Question

A compound has two carboxyls and one aminium group. The pKas of the groups are 2.4, 4.4, and 8.7, respectively. A biochemist has 100 mL of a 0.10 M solution of this compound at a pH of 5.1. She adds 36 mL of 0.1 M NaOH. To the nearest hundredth of a unit, what will be the pH of the solution after addition of the NaOH (assume full stirring to reach a new equilibrium)?

Answer 1

pH = pKa + log [A-]/[HA]

pKa must be closer to pH.

So, pKa = 4.4

[HA]= 100 mL x 0.1 M = 10 mmol

[NaOH] = 36 mL x 0.1 M = 3.6 mmol

HA + NaOH ----------->. NaA + H2O

10 mmol 3.6 mmol

------------------------------------------------------------------

10-3.6 0 3.6 mmol

= 6.4 mmol

Therefore,

[HA] = 6.4 mmol

[NaA] = [A-] = 3.6 mmol

Then,

pH = pKa + log [A-]/[HA]

= 4.4 + log (3.6/6.4)

= 4.15

pH = 4.15

Therefore,

pH of the solution after addition of the NaOH = 4.15