Question

Following the Procedure of this experiment, a student titrated 0.653 g of an unknown weak, monoprotic acid with 0.100 M NaOH and monitored the titration with a pH meter. His titration data were:

Volume of NaOH solution added, mL /// pH
0.00 | 3.30

2.00 | 4.22

4.00 | 4.55

6.00 | 4.76

8.00 | 4.92

10.00 | 5.06

12.00 | 5.18

14.00 | 5.29

16.00 | 5.40

18.00 | 5.51

20.00 | 5.62

22.00 | 5.74

24.00 | 5.88

26.00 | 6.03

28.00 | 6.24

30.00 | 6.57

31.00 | 6.89

32.00 | 8.99
33.00 | 11.08
34.00 | 11.38
36.00 | 11.67
38.00 | 11.83

b) Calculate the number of equivalents of NaOH required to reach equivalence point
c) Determine the number of equivalents of acid titrated
d) Calculate the equivalent mass of the unknown acid
e) Determine pKa of unknown acid
f) Determine Ka of unknown acid
g) Identify unknown acid.

I will give best answer!!

Answer 1

number of equivalents = number of moles needed/ titrant volume in L

0.1M = X / .032 L

0.0032 equivalents NaOH (also equals number of equivalents of acid titrated)

equivalent mass = 0.653/0.0032= 204.06 g

pKa=pH at half equivalence point, so 5.4

pKa= -log Ka, so Ka=10^-pKa or 3.98*10^-6

Unknown acid is potassium hydrogen phthalate or KHP whichever your teacher prefers!