Changing Color in a Cobalt Chloride Equilibrium
Five test tubes each contain 5.0 mL of 0.10 M CoCl2 in a water-ethanol solvent. The equilibrium is [Co(H2O)6]2+ (pink) + 4 Cl− ⇌ [CoCl4]2− (blue) + 6 H2O, and the forward reaction is endothermic. The teacher adds concentrated HCl to tube 1, producing a blue solution. Tube 2 receives 5.0 mL of distilled water and turns pink. Tube 3 receives a few drops of 0.10 M AgNO3 and turns pink, with a white precipitate. Tube 4 is placed in a 60 °C water bath and becomes a deeper blue. Tube 5 is cooled in ice and becomes pink. A student says, 'Adding water to tube 2 shifted the equilibrium left because water is a product, and adding AgNO3 to tube 3 shifted it left because Ag+ reacts with the blue complex.' The teacher must decide which explanation to accept and how to correct the rest.
1. Predict and explain the direction of shift in tubes 1, 3, and 4 using Q versus K or Le Châtelier, and identify the species added or removed.
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In tube 1, concentrated HCl adds Cl−, a reactant. The reaction quotient Q falls below K because the denominator of K contains [Cl−]^4, so the system shifts forward and becomes bluer. In tube 3, Ag+ removes Cl− by precipitating AgCl(s). Removing a reactant makes Q greater than K, so the system shifts in the reverse direction and becomes pinker. In tube 4, heating favors the endothermic direction. Because the forward reaction is endothermic, the equilibrium constant increases with temperature and the system shifts forward, producing more blue [CoCl4]2−. This is a temperature effect on K, not just a concentration effect. The teacher should require students to name the species added or removed and to distinguish a change in Q from a change in K.
2. Critique the student's statement about tube 2. Is 'water is a product' a valid explanation? What is the dominant effect?
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The statement is partly right in direction but wrong in mechanism. Water is the solvent, so its activity is approximately 1 and it is not included in the equilibrium expression. Adding 5.0 mL of water dilutes all dissolved species, but the effect is not equal because Cl− appears to the fourth power in the denominator. If the volume doubles, the reaction quotient becomes Q = (1/2) / [(1/2)(1/2)^4] times its original value, which is 16 times K. Because Q > K, the reaction shifts reverse and the solution turns pink. The dominant cause is dilution of Cl−, not consumption of water as a product. The teacher should have students write the K expression and test the dilution quantitatively. This also explains why adding water to a blue CoCl4^2− solution can look like a simple 'product' effect but is really a concentration effect on a reactant that appears with a high power.
3. Tube 3 forms a white precipitate. Explain why the observation supports the proposed chloride-removal mechanism rather than direct reaction with [CoCl4]2−, and state one control.
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The white solid is consistent with AgCl, which forms when Ag+ combines with free Cl−. That removal of chloride lowers [Cl−] and shifts the equilibrium toward the pink aquo complex. If Ag+ were reacting directly with [CoCl4]2−, the product would not be expected to be simple white AgCl, and the solution might lose the blue complex without the same chloride-dependent shift. A control would be to add AgNO3 to a solution containing chloride but no cobalt complex, or to add a chloride source such as NaCl to a pink sample and observe a shift toward blue. The teacher should ask students to connect the precipitate to Ksp of AgCl and to the equilibrium expression. The observation supports the mechanism only if students also account for the disappearance of Cl−, not merely for the appearance of a precipitate.