Question

A mixture is prepared by combining 10.20 mL of 1.80 x 10^-3 M Fe(NO₃)₃ with 10.00 mL of 1.68 x 10^-3 M KSCN. The solution turns red due to the formation of FeSCN²⁺. The absorbance is measured and, using a calibration plot, the [FeSCN²⁺] at equilibrium is found to be 2.20 x 10^-4 M.

Complete the following ICE table

	Fe3+   +	SCN-     ⇌	FeSCN2+
initial (M)	(A)	(B)	(C)
change (M)	(D)	(E)	(F)
equilibrium (M)	(G)	(H)	(I)

What is the value of K_c?

Answer 1

First we will determine the initial concentrations of Fe3+ and SCN- in the reaction mixture.

We have taken 10.20 mL of .

Each molecule of will dissociate into one   and three nitrate ion.

Also, we have taken 10.00 mL of .

Each molecule of KSCN dissociates into 1 molecule of K+ and 1 molecule of SCN-.

Hence, the total volume of reaction mixture = 10.20 mL + 10.00 mL = 20.20 mL

Now, using the dilution mole equivalence formula, we can calculate the initial concentration of Fe3+ in the reaction mixture

Where M1 and V1 are concentration and volume before the solution are mixed and M2 and V2 are concentration and volume after the solutions are mixed.

Hence,

Similarly,

Now, the equilibrium reaction that takes place is

Note that equal moles of Fe3+ and SCN- combine to form same number of moles of the complex in the right.

Initially, before any reaction happens, the initial concentration of will be zero.

Hence, we can fill the ICE table as follows

Initial, M
Change, M	-x	-x	+x
Equilibrium, M			x

Note that a change in x M each of Fe3+ and SCN- will result in a change of +x M for .

Hence, the equilibrium concentration of is x M.

Now, from absorbance calibration experiment, equilibrium concentration of FeSCN2+ is found to be .

Hence, we can fill the ICE table as follows:

Initial, M
Change, M
Equilibrium, M

Note that the equilibrium constant is calculated by taking the concentrations of reactant and products at equilibrium.

Now, we can calculate the equilibrium constant as follows: