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How many milliliters of 0.05M K(4)[Fe(CN...

How many milliliters of 0.05M `K_(4)[Fe(CN)_(6)]` solution are required for titration of 60 ml of 0.01M `ZnSO_(4)` solution, when the product of reaction is `K_(2)Zn_(3)[Fe(CN)_(6)]_(2)`?

Text Solution

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The correct Answer is:
8
`underset(n_(f)=2)(3ZnSO_(4))+2K_(4)[underset(n_(f)=3)(Fe(CN)_(6))]rarr K_(2)Zn_(3)[Fe(CN)_(6)]_(2)+3K_(2)SO_(4)`
Milliequivalents of `ZnSo_(4)=` miliequivalents of `K_(4)[Fe(CN)_(6)]`
Volume of `K_(4)[Fe(CN)_(6)](V_(2))=(60xx0.01xx2)/(0.05xx3)=8ml`
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