Question

If 26.23 mL of potassium permanganate solution is required to titrate 1.041 g of ferrous ammonium sulfate hexahydrate, `FeSO_4(NH_4)_2SO_4 * 6H_2O`, chow do you calculate the molarity of the `KMnO_4` solution.?

Answer 1

You can do it like this:

Explanation:

We have 1.041 g of ferrous ammonium sulfate hexahydrate.

The `sf(M_r=392.14)`

`:.` `sf(n_(Fe^(2+))=m/M_(r)=1.041/392.14=0.002654color(white)(x)"mol")`

Set up the equation from the two 1/2 equations:

`sf(MnO_4^(-)+8H^++5erarrMn^(2+)+4H_2O" "color(red)((1)))`

`sf(Fe^(2+)rarrFe^(3+)+e" "color(red)((2)))`

To get the electrons to balance we need to X `sf(color(red)((2)))` by 5 then add to `sf(color(red)((1))rArr)`

`sf(MnO_4^(-)+8H^++cancel(5e)+5Fe^(2+)rarrMn^(2+)+4H_2O+5Fe^(3+)+cancel(5e))`

This tells that 5 moles of `sf(Fe^(2+))` react with 1 mole of `sf(MnO_4^-)`

`:.` `sf(n_(MnO_4^(-))=0.002654/5=5.3093xx10^(-4))`

`sf(c=n/v)`

`:.` `sf([MnO_4^(-)]=(5.3093xx10^(-4))/(26.23/1000)=0.02024color(white)(x)"mol/l")`