A student doing the experiment in this module extended the study of the [CoCl_4]

Question

A student doing the experiment in this module extended the study of the [CoCl_4]^2 ion/[Co(H_2O)_6]^2+ ion equilibrium. In one test tube, the student added silver nitrate (AgNO_3) to a blue equilibrium mixture prepared from CoCl_2. The reaction mixture became pink and cloudy, and a white precipitate settled out leaving a clear, pink solution. The student identified the precipitate as silver chloride AgCl. Briefly explain how these observations are consistent with lessthanorequalto Chatelier's principle. In a second test, the student placed a test tube containing a pink equilibrium mixture in a hot-water bath. The solution turned blue. When the student removed the test tube from the hot-water bath and placed it in an ice-water bath, the solution turned pink. (2) Is the forward reaction in the [CoCl_4]^2- ion/[Co(H_2O)_6]^2+ ion equilibrium exothermic or endothermic? Briefly explain how the student's observations support your answer to(2). Write the net ionic equation for this equilibrium, including heat.

Explanation / Answer

1. From the reaction of AgNO3 with blue CoCl4^2- solution,

(1) In CoCl4^2- solution we have Cl- ion present in the solution. Upon addition of AgNO3 to it, Cl- is used up by the Ag+ added to form AgCl precipitate (white) and the excess Co2+ now reacts with H2O presen to form pin color Co(H2O)6^2+ solution. The excess Co2+ in solution reacts with excess H2O to form the pink complex. Thus this is according to Lechatellier's principle.

The second test tube is heated and cooled and color change is observed

(2) the Co(H2O)6^2+/CoCl4^2- reaction is endothermic in nature.

(3) when Co(H2O)6^2+ solution (pink) is heated we get blue color CoCl4^2- solution. Thus the reaction is endothermic in nature. When blue CoCl4^2- is cooled in ice it gives off excess heat to form pink color Co(H2O)6^2+ back in solution. thus heat is given off, reaction is exothermic.

(4) Net ionic equation

Co(H2O)6^2+ + 4Cl- <==(heat above the arrow and cool below the arrow)==> CoCl4^2- + 6H2O

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.