The reaction of thiocyanate (HSCN) with Fe3+ produces a 1:1 colored complex with

Question

The reaction of thiocyanate (HSCN) with Fe3+ produces a 1:1 colored complex with a maximum absorbance at 447 nm. Fe^3+ + HSCN = FeNCS^2+ + H+ The equilibrium concentration was monitored with a Spectronic 20. Exactly 5.00 mL sample of 1.02 x10-3 M solution of buffered HSCN was mixed with a 5.00 mL of a 1.96 x10-3 M buffered solution of Fe3+. The solution was buffered at a pH = 0.30. A transmittance of 31.1% was measured for this mixture at room temperature. Assuming the molar absorptivity for the FeNCS2+complex is 4700 L/cm-mol and the path length of 0.90 cm, calculate the value of the equilibrium constant for this reaction at room temperature.

Explanation / Answer

Ans. #1. Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                       A = Absorbance

                       e = molar absorptivity at specified wavelength (M-1cm-1)

                        L = path length (in cm)

                        C = Molar concentration of the solute

Given,

            Path length, L = 0.9 cm

            e (FeNCS2+) = 4700 L cm-1 mol-1

            Transmittance, T = 31.1 %

Using, Absorbance, A = 2 – log (%T) = 2 – log 31.1 = 2 – 1.493 = 0.507.

Putting the vales in equation 1-

            0.507 = (4700 L cm-1 mol-1) x C x 0.9 cm = C x 4230 L mol-1

            Or, C = 0.507 / (4230 L mol-1) = 1.199 x 10-4 mol L-1

            Or, C = 1.199 x 10-4 M                                             ; [mol/ L = M]

Therefore, [FeNCS2+] in final reaction mixture at equilibrium = 1.199 x 10-4 M

#2. Given,

[HSCN] in its original solution = 1.02 x 10-3 M

[Fe3+] in its original solution = 1.96 x 10-3 M

Total volume of reaction mixture = 5.0 mL (of HSCN) + 5.0 =mL (of Fe3+) = 10.0 mL

Now,

Ans. Using     C1V1 = C2V2                        - equation 2

C1= Concentration, and V1= volume of initial solution 1         ; Original solution

C2= Concentration, and V2 = Volume of final solution 2         ; Reaction mixture

So,

Initial [HSCN] in reaction mixture, C2 = (C1V1)/ V2

                                                            = (1.02 x 10-3 M x 5.0 mL) / 10.0 mL

                                                            = 5.1 x 10-4 M

Initial [Fe3+] in reaction mixture, C2 = (C1V1)/ V2

                                                            = (1.96 x 10-3 M x 5.0 mL) / 10.0 mL

                                                            = 9.8 x 10-4 M

#3. Equilibrium concentrations.

[FeNCS2+] at equilibrium = 1.199 x 10-4 M = 1.20 x 10-4 M

1 mol Fe3+ reacts with 1 mol HSCN to form 1 mol FeNCS2+. Therefore, [FeNCS2+] formed is equal to [Fe3+] as well as [HSCN] consumed during the reaction.

Thus,

            [Fe3+] consumed = [HSCN] consumed = [FeNCS2+] formed = 1.199 x 10-4 M

Now,

[Fe3+] at equilibrium = Initial [Fe3+] - [Fe3+] consumed

                                    = 9.8 x 10-4 M - 1.20 x 10-4 M = 8.60 x 10-4 M

[HSCN] at equilibrium = Initial [HSCN] - [HSCN] consumed

                                    = 5.1 x 10-4 M - 1.20 x 10-4 M = 3.90 x 10-4 M

#4. Equilibrium constant, K = [FeNCS2+] / ([Fe3+] [HSCN])

                                                = (1.20x 10-4 ) / [(8.60 x 10-4 ) (3.90 x 10-4 )]

                                                = 3.55 x 10-6

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