# Experiment 3: Application of Solubility and Formation Equilibria - Preparation of a Black and White Polaroid Negative

Background

Chemical reactions are either endothermic (taking heat from the surroundings) or exothermic (giving off heat). The effects of temperature in chemical equilibrium can be predicted in terms of Le Châtelier's principle. If the reaction is endothermic, a rise in the temperature will shift the equilibrium to the right. In an exothermic reaction, the shift occurs in the opposite direction. The dependence of the equilibrium constant on the temperature is given by the van't Hoff equation:

$\ln \left(\frac{K_{2}}{K_{1}}\right)=-\frac{\Delta H^{o}}{R}\left[\frac{1}{T_{2}}-\frac{1}{T_{1}}\right] \nonumber$

where $$\Delta H^o$$ is the change in enthalpy of the reaction, $$R$$ the universal gas constant (8.31 J K-1mol1), and $$K_1) and \(K_2$$, the equilibrium constants at temperatures $$T_1$$ and $$T_2$$, respectively.

Materials

0.5g Cobalt(II) chloride hexahydrate,$$\mathrm{CoCl}_{2} \bullet 6 \mathrm{H}_{2} \mathrm{O}$$
12 mL 2-propanol, $$C_3H_7OH$$
Distilled water
Test tube with stopper
Hot water bath
Ice-salt bath

Safety

Cobalt chloride is a suspected carcinogen in animals. The dust is a respiratory tract irritant. Isopropanol is a flammable liquid and vapor (may cause a flash fire). REMOVE stopper before heating (never heat a closed system). When heating the testtube be sure it is not pointed towards yourself or colleagues.

Procedure

1. Place 0.5g of $$\mathrm{CoCl}_{2} \bullet 6 \mathrm{H}_{2} \mathrm{O}$$ in a large test tube
2. Add 12 mL of 2-propanol
3. Stopper and shake to dissolve solid. Note the color of the solution.
4. Carefully add water dropwise, with mixing, until a color change is observed.
5. Place the unstoppered test tube in a boiling water bath. Observed any color change.
6. Transfer the test tube to an ice bath. Record your observations.

Clean up: Discard the cobalt solution in the designated containers.

Discussion

Identify the colored species. Interpret your results in terms of Le Châtelier's principle.

$CoCl_2(s) \stackrel{2-propanol}{\longrightarrow} CoCl^{2-}_4 \stackrel{H_2O}{\longrightarrow} [Co(H_2O)_6]^{2+} + heat \nonumber$