1. How do I calculate the concentration of Cu(NH3)42+? I am only given that, in
ID: 496433 • Letter: 1
Question
1. How do I calculate the concentration of Cu(NH3)42+? I am only given that, in a test tube, there is
3.00 mL of 0.05 M CuSO4
mixed with
7.00 mL of 0.20 M NH3/0.20M NH4NO3
No DI water was added to this test tube.
2. Once I have the concentration of Cu(NH3)42+, I understand how to do part 2
3. I understand how to do part 3
4. This is where I get stuck again, I don't know how to solve for this
I am totally clueless on how to do 5-8
Data:
Standard Solution:
Wavelength (nm)
Absorption
580
0.706
585
0.720
590
0.729
595
0.735
600
0.739
605
0.735
610
0.730
615
0.724
620
0.709
625
0.699
630
0.688
Experiment Values
Test Tube #
NH3/NH4+ (mL)
H2O (mL)
Wavelength (nm)
Absorption
1
0.5
4.5
600
0.020
2
1.0
4.0
600
0.063
3
1.5
3.5
600
0.127
4
2.0
3.0
600
0.239
5
2.5
2.5
600
0.454
6
3.0
2.0
600
0.620
Beer-Lambert Law: A = (el)*C
Wavelength (nm)
Absorption
580
0.706
585
0.720
590
0.729
595
0.735
600
0.739
605
0.735
610
0.730
615
0.724
620
0.709
625
0.699
630
0.688
CALCULATIONS l. Calculate the concentration of Cu(NH3)42+ in the standard solution. (Assume all the Cuso4 has been converted to Cu(NH3)42 and take into consideration the dilution you made when preparing the solution.) 2. Calculate the molar absorption coefficient times the length of the path (gl) for this experiment by rearranging the Beer-Lambert equation (see p. 2.1 for review) and plugging in the concentration of the standard solution and its absorbance at the wavelength of maximum absorbance. 3. Using the (el) value determined above, calculate the concentration of Cu(NH3)42 in each of the 6 test tubes. 4. Next, determine the initial concentrations of NH3 and NH4 (before the reaction with Cu(OH)2 takes place) in each of the 6 test tubes Spring 2017 SDSU Chem 201 Page 5.6Explanation / Answer
1. Determination of concentration of [Cu(NH3)4]2+ in solution,
1 mole of Cu2+ reacts with 4 moles of NH3
initial moles of Cu2+ = 3 ml x 0.05 M = 0.15 mmol
initial moles of NH3 = 7 ml x 0.2 M = 1.4 mmol
[Cu(NH3)4]2+ formed = 4 x 0.15 mmol/10 ml = 0.06 M
4. Initial concentration of NH3/NH4+ in solution
Test tube NH3/NH4+ (M)
1 0.2 M x 0.5 ml/5 ml = 0.02
2 0.2 M x 1.0 ml/5 ml = 0.04
3 0.2 M x 1.5 ml/5 ml = 0.06
4 0.2 M x 2.0 ml/5 ml = 0.08
5 0.2 M x 2.5 ml/5 ml = 0.10
6 0.2 M x 3.0 ml/5 ml = 0.12
5. From absorbance data and given molar absorptivity (check you data) calculate equilibrium concentration of [Cu(NH3)4]2+ in each test tube
[Cu(NH3)4]2+ concentration (M) = absorbance/molar absorptivity
Equilibrium concentration of NH3/NH4+ (M) = initial NH3/NH4+ - equilibrium [Cu(NH3)4]2+
From these data we would calculate further values in 6,7 and 8.
required molar absorptivity.
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