What amount of oxalic acid must be used to prepare 10 mL of .130 M oxalic acid solution?

Question

oxalic acid- H2C2O4 • 2 H2O, MW=126.0 g/mol
-This acid has 2 replaceable acidic protons.
-what would be the normality of the acid?

Answer 1

0.13 M oxalic acid soln => in each 1 L= 1000 ml of the soln 0.13 eqs of oxalic acid ispresent.
=> in 10 ml of the soln, 0.13*10/1000 mol of oxalic acid is present
= 0.0013 mol

=> reqd wt of oxalic acid = 126 * 0.0013 = 0.1638 g

since, 2*eq wt of the acid = mol wt of the acid,
=> normality of the soln will be = 0.13*2 = 0.26 N

Answer 2

Normality = Molarity x 2 (replaceable H+)
Normality = Equivalents / Liter
Think of equivalents as the MW divided by the number of replaceable Hydrogens or hydroxides.
So since the MW is 126.0 g/mole, the equivalent weight is 126/2 or 63 g/equivalents.

Sometimes you will be asked about normality in an oxidation/reduction equation. In this case is would refer to the MW divided by the electron change (gain or lost). For example, if you had a solution of 0.10 molar KMnO4 (Mn=+7) and it is reduced to Mn+2 in a reaction, then the Normality of the KMnO4 would be 5x the molarity because the equivalents would equal moles divided my 5.

They may have stopped teaching Normality in the US also...my last chemistry course was in 1965.

Answer 3

No of moles = 10 x 0.13/1000 = 0.00130.

Now multiply by 126 for your answer.

I'm afraid we stopped teaching normality in the UK about 30 years ago... (but I expect it's half the molarity).