Le Chatelier and Kc in the Cobalt Chloride Equilibrium
A student dissolves CoCl2·6H2O in deionized water, producing a pink solution. The relevant equilibrium is [Co(H2O)6]2+(aq) + 4 Cl−(aq) ⇌ [CoCl4]2−(aq) + 6 H2O(l). The hydrated cobalt(II) ion is pink and the tetrachlorocobaltate(II) ion is blue. Adding concentrated HCl turns the solution blue. The student divides the blue mixture into two test tubes: one is placed in a 60 °C water bath and becomes more intensely blue; the other is placed in an ice bath and becomes pink. A separate equilibrium mixture at 25 °C has [Co(H2O)6]2+ = 0.085 M, [CoCl4]2− = 0.015 M, and [Cl−] = 0.40 M. The student must explain the shifts, determine the sign of ΔH for the forward reaction, and calculate Kc at 25 °C. The student has no spectrometer.
1. Predict and explain the color changes when concentrated HCl is added and when water is added to the blue mixture.
Read the worked answer
Adding concentrated HCl increases [Cl−]. The reaction quotient Q = [CoCl4^2−]/([Co(H2O)6^2+][Cl−]^4) becomes smaller than K, so the system shifts right, forming more blue CoCl4^2−. Adding water dilutes the solution and adds solvent on the product side. Dilution lowers the concentrations of dissolved species; because the left side has more dissolved particles (1 + 4 = 5) than the right side (1), the equilibrium shifts left, toward pink Co(H2O)6^2+. Water is the solvent and is omitted from the K expression, but dilution still changes the relative concentrations and drives the shift.
2. Use the temperature data to determine the sign of ΔH for the forward reaction. Explain.
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Heating the blue mixture makes it more intensely blue, so the forward reaction is favored at higher temperature. For an endothermic reaction, increasing temperature increases K and shifts the equilibrium toward products. Therefore ΔH° > 0 for the forward reaction as written. Cooling favors the reverse, exothermic direction, producing pink. This is van't Hoff behavior: ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1).
3. Calculate Kc at 25 °C from the given equilibrium concentrations and predict whether K at 60 °C is larger or smaller.
Read the worked answer
Kc = [CoCl4^2−]/([Co(H2O)6^2+][Cl−]^4), with water omitted because it is the solvent. Substituting: Kc = 0.015/[(0.085)(0.40)^4] = 0.015/[(0.085)(0.0256)] = 0.015/0.002176 = 6.9. At 60 °C the blue product is favored more strongly, so K at 60 °C is larger than 6.9. The equilibrium constant is temperature-dependent; adding HCl or water at fixed temperature changes Q and the position of equilibrium but does not change K.