Question

BEER's LAW                  A = 2 – log (% T) {Eqn # 1}               A = Constant x [Absorbing Molecule] {Eqn # 2}

Hi, I need help with the following Calculations for my Lab Report. The professor only gaves us these two Equations to solve them but I don't know how to start it or what to plug in. If you can please help me out. Thank you in advance. My Raw Data from Lab is below with the questions. Also on the very bottom is the information for the procedure and the concentration amounts.

CALCULATIONS

1. The proportionality constant in Beer's Law

In Solution 1, the concentration of iron (Fe3+) is so large that it is assumed, that the equilibrium has
COMPLETELY SHIFTED TO THE RIGHT. This means that the concentration of SCN– is zero, and the
concentration of the complex-ion (FeSCN2+) is equal to the initial concentration of the SCN–.

If ….A = Beer’s Law Constant x [FeSCN2+],

Re-arranging…. Beer’s Law Constant = A / [FeSCN2+]

2. Calculating the equilibrium concentration of FeSCN2+

Use the calculated Beer's Law Constant from solution 1, and the calculated absorbance of Solutions 2, 3,
and 4, to calculate the equilibrium concentrations of the FeSCN2+ in each solution.

3. Calculating the equilibrium constant, K

Use ICE tables to set-up the equilibrium expression and calculate the K of the complex-ion in solutions 2,
3, and 4.

RAW DATA FROM My LAB

Solution #1     Transmission = 10.5       Absorbance = 0.98

Solution #2     Trans. = 18         Absorbance =    0.75

Solution #3     Trans. = 34         Absrobance = 0.48

Solution #4     Trans. = 57        Absorbance = 0.25

Dilute nitric acid (NO3):                              0.50 M

Potassium Thiocianate (KSCN) sol’n:    0.00050 M

Ferric nitrate solutions (Fe(NO3)3):    0.050 M, 0.020 M, 0.0080 M, 0.0032 M

3. Use two small graduated cylinders to prepare the following four solutions. Use one cylinder for the ferric (Fe(NO3)3) solution, and one for the thiocyanate (KSCN) sol’n. Start with the least concentrated. Then calculate the initial concentrations of the KSCN and ferric ion solution in the new volume and record.

Solution 1: 4 mL of KSCN + 4 mL of the 0.050 M Fe(NO3)3

Solution 2: 4 mL of KSCN + 4 mL of the 0.020 M Fe(NO3)3

Solution 3: 4 mL of KSCN + 4 mL of the 0.0080 M Fe(NO3)3

Solution 4: 4 mL of KSCN + 4 mL of the 0.0032 M Fe(NO3)3

Answer 1

Reaction,

Fe3+(aq) + SCN-(aq) <==> Fe(SCN)2+(aq)

From the given data,

Solution          Fe3+ initial (M)                                     SCN- initial (M)

   #1         0.05 M x 4 ml/10 ml = 0.02                 0.0005 M x 4 ml/10 ml = 0.0002

   #2         0.02 M x 4 ml/10 ml = 0.008               0.0005 M x 4 ml/10 ml = 0.0002

   #3         0.008 M x 4 ml/10 ml = 0.0032           0.0005 M x 4 ml/10 ml = 0.0002

   #4         0.0032 M x 4 ml/10 ml = 0.00128       0.0005 M x 4 ml/10 ml = 0.0002