Question

25.00 mL of a solution of oxalic acid are titrated with 0.2586 m NaOH (aq). The stoichiometric eend point is reached when 43.42 mL of the solution of base is added. What is the molarity of the oxalic acid solution? Oxalic acid reacts with sodium hydroxide as shown:
H2C2O4(aq) + 2 NaOH(aq) --> Na2C2O4(aq) + 2 H2O (l)

I thought I had this figured out last night, but now as many times as I try I can only come up with .044
I had .01123

Answer 1

And I don't think 0.044 is it.
I would call your attention to the 0.2586m NaOH. If that should be 0.2586 M and I assume it should, you should be aware that m stands for molality while M stands for molarity. They are two different animals. A word to the wise?
mols NaOH = M x L = 0.2586M x 0.04342L = about .011.
Use the coefficients in the balanced equation to convert mols NaOH to mols H2C2O4.
mols H2C2O4 = 1/2 mol NaOH = about 0.005.
M H2C2O4 = mols H2C2O4/L H2C2O4 = about 0.005/0.025 = about 0.22M

Answer 2

Thank you!
Yes, I meant M, sorry :)