Question

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?

Answer 1

I got `"280. L"`, if the oxalic acid was anhydrous.

However, it typically is purchased as a dihydrate, `"H"_2"C"_2"O"_4cdot2"H"_2"O"`, so if that is what you are looking for, then it would be `ul"200. L"`.

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

Oxalic acid has a molar mass of `"90.03 g/mol"`, but the dihydrate equivalent would have a molar mass of `"126.06 g/mol"`, so the mols we have are:

`126 cancel"g OA" xx "1 mol OA"/(126.06 cancel"g") = "0.9995 mols"`

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

`"0.9995 mols"/V = "0.005 mols"/"L" " oxalic acid"`

Therefore, the volume the solution must reach by the time the `"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...

`color(blue)(V = "200. L")`