this is what i have Cup Volume of stock CaCl2 solution Volume of 2propanol ml So
ID: 527630 • Letter: T
Question
this is what i have
Cup
Volume of stock CaCl2 solution
Volume of 2propanol ml
Solution color
Absorbance
1
.50
9.50 ml
Light blue
.0259
2
1.00
9.00
Blue
.504
3
1.50
8.50
Medium dark blue
.797
4
2.00
8.00
Dark blue
1.011
Mass of CoCl2*6H2o in 50ml 2propanol. G
.01414
Cup
Volume of CoCl2
Volume of M ml
Volume of p ml
Solution
Abs
T in C
A
1.00
1.00
8.00
Light blue
0.439
21
B
1.00
1.50
7.50
Light blue
0.356
21
C
1.00
2.00
7.00
Clear blue
0.285
21
D
1.00
2.50
6.50
Pale blue
.144
20
E
1.00
3.00
6.00
Very pale blue
.044
20
f
1.00
2.50
7.00
Light blue
.084
10.5
)
Co^2+ concentration of stock CoCl2 solution
0.0139
Cups
[Co(tet)],M
Abs
1
.006
.0367
2
.0012
0.631
3
.0018
0.761
4
.0024
1.299
Slope
Cup
[Co(total)
Abs
[Co(tet)]
[Co(oct)]
[p]
[M]
Keq
T
a
.0012
0.498
.0008107
.0003893
10.45
10.45
.333
20.05
b
.0012
.380
.0007018
.0004982
9.74
9.74
.137
.20.05
c
.0012
.257
.000475
.000725
9.02
9.02
.114
20.05
d
.0012
.143
.0002641
.0009359
8.53
8.53
.131
20.05
e
.0012
.0690
.000127
.0001073
7.84
7.41
.0163
20.05
f
.0012
.156
.0002881
.0009119
9.02
4.94
.237
15.90
Cup
Volume of stock CaCl2 solution
Volume of 2propanol ml
Solution color
Absorbance
1
.50
9.50 ml
Light blue
.0259
2
1.00
9.00
Blue
.504
3
1.50
8.50
Medium dark blue
.797
4
2.00
8.00
Dark blue
1.011
Mass of CoCl2*6H2o in 50ml 2propanol. G
.01414
x MCI DON'T KNOW WHAT To x MC Reward x Mcchemistry x y 102 Manual.pdf C www.ars-chemia.net Manual.pdf 103 131 102 Manual.pdf 103 2. An alternative equilibrium that could be used to explain the conversion of tetrahedrally coordinated Coz in propan-2-ol o the octahedral form upon addition of methanol is represented by the following equation In this equilibrium, five methanol molecules replace the three propan-2-ol molecules in the complex ion. (a Write an expression for Ke for this equilibrium. of Keq for the solutions in cups A, C, and E. (c) Based on your answers to (b), confirm or reject this alternative equilibrium. Briefly explain. P 6 7:05 PM 5/8/2017Explanation / Answer
The alternate equilibrium is given as:
[CoClP3]+(tetrahedral Co2+) + 5M (methanol) <-----> [CoClM5]+ (octahedral Co2+) + 3P (2-propanol)
Keq = [Cooct] [P]3 / [Cotet] [M]5
For Cup A,
Total [Co2+] = 0.0012 M
[Cotet] = 8.107 X 10-4 M
[Cooct] = = 3.893 X 10-4 M
[M] = number of moles of methanol / Total volume
= mass of methanol / Molar mass X Total volume
= Volume of methanol X Density of methanol / Molar mass X Total volume
= 1 X mL• 0.791 g mL-1 /32.04 g mol-1 • 10 X 10-3 L
= 2.47 M
Similarly, [P] = Volume of propanol X Density of propanol / Molar mass X Total volume
= 8 X mL• 0.785 g mL-1 /60.09 g mol-1 • 10 X 10-3 L
= 10.45 M
NB: The volumes of propanol and methanol are different, hence concentrations cannot be same. If you have used same concentrations in the previous Keq calculations, you might want to recheck your Keq calculations.
Keq = [Cooct] [P]3 / [Cotet] [M]5
= (3.893 X 10-4)•(10.45)3 / 8.107 X 10-4 • (2.47)5
= (3.893 X 10-4)•1141.2 / 8.107 X 10-4 • 91.93
= 4442.7 / 745.3
Keq = 5.96
For Cup C,
Total [Co2+] = 0.0012 M
[Cotet] = 4.75 X 10-4 M
[Cooct] = 7.25 X 10-4 M
[M] = number of moles of methanol / Total volume
= mass of methanol / Molar mass X Total volume
= Volume of methanol X Density of methanol / Molar mass X Total volume
= 2 X mL• 0.791 g mL-1 /32.04 g mol-1 • 10 X 10-3 L
= 4.94 M
Similarly, [P] = Volume of propanol X Density of propanol / Molar mass X Total volume
= 7 X mL• 0.785 g mL-1 /60.09 g mol-1 • 10 X 10-3 L
= 9.14 M
Keq = [Cooct] [P]3 / [Cotet] [M]5
= (7.25 X 10-4)•(9.14)3 / 4.75 X 10-4 • (4.94)5
= (7.25 X 10-4)•764.7 / 4.75 X 10-4 • 2941.9
= 5544 / 13974
Keq = 0.39
For Cup E,
Total [Co2+] = 0.0012 M
[Cotet] = 1.27 X 10-4 M
[Cooct] = 1.073 X 10-4 M
[M] = number of moles of methanol / Total volume
= mass of methanol / Molar mass X Total volume
= Volume of methanol X Density of methanol / Molar mass X Total volume
= 3 X mL• 0.791 g mL-1 /32.04 g mol-1 • 10 X 10-3 L
= 7.40 M
Similarly, [P] = Volume of propanol X Density of propanol / Molar mass X Total volume
= 6 X mL• 0.785 g mL-1 /60.09 g mol-1 • 10 X 10-3 L
= 7.83 M
Keq = [Cooct] [P]3 / [Cotet] [M]5
= (1.073 X 10-4)•(7.83)3 / 1.27 X 10-4 • (7.40)5
= (1.073 X 10-4)•480.0 / 1.27 X 10-4 • 22190
= 515 / 28181
Keq = 0.018
Thus you can see that the calculated Keq values for cups A, C and E are not constant (within experimental error). Hence, the proposed equilibrium is inappropriate.
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