Question

the titration of 25.0 mL of a 0.040 M aqueous solution of maleic acid C2H2(COOH)2, with aqueous 0.10 M NaOH solution. For maleic acid, Ka1 = 0.013 and Ka2 = 8.5 x 10-7

How many milliliters of the 0.10 M NaOH solution are needed to completely react all the maleic acid in 25.0 mL of a 0.040 M solution of maleic acid?
If we write maleic acid as H2M, then
H2M + 2NaOH ==> Na2M + 2H2O
mols H2M = M x L
\mols NaOH = twice that (look a the coefficients)
\Then M NaOH = mols NaOH/LNaOH
You know M and mols, solve for L NaOH and convert to mL.

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