Name
Titration Problems
1.
In a titration, what is the difference between the end-point and the equivalence point?
2.
Consider the titration of 30.00 ml of HCl of unknown molarity. It required exactly 42.50 ml of 0.1200 M NaOH to reach the equivalence point. a.
A. Write the net ionic equation for the neutralization reaction. b.
Calculate the molarity of the acid. c.
What is the pH of the hydrochloric acid solution (the analyte) before any NaOH is added to it? d.
What is the pH of the analyte solution after exactly 42.25 ml of NaOH is added to it? At this point, only one drop (0.05 ml) of NaOH is needed to reach the equivalence point. e.
What is the pH at the equivalence point? f.
What is the pH of the analyte solution after exactly 42.55 ml of NaOH is added to it? This is the point one drop after the equivalence point. g.
Draw the titration curve for this experiment being careful to maintain alignment with the pH values calculated in parts (c) through (f).

3.
Many indicators used in acid-base titrations are themselves weak acids and known Ka values. It is necessary that the indicator undergo a color change at or near the equivalence point. To insure this happens, and indicator is chosen that has a pKa close to the pH at the equivalence point. The volumes and molarities of the acids and bases used in two different titrations are given below. Calculate the expected pH at the equivalence point for each titration. a.
A. 20.00 ml of 0.100 M acetic acid (Ka = 1.7 x 10
-9
) + 20.00 ml of 0.100 M
NaOH. b.
50.00 ml of 0.1500 M HOCl (Ka = 3.0 x 10 8) + 46.00 ml of 0.1630 M KOH.
From the list of indicators below, explain which one would be best for teach titration.
Indicator
Bromthymol blue
Thymolphthalein
Alizarian yellow color change yellow to blue colorless to blue colorless to violet pKa
7.1
10.0
11.0
Phenolphthalein
Methyl red colorless to red red to yellow
8.7
5.2

7
11
3 pH
4.
The titration curve for the titration of 25.00 ml of a weak acid with 0.1105 M NaOH is shown below. Using the information given here and from the titration curve, answer the following. a.
What is the molarity of the weak acid solution? b.
c.
Indicate the “buffer region” on the titration curve.
From the titration curve, estimate the Ka of the weak acid.
22.0 ml
0 5 10 15 20 25 30 vol NaOH (ml)

5. A 30.00 ml volume of a weak acid, HA, (Ka = 5.0 x 10
-5
) is titrated with 42.60 ml of 0.1040
M NaOH to the equivalence point.
HA + NaOH

NaA + H
2
O a. What is the pH of the acid solution, the analyte, before any base is added to it? b. What is the pH of the acid solution after exactly 12.50 ml of NaOH is added? c. What is the pH of the acid solution after exactly 21.30 ml of NaOH is added? d. What is the pH of the analyte solutions after exactly 33.10 ml of NaOH is added? e. Write the balanced chemical equation which shows why the pH at the equivalence point in this titration will not be 7.00. f. What is the pH of the analyte solution at the equivalence point?