Citric acid , (COOH)CH2-C(OH)(COOH)-CH2(COOH) is a triprotic acid with three acidic protons. It can be represented as H3A
The equilibrium reaction are:
1) H3A -------- H2A- + H+ pKa1 = 3.14
2) H2A- --------- HA2- + H+ pKa2 = 4.76
3) HA2- -------- A3- + H+ pKa3 = 6.40
The given pH = 4.2 which is closest to pKa2. hence the two ions that will be present are: H2A- and HA2-
As per Henderson Hasselbach equation:
pH = pka + log [HA2-]/[H2A-]-----------(1)
[HA2-]/[H2A-] = 10^(pH-pKa) = 10^(4.2-4.76) = 0.275
i.e. [HA2-] = 0.275 [H2A-]-----------------(2)
It is given that: [HA2-] + [H2A-] = 50 mM ---------------(3)
substituting for HA2- IN eq(3) we get:
1.275 [H2A-] = 50 mM
[H2A-] = 50/1.275 = 39.22 mM
[HA2-] = 0.275(39.22) = 10.79 mM