1. Reactions of Acids & Bases
Neutralization is most important :
Strong Acid + Strong Base
Water + a “salt” (ionic compound)
Two Examples:
NaOH + HCl  H2O + NaCl
Ba(OH)2 + H2SO4  2 H2O + BaSO4
Acids & bases Part II

2. CH3COOH + NaOH  CH3COONa + H2O CH3COONa = CH3COO– + Na+
Neutralization between weak acid & strong base is similar
(weak) (strong)
CH3COOH + NaOH 
CH3COONa + H2O
CH3COONa = CH3COO– + Na+
Acids & bases Part II

3. NH3 + HCl NH4Cl = +NH4 + Cl– Ionic compound
Weak base + strong acid “salt” only
NH HCl
NH4Cl =
+NH4 + Cl–
Ionic compound
Acids & bases Part II

4. Examples of buffer mixtures: CH3COOH + CH3COO– 1 mole 1 mole
Buffers are mixtures of comparable amounts of a weak acid and its conjugate base (or a weak base and its conjugate acid).
Examples of buffer mixtures:
CH3COOH + CH3COO–
1 mole mole
Weak acid conjugate base
Acids & bases Part II

5. NH3 + NH4+ Comparable amounts Weak base conjugate acid
0.5 mole mole
Comparable amounts
Weak base conjugate acid
Acids & bases Part II

6. Why the emphasis on comparable amounts of the conjugate acid-base pair?
If a weak acid HA is placed in solution, most of what is present is HA. There is very little A- in solution because very little dissociation occurs for a weak acid.
Acids & bases Part II

7. In the same way, when a weak base :B is put in solution, most of the material present is :B ; not much H-B+ forms because little reaction with water occurs.
Acids & bases Part II

8. So why is this important?
When a buffer is prepared we have to add extra conjugate base to a weak acid in order to make the amounts present comparable (meaning, roughly the same size).
So why is this important?
Acids & bases Part II

9. Scientists are interested in buffers because
the pH of a buffer solution does not change very much when a small amount of another acid or base is added.
Acids & bases Part II

10. The first issue is “what is the pH of the buffer?”
We can answer this question by looking at the dissociation of a weak acid in water:
for H-A + H2O A- + H3O+
Ka = [A-][H3O+]
[HA]
Acids & bases Part II

11. The expression for Ka can be rearranged by cross-multiplying to give
Ka [HA]
= [H3O+]
[A-]
This tells us that the acidity, [H3O+], depends on Ka multiplied by the ratio of [HA] and [A-].
Acids & bases Part II

12. Now if the concentrations of HA and A- in solution are comparable, this means that [HA]  [A-].
Then the ratio [HA]
 1
[A-]
So, Ka [HA]
 [H3O+] means….
[A-]
Acids & bases Part II

13. Ka  [H3O+].
This means the acidity in a buffer is about the same as the dissociation constant of the weak acid.
Since pH = - log [H3O+] this also means pH = - log Ka = pKa of the weak acid.
Acids & bases Part II

14. Example: for acetic acid, CH3COOH,
Ka = 1.75 x (value from reference, do not memorize!)
So the pH of a buffer that contains comparable amounts of CH3COOH and CH3COO- (acetate), is about 5.
The actual value is pH = 4.75, so this is a good estimate.
Acids & bases Part II

15. The second issue is “why does the pH of the buffer not change very much when a small amount of acid or base is added?”
Let’s see…
Acids & bases Part II

16. Consider a buffer mixture of acetic acid and acetate ion:
CH3COOH + H2O CH3COO– + H3O+
0.25 mole mole
Now suppose 0.02 mole HO- is added to the buffer. It reacts with and uses up 0.02 mole CH3COOH, leaving 0.23 mole. It also makes 0.02 mole CH3COO-.
Acids & bases Part II

17. Before HO- added Ka (1/1)  [H3O+]. So: pH changes, but not very much.
So now the mixture has
CH3COOH + H2O CH3COO– + H3O+
0.23 mole mole
This means for Ka [HA]
 [H3O+],
[A-]
We have Ka(0.23/0.27)  [H3O+].
Before HO- added Ka (1/1)  [H3O+].
So: pH changes, but not very much.
Acids & bases Part II

18. This is why a buffer must contain both conjugate acid-base forms.
Any HO- added to a buffer reacts with the weak acid HA and makes an equal amount of conjugate base A-. The ratio [weak acid]/[conjugate base] = [HA]/[A-] changes only a little. So [H3O+] and pH do not change much as a result.
Acids & bases Part II

19. Any H+ added to the buffer reacts with the conjugate base A- and makes an equal amount of HA.
The ratio [HA]/[A-] does not change much and so neither does [H3O+] or pH.
Acids & bases Part II