An aqueous solution of 6.3 g oxalic acid dehydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is a. 40 mL b. 20 mL c. 10 mL d. 4 mL

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.

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 = 0.4For req volume of NaOH0.4 x 10 = 0.1 x V2=> V2 = 40 ml

FIRST FIND MOLARITY 6.3×1000/126×250= 0.2then find NORMALITY BY MULTIPLING BY 2 { because for normality we should multiply in the in the molarity by (n) factor} then For req volume of NaOH0.4 x 10 = 0.1 x V2=> V2 = 40

Hello Hrishant,
The solution is as follows
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.
Thanks
I hope above solution will clear your all doubts.
Please feel free to post and ask as much doubts as possible.
All the best.