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

Answer

8

Explanation

`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`