1. Acid Strength: STRONG ACIDS
REVIEW: Strong acids dissociate completely in water (strong electrolytes)
HA(aq) + H2O(l) <-> H3O+(aq) + A-(aq)
Acid Base Conjugate A. C.B.
Equilibrium lies far to the RIGHT (almost all of the HA dissociates at equilibrium)
A STRONG acid has a WEAK conjugate base (a much weaker base than water)
Water wins the competition for H+ ions

2. ACID STRENGTH: WEAK ACIDS
HA(aq) + H2O(l) <-> H3O+(aq) + A-(aq)
Acid Base Conjugate A. C.B.
Very little dissociation of HA
Equilibrium lies far to the LEFT
Conjugate base is a much stronger base than water
Water loses the competition for H+ ions

3. Common Strong Acids: MEMORIZE!!
Sulfuric Acid: H2SO4(aq)
Hydrochloric Acid: HCl(aq)
Nitric Acid: HNO3(aq)
Perchloric Acid: HClO4(aq)
Diprotic acids have two acidic protons (H2SO4)
H2SO4(aq) -> H+(aq) + HSO4-(aq) strong
HSO4-(aq) <-> H+(aq) + SO42-(aq) weak
Monoprotic acids have one acidic proton

4. Acids…
Oxyacids: acidic proton is attached to an oxygen atom (weak and strong)
Organic Acids: acids with a carbon atom backbone, commonly contain the carboxyl group (usually weak acids)

5. Ka For monoprotic acids, Ka can be used to determine strength
Larger Ka = stronger acid = equilibrium lies farther RIGHT
BASE STRENGTH: opposite of acid strength…larger Ka = stronger acid = weaker conjugate base
Water acts stronger than weak, but weaker than strong :)

6. Monoprotic Ka Values

7. H2O(l) + H2O(l) -> H3O+ + OH-
Water
Amphoteric substances can act as either an acid or a base (ex: water)
H2O(l) + H2O(l) -> H3O+ + OH-
Equilibrium: Kw = [H3O+][OH-]
Kw = ion-product constant (dissociation constant for water)
At 25°C in pure water [H3O+] = [OH-] = 1.0 X 10-7 M, so Kw = 1.0 X 10-14

8. About Kw
In any aqueous solution at 25°C, no matter what it contains, [H+]*[OH-] must always equal 1.0 X 10-14
Neutral solutions, [H+] = [OH-]
Acidic solutions, [H+] > [OH-]
Basic solutions, [OH-] > [H+]
NO MATTER WHAT, at 25°C,
Kw = [H+][OH-] = 1.0 X 10-14

9. Examples
Calculate [H+] or [OH-] as required for each of the following solutions at 25°C, and state whether the solution is neutral, acidic, or basic.
1.0 X 10-5 M OH-
1.0 X 10-7 M OH-
10.0 M H+

10. Temperature The equilibrium constant Kw varies with temperature…
If Kw increases with temperature, energy is a reactant (endothermic)
I will not ask you questions about these

11. 2H2O(l) <-> H3O+(aq) + OH-(aq)
More Complex Example
At 60°C, the value of Kw is 1 X 10-13
Using Le Chatelier’s principle, predict whether the following reaction is exothermic or endothermic
2H2O(l) <-> H3O+(aq) + OH-(aq)
Calculate [H+] and [OH-] in a neutral solution at 60°C

12. pH Scale
Ranges from 0-14 and represents the small concentrations used in the Kw expression for [OH-] and [H+]
pH = -log[H+]
pOH = -log[OH-]
pK = -log K
Log scale is based on 10, so the pH changes by 1 for every power of 10 change in [H+]
ACIDS: low pHs, BASES: high pH, NEUTRAL: pH = 7

13. Example Fill in the following table:
Significant figures: The # of decimal places in the log = # SF in the original # pH
pOH
[H+]
[OH-]
Acid, base, neutral
Sol’n A
6.88
Sol’n B
8.4 X 10-14
Sol’n C
3.11
Sol’n D
1.0 X 10-7

14. Polyprotic Acids
Acids can break up into more than one proton (ex. H2SO4 = diprotic or H3PO4 = triprotic)
Ka describes the first proton, Ka2 describes the second, etc.
For a typical weak polyprotic acid, Ka1 > Ka2 > Ka3…the acid will get successfully weaker
*For a typical polyprotic acid in water, only the first dissociation step is important in pH calculation

15. Example
Using table 14.4 in your book, calculate the pH of a 1.40 M H2C2O4 (oxalic acid) solution and the equilibrium concentrations of H2C2O4, HC2O4-, C2O42-, and OH-.
ICE table necessary…

16. Example #2
Using data from Table 14.4 in your textbook, calculate the pH, [PO43-], and [OH-] in a 6.0 M phosphoric acid (H3PO4) solution.