Question

The following data were taken to determine the equilibrium constant of the Iron(III)-Thiocyanate Complex Ion: Tube .00200 M KSCN (ml) .00200 M Fe(NO3)3 (ml) Absorbance 2 8 2 .480 4 5 5 .720 6 2 8 .448 The product of eb = 6120 L/mol.

Determine the [Fe(SCN] in tube 2

Determine what the concentration of the Iron(III) in tube 2 would be if none of it reacted.

Determine what the concentration of the thiocyanate ion in tube 2 would be if none of it reacted.

Determine the concentration of the unreacted Iron(III) in tube 2.

Determine the concentration of the unreacted thiocyanate ion in tube 2.

Determine the equilibrium constant of the Iron(III)-Thiocyanate Complex Ion

Answer 1

${Fe^{3+}+SCN^{-}}={FeSCN^{2+}}$

This is the reaction for the formation of Iron_(III) Thiocyanate complex. The product is blood red in colour and hence its absorbance can be measured.

To determine the concentration of Iron_(III) Thiocyanate complex in tube 2:

From Beer Lambert's Law,

A=ebc

where, A - Absorbance of the sample

e - Molar Extinction Coefficient of the given sample

b - length of the cuvette used in the absorbance measurement

c - concentration of the absorbing species.

Hence, to determine the concentration, c=A/eb

eb = 6120 L/mol ( given in the question)

[FeSCN²⁺] = 0.480/6120L/mol = 7.84 x 10^-5mol/L

Concentration of Iron_(III) in tube 2 if none reacted

Total volume in tube 2 = 10ml

Concentration of Iron_(III) = (MV)_Iron(III) / Total volume

                                                   ₌ 0.002M x 2ml / 10ml = 0.0004M = 4.0 x 10^-4M

Concentration of SCN^-in tube 2 if none reacted

Total volume in tube 2 = 10ml

Concentration of Iron_(III) = (MV)_SCN^- / Total volume

                                                   ₌ 0.002M x 8ml / 10ml = 0.0016M = 1.6 x 10^-3M

Concentration of unreacted Iron_(III) in tube 2

Concentration of unreacted Iron_(III) in tube 2 = Initial concentration of Iron_(III) - Concentration of FeSCN²⁺ formed

                                                                      = 4.0 x 10^-4M - 7.84 x 10^-5M

                                                                      = 3.21 x 10^-4M

Concentration of unreacted SCN^-in tube 2

Concentration of unreacted SCN^-in tube 2 = Initial concentration of SCN^- - Concentration of FeSCN²⁺ formed

                                                                      = 1.6 x 10^-3M - 7.84 x 10^-5M

                                                                      = 1.52 x 10^-3M

To determine the equilibrium constant of the reaction :

From the equation given above,

$K_{eq} = \frac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^{-}]}$

Here, the unreacted concentration of Iron_(III) and SCN^- should be considered.

K_eq = 7.84 x 10^-5/ (3.21 x 10^-4)(1.52 x 10^-3)

      = 1.60 x 10²