1. Introduction to Acids and Bases
Chapter 14
Sections 1-5, 7, 11
AP Chemistry
Dr. Knorr

2. Arrhenius Acids and Bases
Acids produce H+ in aqueous solutions
water
HCl H+(aq) + Cl- (aq)
Bases produce OH- in aqueous solutions
NaOH Na+(aq) + OH- (aq)

3. Brønsted-Lowry Acids and Bases
Acids donate H+ in aqueous solutions
water
HCl H+(aq) + Cl- (aq)
Bases accept H+ in aqueous solutions
NH3 + H2O NH4+(aq) + OH- (aq)
All Arrhenius A/B are B-L A/B

4. Lewis Acids and Bases Acids accept electron pairs in aqueous solutions
Bases donate electron pairs in aqueous solutions
H F H F
H N B F H N B F

5. Conjugate Acids and Bases
A conjugate acid-base pair is related by
loss/gain of one hydrogen ion.
H2O + HA H3O+(aq) + A- (aq)
gain H+
Ka = [H+][A-]
[HA]
base
acid
conjugate acid
conjugate base
lose H+

6. Acid Strength H2O + HA H3O+(aq) + A- (aq)
B A CA CB
Strength is related to position at equilibrium
Strong acids have weak CBs and large Ka
Weak acids have strong CBs and small Ka

7. Strong Acids Strong Acids (the big 7): HCl, HBr, HI, HNO3, H2SO4,
HClO4, HClO3
Example:
HCl(aq) H+ (aq) + Cl- (aq)
**Strong acids completely ionize, so equilibrium constants >>1 and cannot be measured accurately
Ka = [H+][Cl-] >>1
[HCl]

8. Water! Autoionization: H2O H+ + OH-
Keq = Kw = [H+] x [OH-] = 1 x 10-14
Therefore [H+] = [OH-] = 1 x 10-7

9. pH [H+]>[OH-] [H+] = [OH-] [OH-]>[H+] pH = - log [H+]
[H+] pH State
1 x 10-5 M 5 Acidic
1 x M 11 Basic
[H+]>[OH-] [H+] = [OH-] [OH-]>[H+]
Neutral
Acidic
Basic

10. pOH Indicates the basicity (aka alkalinity) [OH-] of the solution
pOH = - log [OH-]
Another important relationship: pH + pOH = 14
[OH-] pOH pH State
1 x 10-5 M Basic
1 x M Acidic

11. pH of Strong Acids Example: what is pH of 0.01M HNO3? HNO3H+ + NO3-
What are the major species in solution?
Major: H+, NO3-, and H2O
Minor: HNO3, OH-
pH = -log[H+ ] = -log(0.01) = 2.0

12. pH of Weak Acids
Example: What is pH of a 1.0M sol’n of HOCl if Ka = 3.5 x 10-8?
What are the species in solution?
HOCl(aq) H+ (aq) + OCl-(aq)
H2O(l) H+ (aq) + OH-(aq)
Major species: HOCl & H2O

13. pH of Weak Acids (cont’d)
HOCl(aq) H+ (aq) + OCl-(aq)
Set up an ICE table!
HOCl H+ OCl- I
E x x x
x = 1.9x10-4M = [H+ ], so pH = -log(1.9x10-4) = 3.72
Remember Kw = 10-14, so most H+ comes from HOCl
Ka = [H+][OCl-]
[HOCl]
= __x2__ = 3.5 x 10-8
(1-x)

14. pH of Weak Acid Mixtures
Example: What is pH of a sol’n of 0.10M HCN (Ka = 6.2 x 10-10) and 0.50M HNO2 (Ka = 4.0 x 10-4)
What are the major species in solution?
HCN(aq) H+ (aq) + CN-(aq) Ka = 6.2x10-10
HNO2(aq) H+ (aq) + NO2-(aq) Ka = 4.0x10-4
H2O(l) H+ (aq) + OH-(aq) Kw = 1.0x10-14
Major species: HCN, HNO2 & H2O
HNO2 contributes the most H+ because of largest Ka

15. pH of Weak Acid Mixtures (cont’d)
HNO2(aq) H+ (aq) + NO2-(aq)
Set up an ICE table!
HNO2 H+ NO2- I
E 0.5-x x x
x = 0.014M = [H+ ], so pH = -log(0.014) = 1.85
Ka = [H+][NO2-]
[HNO2]
= x2___ = 4.0 x 10-4
(0.5-x)

16. Weak Acid Mixtures (cont’d)
HCN(aq) H+ (aq) + CN-(aq)
What is [CN-] in the same mixture?
This is a small amount of HCN that dissociates when you consider [HCN]o
Ka = [H+][CN-]
[HCN]
6.2x10-10 = [0.014][CN-] … [CN-] = 4.4x10-9 [0.10]

17. Percent Dissociation (PD)
PD = 100 x amount dissociated (M)
initial concentration (M)
For our previous mixture,
HCNPD = 100 x 4.4x10-9 / = 4.4x10-6 %
HNO2PD = 100 x / = 2.8 %

18. PD Example
In a 0.100M aqueous solution of lactic acid, 3.7% is dissociated. Calculate Ka.
HC3H5O3 H+ C3H5O3-
0.1-x x x
Ka = x2/(0.100-x)
Ka = (3.7x10-3)2/( x10-3)
Ka = 1.4x10-4
PD = 3.7 = 100 * x/0.1
x = 3.7 x 10-3M

19. Polyprotic Acids Supply more than one proton, one at a time
Conjugate base of 1st reaction becomes the acid in the 2nd reaction
Usually Ka1 > Ka2 > Ka3, meaning loss of 2nd or 3rd proton is not as easy as 1st
Examples: H3PO4; H2SO3

20. Polyprotic Acids (cont’d)
Example: calculate the pH of a 5.0M H2SO3 solution and the concentration of all S species at equilibrium
Ka1 = and Ka2 = 1.0 x 10-7
The dominant equilibrium is H2SO3 H+ + HSO3- I
E x x x
Ka1 = [H+][HSO3-]
[H2SO3]
0.015 = (x)(x)
(5.0-x)
x = = [H+] =[HSO3-]
pH = -log[H+] = -log (0.266) = 0.58

21. Polyprotic Acids (cont’d)
If x = 0.266, [H2SO3] = 5-x, or ~4.7M [SO3-2] can be obtained for the expression for Ka2:
Ka2 = [H+][SO3-2]
[HSO3-]
HSO3- H+ + SO3-2
Dissociation of HSO3-
to form SO3-2 only negligibly
affects [H+] and [HSO3-]
because Ka2 is so small.
1.0 x 10-7 = [0.266][SO3-2]
[0.266]
[SO3-2] = 1.0 x 10-7M

22. Sulfuric Acid
Ka1 is large (“strong”), Ka2 is small (“weak”) ~0.012 Major species are H+, HSO4-, and H2O Calculate the pH of a 2.0M H2SO4 solution and all ion concentrations at equilibrium After 1st dissociation, H+ ~ 2.0M, HSO4- ~ 2.0M HSO4-  H+ + SO x 2.0+x x
Ka2 = [H+][SO4-2]
[HSO4-]
0.012= (2+x)(x) … x = 0.012
(2-x)
[H+] ~2.0M, so pH = H2SO4 ~0, [HSO4-] ~2.0M, [SO4-2] = 0.012M