I am filling out a data table for le Chatelier\'s Principle lab for Co (H2O)6 ^2

Question

I am filling out a data table for le Chatelier's Principle lab for Co (H2O)6 ^2+ / CoCl4 ^2-. There is a Solution A. I created this solution by pouring 25 mL of 6.0 M HCL into a beaker with 1.2 g Co (H2O)6 Cl2. When I mix it together, what would the color be? Would it be dark blue /violet color and this would produce the CoCl4 ^2- ion complex? Then, there is a Solution B. This is solution A with a mixture of 5ml ddH2O. Would Solution B be light pink and this will form the Co (H2O)6^2+ complex ion. Then I have another table. If I add water to solution A, what color would it be?. If I add concentrated hcl to solution B, what color would it be? If I put solution A in a boiling water bath, what color would it be? if I put solution A into a cool ice bath, what color would it be? if I heat solution b into a water bath, what color would it be? if I put solution b into a cool bath, what color would it be?

Explanation / Answer

The following equilibrium is being studied

Co(H2O)62+(aq) + 4 Cl-(aq) <=> CoCl42-(aq) + 6 H2O(g)

Pink blue

1) Hence, solution A originally would be blue in colour due to CoCl42-

2) Solution B would be pink due to Co(H2O)62+ since addition of water would shift the equilibrium in the left direction

3) addition of water to solution A would change the color to pink as explained in 2)

4) adding HCl increases the concentration of Cl- shifting the equilibrium towards right and color changes to blue.

6) the forward reaction is endothermic and raising the temperature shifts it towards right leading to no change in blue color of solution A

7) cooling solution A on the contrary would change the color to pink.

8) heating solution B would lead to change in color to blue.

9) cooling solution B would lead to no change in pink color.

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