Question

If the initial amount of Fe(NO3)3 transferred to the cuvette is 0.026 mol, and the absorbance measurements indicate that 0.01 mol of FeSCN2 are present at equilibrium, what must be [Fe3 ] at equilibrium

Answer 1

Concentration of
`[Fe^(3+)]` at equilibrium is 7.394 M.

Step-by-step explanation:

Moles of ferric nitrate = 0.026 mol

Concentration of ferric nitrate =
`(0.026 mol)/(2.164* 10^(-3) L)=12.015 M`

1 mole of ferric nitrate gives 1 mole of ferric ions, then 12.015 M of ferric nitrate solution will have:

[Fe^{3+}]=12.015 M[/tex]

Moles of
`Fe(SCN)]^(2+)` at equilbrium = 0.01 mol

Concentration of
`Fe(SCN)]^(2+)` at equilbrium =

`[Fe(SCN)]^(2+)=(0.01 mol)/(2.164* 10^(-3) L)=4.621 M`

`Fe^(3+)+SCN^-\rightleftharpoons [Fe(SCN)]^(2+)`

Initially

12.015 M 0

At equilibrium

(12.015 - 4.621 )M 4.621 M

Concentration of
`[Fe^(3+)]` at equilibrium

= (12.015 - 4.621 )M = 7.394 M