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.

Answer 1

Answer :

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.