Question

Ferris & Mona used the ORP sensor to titrate a ferrous ammonium sulfate solution, (NH4)2Fe(SO4)2 with KMnO4 titrant. They titrated a 12.50 mL aliquot of the Fe+2 solution with 0.0215 M MnO4- solution and determined that the equivalence point was at 10.8 mL. What is the molarity of the Fe+2 solution?

Answer 1

Molarity of
`Fe^(2+)` solution is 0.0928 M.

Step-by-step explanation:

Balanced equation:
`5Fe^(2+)+MnO_(4)^(-)+8H^(+)\rightarrow 5Fe^(3+)+Mn^(2+)+4H_(2)O`

Number of moles of
`MnO_(4)^(-)` in 10.8 mL of 0.0215 M
`MnO_(4)^(-)` solution

=
`(0.0215)/(1000)* 10.8moles` = 0.000232 moles

Let's assume molarity of
`Fe^(2+)` solution is S(M)

Number of moles of
`Fe^(2+)` in 12.50 mL of S(M)
`Fe^(2+)` solution

=
`(S)/(1000)* 12.50moles` = 0.0125S moles

According to balanced equation, 5 moles of
`Fe^(2+)` react with 1 mol of
`MnO_(4)^(-)`

So, 0.0125S moles of
`Fe^(2+)` react with
`(0.0125S)/(5)` moles of
`MnO_(4)^(-)`

Hence,
`(0.0125S)/(5)=0.000232`

or, S = 0.0928

So, molarity of
`Fe^(2+)` solution is 0.0928 M.