# What volume must the final solution reach if you want to make a "0.01 N" oxalic acid solution in water using "126 g" of oxalic acid solid?

Aug 12, 2017

I got $\text{280. L}$, if the oxalic acid was anhydrous.

However, it typically is purchased as a dihydrate, $\text{H"_2"C"_2"O"_4cdot2"H"_2"O}$, so if that is what you are looking for, then it would be $\underline{\text{200. L}}$.

Normality for acids is defined with respect to the ${\text{H}}^{+}$ the acid gives into solution. $\text{0.01 N}$ oxalic acid therefore would indicate a $\text{0.01 M}$ concentration for ${\text{H}}^{+}$ given to solution, and so, it would actually be $\text{0.005 M}$, as it is a diprotic acid.

Oxalic acid has a molar mass of $\text{90.03 g/mol}$, but the dihydrate equivalent would have a molar mass of $\text{126.06 g/mol}$, so the mols we have are:

$126 \cancel{\text{g OA" xx "1 mol OA"/(126.06 cancel"g") = "0.9995 mols}}$

These mols are dissolved in the appropriate volume $V$ such that

$\text{0.9995 mols"/V = "0.005 mols"/"L" " oxalic acid}$

Therefore, the volume the solution must reach by the time the $\text{126 g}$ of oxalic acid is dissolved is...

V = (0.9995 cancel"mols OA")/(0.005 cancel"mols""/L") = "199.9 L"

To three sig figs, it would be...

$\textcolor{b l u e}{V = \text{200. L}}$