What are the concentrations of the main two citric acid ions in a 50 mm citric acid solution at ph 4.2?

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

that's what i got you can count the number of atoms

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.