Question

An aqueous solution of 6.3 gram of oxalic acid dihydrate is made up of to 250 ml.the volume of 0.1 N naoh required to completely neutralise 10 ml of this solution is?

Answer 1

Sol. Oxalic acid dehydrate H2C2O4 . 2H2O : mw = 126 It is a dibasic acid, hence equivalent weight = 63 ⇒ Normality = 6.3/63 × 1000/250 = 0.4 N ⇒ N1V1 = N2V2 0.1 ×V1 = 0.4 × 10 Hence, V1 = 40 mL. Full explanation This question involves concept of volumetryWe can get the answer using the formulaN1V1=N2V2 ie. Principle of neutralizationHereOxalic acid is a dibasic acid as it can release two H+Hence Molarity = (6.3 x 1000)/( 126 x 250) = 0.2 => normality = 0.2 x 2 = 0.4 For req volume of NaOH0.4 x 10 = 0.1 x V2 => V2 = 40 ml