Question

A compound has two carboxyls and one aminium group. The pKas of the groups are 2, 4.6, and 8.6, respectively. A biochemist has 100 mL of a 0.10 M solution of this compound at a pH of 5.2. She adds 38 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)? Hint: See the Acids and Bases handout on Canvas.

Answer 1

At pH 5.2 the major components will be related to the PKa = 4.6

First calculate acid base ratio to know what fraction is present in the form of the acid at pH 5.2 and what fraction of the total is present in the form of the base pH 5.2

PH = PKa + log (base/acid)

5.2 = 4.6 + log (base/acid)

log (base/acid) = 0.6

(Base/acid) = 10^0.6 = 3.98: 1

Now calculate moles of compound.

n= c v (in L)

n = 0.10M x 100/1000L = 0.01 moles

Calculate moles of base by using molar ratio

1 mol of acid, then 3.98 moles of base

So, when 0.01 mol of acid, mol of base = 0.0398 moles

Now calculate the moles of base added

n= c v (in L)

n = 0.10M x 38/1000L = 0.0038 moles

0.0038 moles NaOH neutralizes 0.0038 moles acid,

So, new moles acid = 0.01– 0.0038mol = 0.0062 mol

New moles of base 0.0398 + 0.0038 = 0.0436

Now calculate the PH by using Henderson–Hasselbalch equation

PH = PKa + log (base/acid)

     = 4.6 + log (0.0436 /0.0062)

    = 4.6 + 0.85 = 5.45

The pH of the solution will be 5.45 after addition of the NaOH