Question

A solution of a theoretical triprotic acid was prepared by dissolving 4.251 g of solid in enough DI water to make 500.0 mL of solution. 13.65 mL of a 0.572 M solution was required to titrate 20.00 mL of this acid's solution.

What is the molar mass of the acid?

Answer 1

The number of moles of NaOH used for titration was = 0.01365 mL x 0.572 mol/L = 0.00781 mol

So 0.00781 mol [H+] were titrated in a 20 mL sample.

This values corresponds to 0.00781 mol x 500 mL/20mL = 0.195 mol [H+] in 4.251 g sample.

                                                                          

The question is not clear about the acidity of H3A.

1. If pK_a1 < pK_a2 – 3

The titration with a net EP for the first dissociation step is possible. In this case you may consider that the acid reacts as HA.

In this case molar mass is                                 

4.251 g / 0.195 mol = 21.8 g/mol (this value is not plausible for a triprotic acid).

2. If pK_a3 < 10 , the acid can be titrated completely (as triprotic acid).

In this case molar mass is

4.251 g / (0.195 mol /3 ) = 65.4 g/mol

I suggest you to answer 65.4 g H₃A /mol (if completely neutralized at titration EP, pH >10)