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 whereas they

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 12.0 mL 33.3 mL 35.0 mL Solution : INACIDICMEDIUM 5 volof `Fe^(2+)` requires1 volof `MnO_(4)^(-)` inacidicmedium ` therefore25Vol` of ` Fe^(2+)` requires` 1/5xx 25vol ` of` MnO_4^(- ) `in acidicmedium `1/5xx 25vol` of `MnO_4 ^(- )~=20volor20 mL` In neutralmedium 3 molof ` Fe^(2+)` requires1volof `MnO_(4)^(-)`in neutralmediumthen25volof ` Fe^(2+)` requires`1/3xx 25vol` of ` MnO_4`inneutralmedium `1/5 XX 25vol` of `MnO_(4)^(- )~=20vol` `therefore1/3x 25vol` of ` MnO_4^(-)=(20) /(5) xx(25)/( 3 )vol`of `MnO_(4)^(-)` `= 33.3volor mL`