Answer :
a) The concentration of [tex]Fe^{3+}[/tex] at equilibrium is 0.08 m.
b) The concentration of [tex]SCN^{-}[/tex] at equilibrium is 0.26 m.
Solution :
The equilibrium reaction is,
                       [tex]Fe^{3+}+SCN^-\rightleftharpoons [FeSCN]^{2+}[/tex]
Initial concentration      0.230     0.410      0
At equilibrium        (0.230 - x)   (0.410 - x)     x
               Â
Given x = 0.150
Therefore,
The concentration of [tex]Fe^{3+}[/tex] = 0.230 - x = 0.230 - 0.150 = 0.08
The concentration of [tex]SCN^{-}[/tex] = 0.410 - x = 0.410 - 0.150 = 0.26
Thus, the concentration of [tex]Fe^{3+}[/tex] at equilibrium is 0.08 m.
The concentration of [tex]SCN^{-}[/tex] at equilibrium is 0.26 m.
