What is the molarity of the oxalic acid (H2C2O4) solution that required 18.29 mL of a 1.75 x10–2 M aqueous solution of NaOH to completely neutralize a 20.00 mL sample of H2C2O4?

1 Answer
anor277
Oct 18, 2017

Approx. #0.01*mol*L^-1#....

Explanation:

For all these sorts of problems a stoichiometric equation is mandatory, so that we can establish the equivalence:

#"HO(O=)CC(=O)OH(aq) +2NaOH(aq)" rarr Na^(+)""^(-)"O(O=)CC(=O)O"^(-)Na^(+) +2H_2O#

And clearly, oxalic acid is a DIACID that reacts with TWO EQUIV of base. We gots an #18*mL# volume of #1.75xx10^-2*mol*L^-1# sodium hydroxide....and this represents a molar quantity of....

#18xx10^-3*Lxx1.75xx10^-2*mol*L^-1=3.15xx10^-4*mol#.

AND given the stoichiometry of the reaction HALF this molar quantity of oxalic acid were present in the original volume.....and so #"concentration"-=(1/2xx3.15xx10^-4*mol)/(20.00*mLxx10^-3*L*mL^-1)#

#=7.88xx10^-3*mol*L^-1#

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