Question

For this equilibrium :

Fe³⁺ + SCN^- <====> [FeSCN]²⁺

If Z = 4.2810³ M ^-1. A solution at equilibrium that initially contained 1.59 x10^-3[SCN] and 8.46x10^-4 M [Fe2+] was found to have an absorbance of 0.402 units. What is the equilibrium constant? please explain

Answer 1

Fe³⁺ (aq) + SCN^- (aq)<====> [FeSCN]²⁺(aq)

We are provided with the following-

Z, slope = 4.28103M^-1

Absorbance, A = 0.402

[Fe³⁺] = 8.46x10^-4M   [SCN^-]= 1.59 x10^-3 M

We are required to find Kc i.e. the equilibrium constant for this reaction-

The equation of Kc for the above reaction is-

Kc = [FeSCN]²⁺ / [Fe³⁺]  [SCN^-]

We already know [Fe³⁺] and [SCN^-] as it has been already provided in the question. We just need to find out [FeSCN]²⁺, put all the data in the Kc equation to get the answer.

To know [FeSCN]²⁺, there exists a relation between absorbance, concentration and slope which is-

Absorbance=slope⋅conc.

conc.of [FeSCN]²⁺ =Absorbance / slope

or, conc.of [FeSCN]²⁺ = 0.402 / 4.28103M^-1

or, conc.of [FeSCN]²⁺ = 0.0939M

Now that we have obtained the conc.of [FeSCN]²⁺ , we can now putt the all the values in the Kc equation to find the value of Kc-

Kc = [FeSCN]²⁺ / [Fe³⁺]  [SCN^-]

Kc = 0.0939M / (8.46x10^-4M x1.59 x10^-3 M)

Kc =  0.0939M / 1.34514 x 10^-6M

Kc = 0.0698 x 10⁶