What form of the complex ion, Co(H_2O)_6^2+ (aq) or CoCI_4^2- (aq), is predomina

Question

What form of the complex ion, Co(H_2O)_6^2+ (aq) or CoCI_4^2- (aq), is predominate in: the 12 M HCI (aq) ____________ the diluted solution ____________ the heated solution __________ Explain why you obtained the observed color in 12 M HCI (aq) (Step 2). Explain the observed color change that occurred when water was added to the solution in Step 3. Consider how water affects the ion concentrations and Q in this system. Consider your observations in the hot water bath in Step 4. In which direction did the equilibrium shift? __________ Did the value of K get smaller or larger? __________ Is the reaction (as written) exothermic or endothermic? Explain. ____________

Explanation / Answer

Predominant form present in solution at,

- the 12 M HCl (aq) : CoCl4^2- (blue)

- The dilute solution : Co(H2O)6^2+ (pink)

- the heated solution : CoCl4^2- (blue)

When 12 M HCl is added, the excess Cl- pushed the equilibrium to the CoCl4^2- side until all the added HCl has been used up and equilibrium is reestablished again. This is according to the LeChatelier's principle.

When the solution was diluted, more H2O is present which pushed the equilibrium to the left towards Co(H2O)6^2+ side untill all the CoCl4^2- has reacted with excess H2O added for dilution. This is according to the LeChatelier's principle. Adding HCl makes gthe endothermic reaction to formation of CoCl4^2- go, that is Q of system has decreased.

In hot water bath,

- the equilibrium has shifted to CoCl4^2- (blue) : excess heat used for endothermic reaction

- The value of K got larger as more product CoCl4 ^2- is formed

- The reaction is endothermic as heat is used to run the reaction in the forward direction and forms CoCl4^2- (blue)

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