Question

A mixture contains Na2C2O4 and KHC2O4 in 1:1 molar ratio. Mixture is neutralized by 100 ml of 0.01 M KOH. Thus the same mixture is oxidised by what volume of 0.01 M KMnO4?

Answer 1

EQUIVALENT CONCEPT: 

Steps:
1) In a mixture of $Na_{2}C_{2}O_{4}$ and $KHC_{2}O_{4}$ ,only $KHC_{2}O_{4}$ will react with KOH. 

=> No. of milli  moles of KOH = No. of milli  moles of  $KHC_{2}O_{4}$  (only 1 H) 
=>  100mL * 0.01 = n  
=> n=1 milli mole  of KHC2O4 . 

 2) As, No. of  milli mole of   $Na_{2}C_{2}O_{4}$ = No. of milli mole of $KHC_{2}O_{4}$ 
=> No. of milli mole of Sodium Oxalate = 1 
 
3)Valency Factor of Sodium Oxalate and Potassium Hydrogen Oxalate is 2.  (each) 
Valency Factor of $KMnO_{4}$ =5 :
 Now in Oxidation , both will be used due to C which is in +3 Oxidation state in both compounds .
=> No. of equivalents of Sodium Oxalate + No. of equvalents of Potassium Hydrogen Oxalate  =   No. of  equivalents of  $KMnO_{4}$ 

=> 2*1 + 2*1 = 5*0.01 *V 
=> V= 20mL . 

Hence,Volume of Potassium Permangante used up is 20mL .