It will be helpful to read the entire experiment first, before attempting the fo

ID: 1009356 • Letter: I

Question

It will be helpful to read the entire experiment first, before attempting the following questions 1. a) What color light is absorbed by the complex? b What color of light has a waveleneh of 480 m?Blue b) What color of light has a wavelength of 480 nm? c Why will the absorbance measurements be done at a wavelength near 480nm? BLue What is your hypothesis for how the equilibrium constant will vary with the trials in Part B? Explain. 2. 3. Calculate the [FeSCN ] in mol/L in each test tube for Part A and enter your results in the table below and in the table you prepared in your notebook. Show a sample calculation for test tube #A2 | 1.25 x 10-4 M Fe(NO)A (mL) 1.00 2.00 3.00 4.00 5.00 | | FeSCN+21 mol/L | 1.0 M KSCN 10.1 M HNO3 Test Tube No Al A2 A3 A4 A5 (mL) A 4.00 3.00 2.00 1.00 0.00 Absorbance (mL) 5.00 5.00 5.00 5.00 5.00 Sample calculation for test tube #A2 (Problem 4 is on the back!)

Explanation / Answer

Main equation

Fe(NO3)3 (aq) + 3KSCN (aq) ---> Fe(SCN)3 (aq) + 3KNO3 (aq)

The complete ionic equation

Fe+3 (aq) + 3NO3-(aq) + 3 K+(aq) + 3 SCN-(aq) à Fe(SCN)+2(aq) + 2SCN-(aq) + 3K+(aq) + 3NO3-(aq)

Net ionic equation

Cancelling out the spectator ions produces:
Fe+3(aq) + SCN-(aq) --> Fe(SCN)+2 (aq)

From reaction stoichiometry we can say that to form 1mol of Fe(SCN)+2 we need 1 mol of Fe(NO3)3 and 3 mol of KSCN.

Given concentration of KSCN= 1 M

And that of Fe(NO3)3 = 0.000125M

Therefore moles of reactant = molarity * volume in L

Moles of 0.000125M Fe(NO3)3 in 1 ml= 0.000125 * 0.001L = 1.25 * 10-7 mol

Moles of 0.000125M Fe(NO3)3 in 2 ml= 0.000125 * 0.002L = 2.5 * 10-7 mol

Moles of 0.000125M Fe(NO3)3 in 3 ml= 0.000125 * 0.003L = 3.75 * 10-7 mol

Moles of 0.000125M Fe(NO3)3 in 4 ml= 0.000125 * 0.004L = 5.0 * 10-7 mol

Moles of 0.000125M Fe(NO3)3 in 5 ml= 0.000125 * 0.005L = 6.25 * 10-7 mol

Number of moles of KSCN are constant in each test tube which is = 0.1M * 0.05L = 0.005 mols

We can see from above calculations that KSCN is in much excess than require, this means this reactant is not a limiting reagent. However Fe(NO3)3 concentration is very less and we can say that all of it would get converted to product at the end of reaction. Hence F(NO3)3 is a limiting reagent in this reaction.

Therefore; concentration of [FeSCN]2+ = Concentration of Fe(NO3)

Final volume of reaction mixture is 10ml therefore concentration of [FeSCN]2+ = Concentration of Fe(NO3) in each test tube would be as follows,

Molarity of product = number of moles of Fe(NO3)3 (as calculated above) / 0.010L

Test tube No.

[Fe(SCN)]+2

g/mol

0.000125M Fe(NO3)3 ml

0.1M KSCN

0.1M HNO3

1.25 * 10-5

2.5 * 10-5

3.75 * 10-5

5.0 * 10-5

6.25 * 10-5

Part b)

Similarly given equation here is

A2+ + 3B1- à AB32+

[A2+]

[B-]

[AB32+]

0.0219M

0.0612M

-x

-3x

0.0219-x

0.0612-3x

Equilibrium constant Kc = [AB3]2+/[A2+] [B-]3

Kc = x/(0.0219-x) (0.0612-3x)3

From the given reaction stoichiometry we can say that to form 1 mol of product we need 1 mol of A and 3 mols of B.

No. of mols of A = Molarity * volume in L = 0.0219M * 0.348L = 0.007621

No. of mols of B = Molarity * volume in L = 0.0612M * 0.525L = 0.03213

From the reaction stoichiometry for 0.007621mols of a we need = (3 * 0.007621) mols of B

Which is 0.02286,

But, B is given in excess so A is limiting reagent here are concentration of product = concentration of limiting reagent.

Concentration of AB32+ = Moles of A/Total volume of reaction in L = 0.007621/0.873 = 0.00872 M

Kc = 6.5133