Question

An unknown compound, X, is known to possess two groups that ionize as the pH changes. One of these groups is a carboxylate with a pKa of 2.0. When 100 mL of 0.1 M NaOH is added to 100 mL of 0.1 M solution of X at pH 2.0, the pH increases to 6.72. What is the pKa of the second ionizable group of X?

A. 4.33

B. 6.72 CORRECT ANSWER

C. 7.20

D. 8.55

E. 9.76

Please show how B. is the correct answer

Answer 1

The first pKa corresponds to a carboxylic acid with pka = 2.0 and other is some group whose pKa2 we have to calculate .

When the pH of the solution = 2.0 = pKa it means the carboxylic acid is 50% ionised.

Let X be a compound with formula HOOC-R - BH+

We need to find pKa of BH+

When the pH of the solution = 2.0 = pKa it means the carboxylic acid is 50% ionised.

That is

HOOC-R - BH+ [I] <-------> ^-OOC-R - BH+ [II] <------------> ^-OOC-R - B [III]

ar pH = 0 pH = pKa1 pH = pKa2

At pH = 0 both the groups are protonated.

At pH = pKa1 it is a buffer which has 50% structure I and 50% structure II

when the pH is raised to pH = pKa2 , the structure of X is 50% II and 50%III, that all COOH protons arelost and 50% of BH+ar lost

by adding NaOH solution , it takes exactly one equivalent of base as the remaining 50% of COOH are removed and 50% of BH+ are lost from pH =pka1 to pH = pKa2

Thus the equivalents of NaOH dded to X are exactly one equivalent [ 100x0.1 = 10 mmoles of base added to 100x0.1 = 10 mmoles of X] .

Thus after adding one equivalent of base from pH = pKa1 , the pH of solution should equal to pKa2

Thus pH = pKa2 = 6.72