Normality and Reactions
You will sometimes need to work out the Normality (or equivalence)
of a solution resulting after a reaction.
For example, consider the following reaction:
(CH2COOH)2 + 2 NaOH ––––» (CH2COONa)2 + 2 H2O
What is the Normality of the solution if 1 mole of Succinic Acid and 2 moles of Sodium Hydroxide are used and the final volume is 1000 ml?
Find the mass of Sodium Succinate that can be produced.
From the amounts of reactants used and the balanced equation, you can deduce that 1 mole of Sodium Succinate can be produced.
Thus Mass of Sodium Succinate produced = Moles x RMM = 1 x 162 g = 162g
From the amounts of reactants used and the balanced equation, you can deduce that 1 mole of Sodium Succinate can be produced.
Thus Mass of Sodium Succinate produced = Moles x RMM = 1 x 162 g = 162g
Find the gram equivalent of Sodium Succinate.
1 molecule of Sodium Succinate can replace 2 atoms of Hydrogen. Thus:
Gram equivalent of Sodium Succinate = RMM/2 = 162/2 = 81 g
1 molecule of Sodium Succinate can replace 2 atoms of Hydrogen. Thus:
Gram equivalent of Sodium Succinate = RMM/2 = 162/2 = 81 g
Find the number of equivalents in 1000 ml of solution.
Number of Equivalents = Total Weight of Succinate = 162 = 2 Eq
Gram Equivalent of Succinate 81
Thus the resulting concentration of Succinate is 2 Eq / litre = 2 Normality.
Number of Equivalents = Total Weight of Succinate = 162 = 2 Eq
Gram Equivalent of Succinate 81
Thus the resulting concentration of Succinate is 2 Eq / litre = 2 Normality.