A MIXTURE OF H2C2O4 AND NAHC2O4 WEIGHING 2.02G WAS DISSOLVED IN WATER AND THE SOLUTION MADE UPTO ONE LITRE .TEN MILLILITRE OF THE SOLUTION REQUIRED 3.0ML OF 0.1N SODIUM HYDROXIDE SOLUTION FOR COMPLETE NEUTRALIZATION . IN OTHER EXPERIMENT, 10.0 ML OF THE SAME SOLUTION IN HOT DILUTE SULPHURIC ACID MEDIUM, REQUIRED 4.0 ML OF 0.1N POTASSIUM PERMANGANATE SOLUTIONS FOR COMPLETE REACTION. CALCULATE THE AMOUNT OF NAH2C2O4

Total Normality of the solution

NTotal = [(x/45) + (2.02 –x/112) ]
Since 10 ml of the solution is titrated, hence the amount (in equivalent) of the mixture in 10ml of the solution will be

10 /1000 [ (x/45) + (2.02 –x/112) ] = 10-2 [ (x/45) + (2.02 –x/112) ] eq......... 1

Since 3ml of 0.1N NaOH is consumed, the amount (in equivalent) of the mixture will be 3 ml / 1000 (0.1) = 3 x 10-4 eq ..........2
Equating 1 and 2 we get
3 x 10-4 = 10-2 [ (x/45) + (2.02 – x /112) ]

Solving x we get, x = 0.9g ( Oxalic acid) 2.02 – x = 2.02 -0.9

= 1.12 g (NaHC2O4)