Question

A compound is known to have a free amino group with a pk of 8.8, and one other ionizable group with a pKo between 5 and 7. To 100 mL of a 0.2 M solution of this compound at pH 8.2 was added 40 mL of a solution of 0.2 M hydrochloric acid. The pH changed to 6.2. The pK, of the second ionizable group is: a) The pH cannot be determined from this information.

Answer 1

Step 1:

Moles of unknown compound = 0.100 L * 0.2 moles/L = 0.020 moles

Moles of HCl added = 0.040 L * 0.2 moles/L = 0.008 moles

Step 2:

Find the conjugate base B and acid HA ratio in the unknown using the Henderson - Hasselbalch equation:

pH = pka + log[B]/[HA]

8.2 = 8.8 + log[B]/[HA]

B/HA = 10^(8.2-8.8) = 0.251 i.e. B = 0.251(HA)

The total moles : B + HA = 0.020

1.251 HA = 0.020

HA = 0.016

B = 0.020 -0.016 = 0.004

Step 3:

Moles of HCl added = 0.008 moles

New base [B] = 0.016 moles

New acid [HA] = 0.004 moles

pH = 6.2

6.2 = pka + log [0.016/0.004] =

pka = 5.59

pka of second ionizable group = 5.6