Question

Chemical reactions occur to reach a state of equilibrium. The equilibrium state can be characterized by quantitatively defining its equilibrium constant, Keq. In this experiment, you will determine the value of Keq for the reaction between iron (III) ions and thiocyanate ions, SCN– . Fe3+ (aq) + SCN– (aq) → FeSCN2+ (aq) The equilibrium constant, Keq, is defined by the equation shown below. ]SCN][Fe[ ]FeSCN[ 3 2 + − + Keq = To find the value of Keq, which depends only upon temperature, it is necessary to determine the molar concentration of each of the three species in solution at equilibrium. You will determine the concentration by measuring light that passes through a sample of the equilibrium mixtures. The amount of light absorbed by a colored solution is proportional to its concentration. The red FeSCN2+ solution absorbs blue light, thus the Colorimeter users will be instructed to use the 470 nm (blue) LED. Spectrometer users will determine an appropriate wavelength based on the absorbance spectrum of the solution. In order to successfully evaluate this equilibrium system, it is necessary to conduct three separate tests. First, you will prepare a series of standard solutions of FeSCN2+ from solutions of varying concentrations of SCN– and constant concentrations of H+ and Fe3+ that are in stoichiometric excess. The excess of H+ ions will ensure that Fe3+ engages in no side reactions (to form FeOH2+, for example). The excess of Fe3+ ions will make the SCN– ions the limiting reagent, thus all of the SCN– used will form FeSCN2+ ions. The FeSCN2+ complex forms slowly, taking at least one minute for the color to develop. It is best to take absorbance readings after a specific amount of time has elapsed, between two and four minutes after preparing the equilibrium mixture. Do not wait much longer than four minutes to take readings, however, because the mixture is light sensitive and the FeSCN2+ ions will slowly decompose. In Part II of the experiment, you will analyze a solution of unknown [SCN– ] by using the same procedure that you followed in Part I. In this manner, you will determine the molar concentration of the SCN– solution. Third, you will prepare a new series of solutions that have varied concentrations of the Fe3+ ions and the SCN– ions, with a constant concentration of H+ ions. The concentrations of Fe³⁺ and SCN^- will be comparable in concentration with neither species in vast excess over the other. You will use the results of this test to accurately evaluate the equilibrium concentrations of each species in vast excess over the other. You will use the results of this test to accurately evaluate the equilibrium concentrations of each species, and thus the equilibrium constant for this reaction.

1. For the solutions that you will prepare in Step I of Part I below, calculate the [FeSCN²⁺]. Pressume that all of the

SCN^- ions react. In part I of th experiment, mol of SCN^-= mol of FeSCN²⁺. Thus, the calculation of [FeSCN²⁺] is: mol FeSCN²⁺/ L of total solution. Record these values in the table below.

Volumetric flask number [FeSCN²⁺]

1

2

3

4

(blank) 0.00M

2. For the solutions that you will prepare in Step 1 of Part 1 below, calculate the [FeSCN²⁺ ]. Pressume that all of the SCN^- ions react. In Part I of the experiment, mol of SCN^- = mole of FeSCN²⁺. Thus, the calculation of [FeSCN2+] is: mol FeSCN²⁺/L of total solution. Record these values in the table below.

Volumetric Flask number [FeSCN²⁺]

1

2

3

4

(blank) 0.00 M

Answer 1

In the reaction, 0.0020 M of SCN- is being used. In 1st beaker, 5 ml of this solution is taken and it is made upto 50 ml by adding Fe3+ solution and rest is water. So, the solution is diluted. To calculate the concentration of SCN- in the prepared solution after dilution, Formula used is

C₁V₁ = C₂V₂

where LHS is for desired solution and RHS is for available solution.

In this question, desired solution is 50 ml of unknown concentration solution.

So, C1 = ?

V1 = 50 ml

Available solution is 0.0020 M solution.

So, C2 = 0.0020 M

V2 = 5 ml

Put all these values in the formula,

C₁V₁ = C₂V₂

C₁*50 = 0.0020*5

C1 = 0.00020 M

[SCN-] = 0.00020 M

It is given that mol of SCN-= mol of FeSCN2+

So, [FeSCN]²⁺ = 0.00020 M

Similarly, in 2nd beaker, 4 ml [SCN-] is used. We can do the same calculations,

Instead of 5 ml, put 4 ml.

C₁V₁ = C₂V₂

C₁*50 = 0.0020*4

C1 = 0.00016 M

[SCN-] = 0.00016 M

It is given that mol of SCN-= mol of FeSCN2+

So, [FeSCN]²⁺ = 0.00016 M

So, we can complete the table given,

Beaker number	[FeSCN2+]
1	0.00020 M
2	0.00016 M
3	0.00012 M
4	0.00008 M
Blank	0.00 M