1. 8.3 Bases
Similar to weak acids, weak bases react with water to a solution of ions at equilibrium.
The general equation is:
B(aq) + H2O(l)  HB+(aq) + OH-(aq)
where, B represents any base
The equilibrium expression for this reaction is:
Kc = [HB+][OH-]
[B][H2O]

2. Base Dissociation Constant, Kb
Just as with weak acids, the concentration of water is almost constant in a solution of weak base.
Therefore, we can group the constant terms in the equation.
[H2O] Kc = [HB+][OH-] = Kb
[B]
Table 8.3 (Pg. 404 lists some common base dissociation constants, Kb)

3. Calculating pH and pOH
An aqueous solution of ammonia has a concentration of mol/L. Calculate the pH of the solution.
1. Write the balanced chemical equation.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
Given: [NH3] = mol/L
2. Set up an ICE table

4. P.P. #29, Continued
Concentration (mol/L)
NH3(aq)
H2O(l)
NH4+(aq)
OH- (aq)
Initial
0.105 ~0
Change -x +x
Equilibrium x
Write the equation for the base dissociation constant, Kb
Kb =[NH4+][OH-]
[NH3]
From the table, Kb(NH3) = 1.8 x 10-5

5. P.P. #29, Continued
Substitute Kb and the equilibrium values from the ICE table into the equation. Solve for x.
1.8 x 10-5 = (x) (x)
(0.105 – x)
Look at [NH3]/Kb
= 0.105/1.8x10-5 = 5833 > 100, so the amount of NH3 that dissociates is negligible compared to the initial [NH3]
Therefore, (0.105 – x), becomes (0.105)  solve for x

6. P.P. #29, Continued Solve for x 1.8 x 10-5 = (x) (x) (0.105)
6. Solve for pOH
pOH = -log[OH-] = -log[1.374 x 10-3] = 2.832

7. P.P. #29, Continued Solve for pH pH = 14 – pOH = 14 – 2.832 = 11.168
Therefore, the pH of the ammonia solution is 11.14

8. The relationship b/n Ka and Kb
For any acid and its conjugate base
Ka(Kb) = [H3O+][OH-] = Kw
What does this mean?
The stronger an acid is, the weaker its conjugate base is.
The conjugate of a strong acid is always a weak base.
The conjugate of a strong base is always a weak acid.

9. Buffer Solutions Contain a mixture of:
A weak acid and its conjugate base OR
A weak base and its conjugate acid
Buffer solutions resist a change in pH when a moderate amount of acid or base is added.
Buffer solutions are made 2 ways:
Use a weak acid and one of its salts
Eg. Mix acetic acid and sodium acetate
Use a weak base and one of its salts
Eg. Mix ammonia and ammonium chloride

10. Making a Buffer

11. How do buffers work?
Consider a buffer made with acetic acid and sodium acetate.
Acetic acid is weak, so [CH3COOH] is high
Sodium acetate is very soluble, so [CH3COO-] is high
Adding and acid or base has only a slight effect on pH because the H3O+ or OH- ions are removed by one of the components of the buffer solution.

12. Importance of Buffers Very important in biological systems
For example:
pH of arterial blood is ~7.4
It must stay b/n 7.0 and 7.5 or the organism may die.
Blood is buffered by the equilibrium between CO3 ions and HCO3 ions. (formed by the reaction of dissolved CO2 and H2O)

13. 8.4 Acid-Base Titration Curves
= Graph of the pH of an acid (or base) against the volume of and added base (or acid)
Titration reactions are used to find the equivalence point
Equivalence point – point where the acid and base completely react with one another.
If,
V1, V2 and C1 are known
C2 can be found (C1V1 = C2V2)

14. How pH indicators show equivalence points
pH changes rapidly near the equivalence point
A single drop of titrant can change the pH by 2 pH units.
The colour change indicates equivalence even if the colour change happens at pH of 8
As long as it is in the steep section of the curve
Strong acid titrated with strong base

15. Selecting a pH Indicator
Weak acid titrated with strong base.

16. Section Review Which indicator should be used for each curve?
Is this the titration curve for a weak acid titrated with a strong base? Explain.