An unknown acid was titrated two times with 0.10 M NaOH. The first titration was done using an indicator to determine the equivalence point volume. A second titration was carried out where the pH was determined throughout the entire titration. The data is given below.
Titration 1: mL unknown acid = 25.0, mL of NaOH needed to reach equivalence point = 30.00
Titration 2: mL unknown acid = 25.0
mL 0.1M pH
NaOH Added
0.0 3.72
5.0 5.82
10.0 6.22
15.0 6.52
6.82 20.0
25.0 7.22
30.0 9.63
35.0 11.92
A. What is the concentration of the unknown acid?
B. What is the Ka of the unknown acid?
C. Which of the indicators given below would be a good choice to use in the titration of this acid with the NaOH?
Indicator pH Range of Color
Change
Thymol Blue 1.2- 2.8
Bromphenol Blue 3.0- 4.6
Alizarin 5.6-7.2
Phenol Red 6.8- 8.2
O-cresolphthalein 8.4- 9.8
Alizarin Yellow R 10.0- 12.0
1) For first titration at equivalence point number of mole of acid = number of moles of base
M1V1 = M2V2
M1 x 25 = 0.10 x 30
M1 = 0.12 M
2) pH = pKa + log [A-] / [HA]
6.52 = pKa + log [1.5] / [1.5]
pKa = 6.52 = -log Ka
Ka = 3 x 10-7
3) In titration 2, the equivalence point was determined to be between the pH 7.22 and 9.63 based on the sharp change in pH. This means that the indicator should change color in that pH range. If this is accurate, then O-cresolphthalein would be the best indicator to use because its range of color change is closest to that of the equivalence point.