Question

How to calculate molarity of `Fe^(3+)`? `(Cr_2O_7)^(2-) + Fe^(2+) = Cr^(3+) + Fe^(3+)`

[ I just need help with the 3rd subquestion, but a confirmation for the other two will be greatly appreciated. ]

A 30.00 mL of 0.025 M sodium dichromate, `Na_2Cr_2O_7` solutions is titrated with iron (II) sulphate, `Fe_2SO_4` solution in acidic conditions, according to the following reaction:

`(Cr_2O_7)^(2-) + Fe^(2+) = Cr^(3+) + Fe^(3+)`

The titration requires 40.00 mL of `Fe_2SO_4` solutions to reach the end point.

(i) Balance the redox equation.
(ii) Calculate the mass of `Na_2Cr_2O_7` needed to prepare a 0.025 M solution in a 50 mL volumetric flask.
(iii) Determine the molarity of the `Fe ^(3+)` solution.

Answer 1

I make `[Fe^(3+)]` to be under `0.2*mol*L^-1`...

Explanation:

We should write the stoichiometric equation....

`underbrace(Cr_2O_7^(2-)+14H^+ +6e^(-) rarr 2Cr^(3+) + 7H_2O)_"orange to green...."`

`Fe^(2+) rarrFe^(3+) + e^(-)`

..we take one of the former and SIX of the latter to retire the electrons...

`6Fe^(2+) +Cr_2O_7^(2-)+14H^+ rarr 2Cr^(3+) + 6Fe^(3+) +7H_2O`

...the which is balanced with respect to mass and charge (is it?), and forms the basis of our calculation.

With respect to dichromate we got...`0.025*mol*L^-1xx50xx10^-3*L=0.00125*mol`

And given the equation, there are thus SIX equiv with respect to ferrous ion...

`[Fe^(2+)]=[Fe^(3+)]=(6xx0.00125*mol)/(40.00xx10^-3*L)=??*mol*L^-1`