0.0333 M `KMnO_4` is used for the following titration.What volume of the solution in ml will be required to read with 0.158 g of `Na_2S_2O_3`? (Mol. Wt. of `Na_2S_2O_3=158`)
`S_2O_3^(2-)+MnO_4^(-)+H_2O to MnO_2(s)+SO_4^(2-)+OH^(-)`
(not balanced)

Answer

80

Explanation

`S_2O_3^(2-)to2SO_4^(2-)`
Change in oxidation number of sulphur per molecule of `S_2O_3^(2-)=2xx(6-2)=8`
Change in oxidation number of Mn per molecule of `MnO_4^(-)=7-4=3`
No of moles in 0.158 g of `Na_2S_2O_3=0.158/158=1xx10^(-3)`
No of equivalents =`8xx10^(-3)`
Normality of 0.0333 M `KMnO_4` solution =`0.0333xx3=0.1`
Let V ml of volume of `KMnO_4` be required, then
`V/1000xx0.1=8xx10^(-3)`
or V=80 ml