1. Titration & pH curves
[17.3]

2. Titration
A known concentration of base (or acid) is slowly added to a solution of acid (or base).

3. Titration
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

4. Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.

5. Titration of a Strong Acid with a Strong Base
Just before and after the equivalence point, the pH increases rapidly. At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

6. Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.

7. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
The pH at the equivalence point will be >7.
Phenolphthalein is commonly used as an indicator in these titrations.

8. Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

9. Titration Problem Solving

10. 1. Stoichiometry problems; then 2. equilibrium problems to find pH
Consider titration of mL of M acetic acid (Ka = 1.8 x 10-5) by M KOH. Calculate pH after KOH has been added:
0.0 mL b mL c mL
d mL e mL f mL
Only a weak acid is present…
HC2H3O2 ↔ H C2H3O2 - I
.200M
≈ 0 C -x +x E
0.200 –x x

11. b. Added OH- will react completely with the best acid present: HC2H3O2
1.8 x = x2
0.200
x = [H+] = 1.9 x 10-3 M
pH = 2.72
b. Added OH- will react completely with the best acid present: HC2H3O2
HC2H3O2 + OH ↔ C2H3O H2O I
.200 M x .1000L = o.0200 moles
0.100 M x L = moles
---- C E
mols
mols

12. After reaction we have a buffer solution
pH = -log (1.8 x 10-5) + log ( /0.150L)
( /0.150L)
pH = (-0.477) = 4.26
c. HC2H3O OH ↔ C2H3O H2O I
0.200 M x L = mols
M x L = moles C E
mols
pH = -log (1.8 x 10-5) + log (0.0100/0.200L)
( /0.200L)
pH = 4.74

13. d. HC2H3O2 + OH- ↔ C2H3O2- + H2O pH = 4.74 + log (0.0150/0.0050) = 5.22 I
0.100M x L = mols C E
0.0050
0.0150

14. Moles of acid = moles of base pH determined by conjugate base
HC2H3O2 + OH- ↔ C2H3O H2O
**equivalence point
Moles of acid = moles of base
pH determined by conjugate base
Kb = Kw/Ka Kb = 5.6 x 10-10 I
0.0200
0.100 M x L = C E
mols

15. C2H3O2- + H2O ↔ HC2H3O2 + OH- 5.6 x 10-10 = x2 0.0667
X= [OH-] = 6.1 x 10-6M
pOH = 5.21; pH = 8.79 I
0.0200/0.300 L = M C -x +x E x x

16. HC2H3O2 + OH- ↔ C2H3O H2O I
0.0200
0.0250 C E
0.0050
[OH-] = /0.350 L
= M
pOH = 1.85
pH = 12.15 LR
Excess reactant
Since [OH-] is strong, pH is determined by excess, conjugate base effect is negligible

17. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

18. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.

19. Buffers Buffer exists at ½ way titration point pH is stable
pKa value
Figure: 17-08
Title:
Adding a strong acid to a strong base.
Caption:
The shape of a pH curve for titration of a strong base with a strong acid. The pH starts out at a high value characteristic of the base and then decreases as acid is added, dropping rapidly at the equivalence point.
Buffers

20. Weak vs. strong
½ way titration
Equivalence point

21. Titrations of Polyprotic Acids
In these cases there is an equivalence point for each dissociation.
Also buffering at each ½ titration point.