A CHEM 112 student has 750 mL sample of an unknown monoproti

A CHEM 112 student has 75.0 mL sample of an unknown monoprotic acid solution. He titrated it with 0.100 M NaOH and it required 22.12 mL of the NaOH solution to reach the equivalence point of the titration 6. A) What is the concentration of the unknown acid in the original 75.0 mL solution? B) The student monitored the pH throughout the titration, and after he had added only 15.0 mL of the NaOH solution, the measured pH was 6.39. What is the value of Ka for this \"unknown\" acid?

Solution

6)

A)

mmoles of NaOH = 22.12 x 0.100 = 2.212

volume of acid = 75.0 mL

mmoles of acid = mmoles of NaOH

75 x M = 2.212

M = 0.0295

original concentration of acid = 0.0295 M

B)

mmoles of NaOH = 15 x 0.1 = 1.5

mmoles of acid = 75 x 0.0295 = 2.212

HA +   NaOH   ---------------> A-    +   H2O

2.212      1.5                             0            0

0.712         0                          1.5

pH = pKa + log [salt / acid]

6.39 = pKa + log (1.5 / 0.712)

pKa = 6.07

Ka = 10^-pKa

Ka of unknown acid = 8.51 x 10^-7