Respuesta :
Answer : The concentration of [tex]Fe^{3+}[/tex] at equilibrium is 0 M.
Solution : Given,
Concentration of [tex]Fe^{3+}[/tex] = 0.0200 M
Concentration of [tex]C_2O_4^{2-}[/tex] = 1.00 M
The given equilibrium reaction is,
[tex]Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)[/tex]
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}[/tex]
[tex]1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}[/tex]
By solving the term, we get:
[tex]x=0.02M[/tex]
Concentration of [tex]Fe^{3+}[/tex] at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of [tex]Fe^{3+}[/tex] at equilibrium is 0 M.
