1. Acid-Base Titrations
End point and equivalence point
Calculating the concentration of unknown acid or base
Choosing an indicator dye
Titration of a strong acid by a strong base
Titration of a weak acid by a strong base
Calculation of the titration curve - buffer region of a titration curve
Dependence of titration curve on pKa
Using the H-H equation to calculate pH

2. Acid-Base Titrations strong acid titrated with strong basepH
At the equivalence point, mol H+ = mol OH-
strong acid titrated with strong base
strong base titrated with strong acid

3. Titrations - Determination of the concentration of an unknown acid or base
A burette is used to add a precise volume of acid (or base) to the unknown solution. At the end-point of the reaction, the indicator changes color.
1. The flask contains the unknown acid
2. Base is added drop by drop to the end-point.
3. The indicator (in this case phenolphthalein) turns pink at the end-point.

4. Calculating the Concentration of an Unknown Acid or Base
end point: indicator changes color
equivalence point: moles [H+] = moles [OH-]
When the indicator is properly chosen: - end point = equivalence point
within experimental error.
Example: M HCl titrates mL unknown base mL HCl is added at the equivalence point.
(Molarity * Volume)acid = (Molarity * Volume)base
[base] = ( M * L )/ L = M

5. The Common Indicator Dyes

6. Titration of a Weak Acid with a Strong Base
pH = pKa + log10
[base]
[acid]
CH3COOH = CH3COO- + H+
pKa = 4.75
When pH = 4.75, [CH3COOH] = [CH3COO-]
This is the center of the buffering region.

7. Exercises: Calculate the Titration Profile
pH = pKa + log10
[base]
[acid]
What is the pH when 90.0 mL of NaOH has been added?
Ans: 5.70
What is the pH when 30.0 mL of NaOH has been added?
Ans: 4.38

8. Dependence of the Titration Curve on Acid Strength
On this diagram, note that the pH = pKa at the midpoint of the buffering region for each weak acid.

9. Titration of a Weak Base by a Strong Acid
pH = pKa + log10
[base]
[acid]
NH4+ = NH3 + H+
pKa = 9.25
When pH = 9.25, [NH4+] = [NH3]

10. Problem Calculate the pH of a 500 mL solution prepared from:
0.050 mol of acetic acid and mol sodium acetate.
HAc = H+ + Ac pKa = 4.75
0.020 / 0.50
[Ac-]
pH = pKa + log10
[HAc]
4.75
0.050 / 0.50
pH = 4.35

11. Problem (2)
(b)
Suppose that mol NaOH is added to the buffer of part (a). What is the pH?
HAc + OH- = H2O + Ac- pKa = 4.75
( ) / 0.50
[Ac-]
pH = pKa + log10
[HAc]
( ) / 0.50
4.63
4.75
As long as the buffer capacity is not exceeded, the change of pH is small, in this case, to 4.63