Question

A reaction has initial concentration of 0.230 m fe3+ and 0.410 m scn- at equilibrium, the concentration of fescn2+ is 0.150 m.

a.what is the concentration of the fe3+ at equilibrium?
b.what is the concentration of the scn- at equilibrium?

Answer 1

a) The concentration of
`Fe^(3+)` at equilibrium is 0.08 m.

b) The concentration of
`SCN^(-)` at equilibrium is 0.26 m.

Solution :

The equilibrium reaction is,

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

Initial concentration 0.230 0.410 0

At equilibrium (0.230 - x) (0.410 - x) x

Given x = 0.150

Therefore,

The concentration of
`Fe^(3+)` = 0.230 - x = 0.230 - 0.150 = 0.08

The concentration of
`SCN^(-)` = 0.410 - x = 0.410 - 0.150 = 0.26

Thus, the concentration of
`Fe^(3+)` at equilibrium is 0.08 m.

The concentration of
`SCN^(-)` at equilibrium is 0.26 m.