pOH = pKb + log [BH+] = pKb + log [salt]
[B] [base]

53. Acid-Base Properties of Salts
These salts simply dissociate in water:
KCl(s)  K+(aq) + Cl-(aq)
Type of Salt
Examples
Comment
pH of solution
Cation is from a strong base, anion from a strong acid
KCl, KNO3, NaCl, NaNO3
Both ions are neutral
neutral

54. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is from a strong base, anion from a weak acid
NaC2H3O2KCN, NaF
Cation is neutral, anion is basic
basic
The basic anion can accept a proton from water:
C2H3O2- + H2O HC2H3O2 + OH-

55. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is the conjugate acid of a weak base, anion is from a strong acid
NH4Cl, NH4NO3
Cation is acidic, anion is neutral
acidic
The acidic cation can act as a proton donor:
NH NH3 + H+

56. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is the conjugate acid of a weak base, anion is conjugate base of a weak acid
NH4C2H3O2 NH4CN
Cation is acidic, anion is basic
See below
IF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral

57. Acid-Base Properties of Salts
Type of Salt
Examples
Comment
pH of solution
Cation is a highly charged metal ion; anion is from a strong acid
Al(NO3)2; FeCl3
Hydrated cation acts as an acid; anion is neutral
acidic
Step #1:
AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq)
Salt water Complex ion anion
Step #2:
Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq)
Acid Conjugate base Proton