4: Experiment What is the Equilibrium Constant for Cobalt (II) Complex in
Alcohol Solvents at Various Temperatures?
PRE-LABORATORY QUESTIONS
1. In this experiment the stock CoCl2 is CoCl26H2O in 2-propanol. Why must
contamination with water be avoided? Why is the water originally complexed to the
cobalt insignificant?
H2O can replace the alcohols as a ligand. [Co(H2O)6] 2+is an octahedral complex,
water being even smaller than methanol. The stock CoCl2 concentration is in the
millimolar range, so it contributes at most about 10 mM water. Pure water is 55.5
M; 2-propanol is over 10 M and methanol almost 5 M. 10 mM water cannot compete
with such high concentrations of alcohols.
2. Crystals of CoCl26H2O are pink; crystals of CoCl2 are blue. What do these colors
suggest about the coordination of cobalt in each compound?
Pink suggest octahedral coordination and blue suggest tetrahedral coordination.
3. Predict the geometry and color of a solution of CoCl2 dissolved in 2-methyl-2propanol, (CH3)3COH. Write the chemical formula of the coordination complex in
this solution.
2-methyl-2-propanol, also known as tertiary butanol or t-butanol, is even bulkier than
2-propanol and surely would not be able to fit into an octahedral complex.
Co((CH3)3COH)42+will be tetrahedral and blue.
4. If the conversion of a tetrahedral cobalt complex to an octahedral complex (Eq. 4) is
exothermic, H= -31 kJ mol-1, what will be the effect of increasing temperature on
the Keq and on the color of the solution?
Keq will decrease and the solution will become more blue.
5. To determine the extinction coefficient of the tetrahedral complex CoBr2P2 a
student dissolves 0.26 g of CoBr26H2O in 50.00 mL of 2-propanol to make a stock
solution. makes He up four solutions from this stock as follows:
Solution
Volume of stock
CoBr2 solution, mL
Volume of 2propanol, mL
Absorbance
at 670 nm
1
0.50
9.50
0.303
2
1.00
9.00
0.577
3
1.50
8.50
0.855
4
2.00
8.00
1.149
Calculate the concentration of CoBr2P2 in each solution. Plot A670 against
concentration. Determine .
stock [CoP4]=
(0.26 g)/326.7 g/mol)(0.0500
L) = 0.016 M
standard curve
1.2
y = 360.96x
R2 = 0.9992
1
sample [CoP4]=
A670
0.8
(0.016 M)(Volsample
(mL))/10.00 mL
0.6
0.4
0.2
0
0
0.001
0.002
= 3.6 x 102 M-1
0.003
[Co complex]
6. The stock solution in question 5 is used to explore the equilibrium
CoBr2P2 + 3 M CoBr2PM3 + P
or, more concisely,
[Co(tet)] + 3 M [Co(oct)] + P
In a sample containing 1.00 mL stock CoBr2 solution, 2.00 mL methanol and 7.00
mL 2-propanol, the absorbance at 670 nm is 0.181 at 20 oC. The density of
methanol is 0.791 g/mL and that of 2-propanol is 0.785 g/ml.
a. Write the equation for Keq.
K
Co
P
(oct)
eq
Co
M3
(tet)
b. Calculate the total concentration of Co, [Co(tet)] +[Co(oct)].
sample [Cototal]=(0.016 M)(1.00 mL)/(10.00 mL) = 0.0016 M
c. Calculate [Cotet].
[Co(tet)]= A/ = 0.181/(3.6 x 102 M-1)= 0.00050 M
d. Calculate [Co(oct)].
[Co(oct)]= [Cototal] - [Cotet ] = 0.0016 0.00050 = 0.0011 M
e. Calculate the concentrations of methanol and 2-propanol.
[methanol] = (2.00mL)(0.791g/mL)(1 mol/32.04 g)(1/.0100 L) = 4.94M
[2-propano] = (8.00mL)(0.785g/mL)(1 mol/60.09 g)(1/0.0100 L) = 10.5M
f. Calculate Keq.
K eq
0.0011 10.5
0.00050 4.94
0.19
3
g. If Ho is -31 kJ mol-1, what will Keq be at 10 oC?
Ho 1
R T2
vant Hoff equation: ln
K2
K1
ln 0.19
31000
1
8.3145 293.14
ln Keq10C
Keq10C = 0.30
1
T1
1
283.14
0.449