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Q.

A solution is prepared by dissolving a solid mixture of K2C2O4 and KHC2O4. A 10 mL portion of this solution required 10 mL, 0.05 M KOH solution for titration reaction. In a seperate analysis 10 mL, of the same stock solution required 10 mL, 0.06 M acidified KMnO4 solution for titration. Which of the following are correct?

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a

The original mixture contains K2C2O4 and KHC2O4 in 2:1 molar ratio

b

20 mL of the original stock solution requires 6 milli-equivalents of acidified dichromate solution for titration
 

c

 The original mixture contains K2C2O4 and KHC2O4 in 2:1 molar ratio.
 

d

20 mL of the original stock solution require 3 millimoles of acidified dichromate solution for titration

answer is A, C.

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Detailed Solution

milli equivalent of  KHC2O4 in 10mL=10×0.05=0.5 

=mmol of KHC2O4

milli equivalent of K2C2O4 + milli equivalent of KHC2O4  =10×0.06×5 m mol of  K2C2O4×2+0.5×2=3

m mol of  K2C2O4=1

(a) 10.5=2:1

(b) 10 mL requires 0.5×2+1×2= 3 milli equivalent oof acidified

Cr2O72

(c) 20 mL requires 6 milli equivalent of acidified Cr2O72

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