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A quantity of 25.0 mL of solution contai...

A quantity of 25.0 mL of solution containing both `Fe^(2+)` and `Fe^(3+)` ions is titrated with 25.0 mL of 0.0200 `M KMnO_(4)` (in dilute `H_(2)SO_(4)`). As a result, all of the `Fe^(2+)` ions are oxidised to `Fe^(3+)` ions.
Next 25 mL of the original solution is treated with Zn metal finally, the solution requires 40.0 mL of the same `KMnO_(4)` solution for oxidation to `Fe^(3+)`.
`MnO_(4)^(-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_(2)O`
Zinc aded in the second titration wil

A

0.01 M

B

0.02 M

C

0.10 M

D

0.20 M

Text Solution

Verified by Experts

The correct Answer is:
C
eq. `Fe^(+2)` = eq. `KMnO_(4)`
`25xxMxx1=25xx0.02xx5impliesM=0.1`
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