Question

0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(aq) ⇄ [Fe(C2O4)3]3-(aq) Kc = 1.67 × 1020 What is the concentration of Fe3+(aq) when equilibrium is reached? 1.67 × 1020 M 8.35 × 10-19 M 6.9a × 1021 M 1.44 × 10-22 M 0.980 aM 0.940 M 0.0100 atm

Answer 1

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution :  Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

                            $Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.