1. Acids, Bases, and Aqueous Equilibria

2. Nature of Acids and Bases
Sour taste, corrosive to metals
Bases--
Bitter taste, feel slippery, corrosive to fat
Don’t use these to identify acids/bases in lab!

3. Definitions of Acids and Bases
Arrhenius Concept: Acids produce H+ in solution, bases produce OH ion.
Brønsted-Lowry: Acids are H+ donors, bases are proton acceptors.
HCl + H2O  Cl + H3O+
acid base

4. Conjugate Acid/Base Pairs
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
conj conj
acid base acid base 1
conjugate base: everything that remains of the acid molecule after a proton is lost.
conjugate acid: formed when the proton is transferred to the base.

5. Acid Dissociation Constant (Ka)
An equilibrium exists in water solutions of acids :
HA(aq) + H2O(l)  H3O+(aq) + A(aq)
or HA(aq)  H+ (aq) + A- (aq)

6. Example
Give dissociation reactions for these: HCl, HC2H3O2, NH4+, C6H5NH3+
HCl  H+ + Cl-
HC2H3O2  H+ + C2H3O2-
NH4+  H+ + NH3
C6H5NH3+  H+ + C6H5NH2-

7. Acid Strength Strong Acid:
Its equilibrium position lies far to the right. (HNO3, HCl, HBr, HI, HClO4, H2SO4) Ka >> 1 – these are the ONLY strong acids
Yields a weak conjugate base. (NO3, or others from above acids)
H2SO4 is only strong in its 1st H+

8. Acid Strength (continued)
Weak Acid:
Its equilibrium lies far to the left. (CH3COOH, and other organic acids) Ka << 1
Yields a much stronger (it is relatively strong) conjugate base than water. (CH3COO)

9. Types of Acids
Binary Acids– Hydrogen bonded to elements other than water, which has acid characteristics– HCl, HCN, H2S
Oxyacids– Hydrogen bonded to a polyatomic ion containing oxygen—H2CO3, H3PO4
Organic Acids—contain the carboxyl group, OH
-C=O which are all weak acids

10. Strong Acids
Weak Acids

12. Relative acid strength Relative conjugate base strength Very strong
14_323
Relative
acid strength
Relative
conjugate
base strength
Very
strong
Very
weak
Strong
Weak
Weak
Strong
Very
weak
Very
strong

14. Example
From the previous slide, arrange these bases from weak to stronger: H2O, F-, Cl-, NO2-, CN-
Cl- is from strong acid, as is H2O (from H3O+), so both are very weak. CN- is from the weakest acid and is therefore the strongest. HF is a stonger acid than HNO2, so NO2- is stronger than F-, so the ranking from weak to strong is:
Cl- < H2O < F- < NO2- < CN-

15. Water as an Acid and a Base
Water is amphoteric (it can behave either as an acid or a base).
H2O + H2O  H3O+ + OH
conj conj
acid base acid base
Kw = 1  1014 at 25°C= [H+] [OH-]
Must always be a balance between H+ and OH-

16. Example Calculate [H+] and [OH-] in these solutions:
a. 1.0 x 10-5 M OH-
b. 1.0 x 10-7 M OH-
c M H+
[OH-]= 1.0 x 10-5 M [H+]= Kw / [OH-] =1 x 10-14/ 1.0 x = 1.0 x 10-9 M
[OH-]= 1.0 x 10-7 M [H+]= Kw / [OH-] = 1 x 10-14/ 1.0 x = 1.0 x 10-7 M
[H+]= 10.0 M [OH-]= Kw / [H+] = 1 x 10-14/10.0 =1.0 x M

17. The pH Scale There is a more convenient way to indicate [H+]
pH  log[H+]
pH in water normally ranges from 0 to 14, but can extend to negative or >14 values
Kw = 1.00  1014 = [H+] [OH]
pKw = = pH + pOH
As pH rises, pOH falls (sum = 14.00).

18. Example Calculate pH and pOH for each of these
solutions: a. 1.0 x 10-3 M OH- b M H+
-log [OH-] = 3 = pOH pH = 14 – pOH = 11
b. -log [H+] = 0= pH pOH = 14 – pH = 14

20. Example If pH of blood is 7.41, find pOH, [H+], and [OH-]
pOH= 14-pH= [H+] = 10-pH = = 3.9 x M
[OH-]= = 2.6 x 10-7 M
Calculate pH for 0.10 M HNO3 and 1 x M HCl
Since both are strong acids, they are totally dissociated, and [H+] = acid strength of the major species. pH of HNO3 is therefore –log(0.10) = 1. But the HCl solution is so dilute that the water provides most of the [H+], and so the pH=7.

21. Solving Weak Acid Equilibrium Problems
List major species in solution.
Choose species that can produce H+ and write reactions.
Based on K values, decide on dominant equilibrium.
Write equilibrium expression for dominant equilibrium.
List initial concentrations in dominant equilibrium. (I)

22. Solving Weak Acid Equilibrium Problems (continued)
Define change at equilibrium (as “x”). (C)
Write equilibrium concentrations in terms of x. (E)
Substitute equilibrium concentrations into equilibrium expression.
Solve for x the “easy way.”
Verify assumptions using 5% rule.
Calculate [H+] and pH.

23. Example x2 = 3.5 x 10-8(0.1) =3.5 x 10-9 x= 5.9 x 10-5 M = [H+]
pH = -log(5.9 x 10-5) =4.23
Calculate the pH of a M solution of HOCl (Ka=3.5 x 10-8)
HOCl H+ OCl- I
C -x x x
E 0.1-x x x

24. Percent Dissociation (Ionization)
Calculate this from [H+]
This becomes greater as the acid concentration becomes more dilute—Example 14.10

25. Example
Calculate percent dissociation for a M HC2H3O2 and b M HC2H3O Ka= 1.8 x 10-5
Acid H A b. Acid H A- I
C -x x x x x +x
E 1-x x x x x x
1.8 x 10-5 = x2 / x 10-5 = x2 / 0.1
x= [H+] = 4.2 x 10-3 M x= 1.3 x 10-3 M
% diss= 4.2 x 10-3 / % diss = 1.3 x 10-3 / 0.100
= 0.42 % = 1.3 %

26. Acid concentration Percent dissociation H concentration
14_325
Acid concentration
Percent dissociation H +
concentration
More concentrated
More dilute

27. NaOH(s)  Na+(aq) + OH(aq)
Bases
“Strong” and “weak” are used in the same sense for bases as for acids.
strong = complete dissociation (hydroxide ion supplied to solution) Most common are metal hydroxides. Kb is very large.
NaOH(s)  Na+(aq) + OH(aq)

28. Example Calculate pH of 5.0 x 10-2 M NaOH
Strong base means [OH-] = [NaOH]= 0.05 M
pOH = -log(0.05) = 1.30
pH = 14- pOH = 12.70

29. Bases (continued)
weak = very little dissociation (or reaction with water) Usually contain an -NHn group
H3CNH2(aq) + H2O(l)  H3CNH3+(aq) + OH(aq)
Kb for weak bases is usually very small < 10-3

30. Calcualtions Involving Weak Bases
Use the same ICE method as with weak acids
Notice that x will equal [OH-] rather than [H+]
Calculate pOH and from that calculate pH

31. Example Calculate the pH of a 15.0 M solution of NH3
Kb = 1.8 x NH3 NH4+ OH- I
C -x +x x
E 15-x x x
1.8 x 10-5 = x2 / 15 x2= 1.8 x 10-5 (15)=2.7 x 10-4
x= 1.6 x 10-2 M = [OH-] pOH= 1.80
pH= 14-pOH=12.2

32. Polyprotic Acids
. . . can furnish more than one proton (H+) to the solution. Each one comes off separately.

33. Acid-Base Properties of Salts

34. Relationship of Ka to Kb
For acidic/basic salts, Ka or Kb must be calculated from the parent acid/base value
Kb = Kw Ka = Kw
Ka Kb
Example:
Ka of HC2H3O2 = 1.8 x 10-5
Kb = 1 x / 1.8 x 10-5 = 5.6 x 10-10

35. Structure and Acid-Base Properties
Two factors for acidity in binary compounds:
Bond Polarity (high is good)
Bond Strength (low is good)

37. Oxides Acidic Oxides (Acid Anhydrides):
OX bond is strong and covalent.
SO2, NO2, CrO3
Basic Oxides (Basic Anhydrides):
OX bond is ionic.
K2O, CaO

38. Lewis Acids and Bases Lewis Acid: electron pair acceptor
Lewis Base: electron pair donor