Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

IN an experiment 500 ml of 0.01 M of kmno4 solution is added to 500 ml of KI solution in basic medium.The resultant solution was titrated with 0.015 M na2s2o3 solution using starch indicator.what is the volume(in ml) of na2s2o3 solution requied to completely react with liberated iodine?

IN an experiment 500 ml of 0.01 M of kmno4 solution is added to 500 ml of KI  solution in basic medium.The resultant solution was titrated with 0.015 M na2s2o3 solution using starch indicator.what is the volume(in ml) of na2s2o3 solution requied to completely react with liberated iodine?

Grade:12th pass

2 Answers

Vikas TU
14149 Points
4 years ago
On reaction. KmnO4 will reduce to MnO2 and KI to I2.
moles of Mn x v.f = moles of I2 x v.f
0.01*500 x 3 = mmoles x 1
millimoles = 15 mm.
after then,
the titration with  Na2S2O3  ,
Na2S2O3 + I2  ------------------>  Na2S4O6  +  NaI      
                 for na2so3,
moles x v.f of na2so3 = moles of I2 x v.f
frome here u can now easily caluclate the volume.
Iswarya Mahati
29 Points
4 years ago
i am not getting the correct sorry 
i am really sorry my question was wrong ki has 0.1 M THEN HOW ARE WE SUPPOSED TO DO IT NOW

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free