MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_(2). The oxidation of 25 mL of a solution X containing Fe^(2+) ions required in acidic condition, 20 mL of a solution Y containing MnO_(4)^(-) ions. What volume of solution Y would be required to oxidise 25 mL of solution X containing Fe^(2+) ions in neutral condition ?

MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions where

1.	MnO_(4)^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_(2). The oxidation of 25 mL of a solution X containing Fe^(2+) ions required in acidic condition, 20 mL of a solution Y containing MnO_(4)^(-) ions. What volume of solution Y would be required to oxidise 25 mL of solution X containing Fe^(2+) ions in neutral condition ?
Answer» 11.4 mL 12 mL 33.3 mL 35 mL Solution :`MnO_(4)^(-)+5Fe^(2+)+8H^(+) to Mn^(2+)+5Fe^(3+)+4H_(2)O` `(M_(1)V_(1))/(1)=(M_(2)V_(2))/(5)` `(M_(1)xx20)/(1)=(M_(2)xx25)/(5)` `M_(1)=(M_(2))/(4)` Neutral MEDIUM `MnO_(4)^(-)+3Fe^(2+) +4H^(+) to MnO_(2)+3Fe^(3+)+2H_(2)O` `(M_(1)V_(1))/(1)=(M_(2)V_(2))/(3)` `((M_(2))/(4))xxV_(1)=(M_(2)xx25)/(3)` `V_(1)=33.3 mL`