How would you find the concentration of a diprotic acid that was titrated with NaOH?

Lancenigo di Villorba (TV), Italy

You are treating two ACID/BASE NEUTRALIZATIONs.

TOPICs
As you suspected, STOICHIOMETRIC COEFFICIENTs LEAD WORTHY PROPORTION TO REACH NEUTRALIZATION STATUS.
For Your Case

H2A(aq) + 2 NaOH(aq) ---> Na2A(aq) + 2 H2O(aq)

So, once you assured to used only 0.1059 M CONCENTRATED SOLUTION OF STRONG BASE (e.g. NaOH) WHICH OCCURRED IN MEASURE LIKE 32.57 mL AGAINST 24.55 mL OF UNKNOWN ACID.
ONCE I WROTE "NET IONIC REACTION"

2 H+(aq) + 2 OH-(aq) ---> 2 H2O(aq)

I HIGHLIGHTED THAT NaOH IS MONOVALENT BASE SINCE IT GIVES ONE HYDROXYL ION (e.g. OH- which comes from Arrhenius definition of BASEs). ON THE OTHER HAND, H2A IS BIVALENT ACID SINCE IT GIVES TWO HYDROGEN ION (e.g. H+ which comes from Arrhenius definition of ACIDs).
So, YOU PLAYED BY A MONOVALENT BASE AND A BIVALENT ACID.


CONCLUSION
The mathematical relationship is

(n,acid * M,acid) * V,acid = (n,base * M,base) * V,base

where

n,acid = 2
n,base = 1
V,acid = 24.55 mL
V,base = 32.57 mL
M,base = 0.1059 M

so it has to be

M,acid = (n,base * M,base) * V,base / (n,acid * V,acid) =
= (1 * 0.1059) * 32.57 / (2 * 24.55) = 0.070 M

I hope this helps you.