Question

A symmetrical, dicarboxylic acid has two macroscopic ionization constants (pK1 and pK2), the values of which are, respectively, 4.3 and 5.38. Given this information, what is the value of the microscopic ionization constant pKC? Recall that K1 = KA + KB and that K2 = KCKD/(KC + KD). Enter your answer to the nearest hundredths.

Answer 1

We know that

K1 = KA + KB (eqn 1)

K2 = KCKD/(KC + KD). (eqn 2)

KA,KB,KC and KD are microscopic equilibrium constants amd K1 and K2 are two overall macroscopic equlibrium constants.

An equlibria involving these constants also give the relation KA/KB =KC/KD.

In this case,we are dealing with a symmetrical dicarboxylic acid in which the two ionisable protons are identical, KA =KB and KC=KD.

equations 1 and 2 modify as

pK1= pKA - log2

pK2= pKC + log2

pKC =pk2- log 2

=5.38 - log2

=5.0789 (answer)