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?

Question

Cameron Andrews · Accepted Answer

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"#. 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"# Three sig figs' worth of data would be... #color(blue)(V = "200. L")#

Jose Ayala · Answer

undefined To determine the volume of the final solution needed to make a 0.01 N oxalic acid solution using 126 g of oxalic acid solid, we need to calculate the number of moles of oxalic acid first.

Given:
Mass of oxalic acid (H2C2O4) = 126 g
Molar mass of oxalic acid = 126.07 g/mol (C2H2O4)

Number of moles of oxalic acid:
$ \text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} $
$ \text{Number of moles} = \frac{126 \, \text{g}}{126.07 \, \text{g/mol}} $
$ \text{Number of moles} \approx 0.9997 \, \text{moles} $

To prepare a 0.01 N solution, we need 0.01 moles of oxalic acid per liter of solution.

Volume of the final solution:
$ \text{Volume} = \frac{\text{Number of moles}}{\text{Normality}} $
$ \text{Volume} = \frac{0.9997 \, \text{moles}}{0.01 \, \text{N}} $
$ \text{Volume} \approx 99.97 \, \text{L} $

Therefore, the final solution must reach approximately 99.97 liters to make a 0.01 N oxalic acid solution using 126 g of oxalic acid solid.