Answer to Question #141957 in Chemistry for Abhi

Question

Calculate the normality of oxalic acid solution prepared by dissolving 3.15g of oxalic acid crystals in 250ml of distilled water

Answer 1

Oxalic acid chemical formula is C₂H₂O₄. It is the simplest dicarboxylic acid. It molar mass is 90.03 g/mol. As it dissolves with formation of two protons and an oxalate C₂O₄^2- anion, its equivalence factor is 2 and its equivalent weight is 90.03/2 = 45.015 g/equiv. The equivalent concentration of a solution is the molar concentration divided by the equivalence factor. 1L equals 1000 mL. Finally, the normality of oxalic acid solution is:

$\text{Normality} = \frac{m}{eq. weight \cdot V}$

$\text{Normality} = \frac{3.15\text{g}}{45.015 \text{g/equiv}\cdot250\cdot10^{-3} L} =0.28$ N.

Answer: the normality of oxalic acid solution prepared by dissolving 3.15g of oxalic acid crystals in 250ml of distilled water is 0.28 N.

Remark: if the oxalic acid is in its dihydrate form C₂H₂O₄$\cdot$ 2H₂O, its equivalence weight is 126.07/2=63.035 g/equiv and then the normality is 0.20N.