Answer : The concentration of
at equilibrium is 0 M.
Solution : Given,
Concentration of
= 0.0200 M
Concentration of
= 1.00 M
The given equilibrium reaction is,
![Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)](https://tex.z-dn.net/?f=Fe%5E%7B3%2B%7D%28aq%29%2B3C_2O_4%5E%7B2-%7D%28aq%29%5Crightleftharpoons%20%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%28aq%29)
Initially conc. 0.02 1.00 0
At eqm. (0.02-x) (1.00-3x) x
The expression of
will be,
![K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%5BFe%28C_2O_4%29_3%5D%5E%7B3-%7D%5D%7D%7B%5BC_2O_4%5E%7B2-%7D%5D%5E3%5BFe%5E%7B3%2B%7D%5D%7D)

By solving the term, we get:

Concentration of
at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M
Therefore, the concentration of
at equilibrium is 0 M.