Wednesday
August 20, 2014

Homework Help: Chemistry

Posted by Kim on Tuesday, May 29, 2007 at 4:17pm.

I have a few question about an equilibrium lab we performed.

Approximately .1 g of CoCl2 * 6H2O was mixed with 2 mL of 12 M HCl. The solution turned blue. Then, water was added and stirred into the solution in 2 mL incriments until no further color change occured- the final color was coral/pink. The test tube with solution was first put into a hot-water bath- the color turned purple. The solution was then placed in a cool-water bath [regular tap water] and the color returned to the pink/coral color of the diluted solution.

The questions on this lab are:
Give the formula of the Co(II) complex ion in solution in
a) 12 M HCl (I said CoCl4 -2)
b) diluted solution (I said Co(H2O)6 +2)
c) hot solution (I wasn't sure how to answer this one; It didn't make sense to me that the formula would change with temperature, but I assume it's possible. Is the formula, then, CoCl4-2?)
d) cooled solution (I said Co(H2O)6 +2 because it returned to the color of the diluted solution

Explain the color change that occurred when...
a) water was added, forming the dilute solution (Consider how a change in [Cl-] amd [H2O] will shift the reaction
Co(H2O)6^+2 + 4Cl- <==> CoCl^-2 + 6H2O
(I would imagine that an increase in the concentration of H2O would result in a decrease in the concentration of Cl- as the system equilibrates. But I don't get how this results in a color change, unless I mention that the solution becomes less acidic?)

b. the solution was heated. (How would the above reaction shift if Kc went up? How did increasing the temperature affect the value of Kc? What is the sign of delta-H?)

(I could easily answer the question in parentheses if this was dealing with a precipitate-reaction. But I don't really get the color change aspect of this)


Thank you for any help that you can contribute. This was a very long post but I thought that the questions wouldn't make sense if they were broken up.

Approximately .1 g of CoCl2 * 6H2O was mixed with 2 mL of 12 M HCl. The solution turned blue. Then, water was added and stirred into the solution in 2 mL incriments until no further color change occured- the final color was coral/pink. The test tube with solution was first put into a hot-water bath- the color turned purple. The solution was then placed in a cool-water bath [regular tap water] and the color returned to the pink/coral color of the diluted solution.

The questions on this lab are:
Give the formula of the Co(II) complex ion in solution in
a) 12 M HCl (I said CoCl4 -2)
right.
The reaction is
[Co(H2O)6]^2+ + 4Cl^- ==> CoCl4^-2 + 6H2O.
This is all about equilibrium. The additiion of HCl increased the Cl^- and that shifts the equilibrium to the right. The color changes from pink to blue.


b) diluted solution (I said Co(H2O)6 +2) right. Addition of water forces the equilibrium to the left so the solution changes from blue to pink.

c) hot solution (I wasn't sure how to answer this one; It didn't make sense to me that the formula would change with temperature, but I assume it's possible. Is the formula, then, CoCl4-2?)
The idea here is that the equilibrium constant, Kc, changes with T. So if the color changes, you KNOW more CoCl4^- is formed at the expense of the Co(H2O)6^+2, and a mixture of the blue and the pink is purple. Cooling has the opposite effect. Refer back to this discussion when we cover the question on delta H below.

d) cooled solution (I said Co(H2O)6 +2 because it returned to the color of the diluted solution
discussed in part c.

Explain the color change that occurred when...
a) water was added, forming the dilute solution (Consider how a change in [Cl-] amd [H2O] will shift the reaction
Co(H2O)6^+2 + 4Cl- <==> CoCl^-2 + 6H2O
(I would imagine that an increase in the concentration of H2O would result in a decrease in the concentration of Cl- as the system equilibrates. But I don't get how this results in a color change, unless I mention that the solution becomes less acidic?) refer to the earlier discussion about the shift of equilibrium. Increasing Cl^- moves the equilibrium to the right, increasing H2O moves the equilibrium to the left. It isn't necessary to go into ore or less acidic or diluting the HCl. A shift is a shift.

b. the solution was heated. (How would the above reaction shift if Kc went up? How did increasing the temperature affect the value of Kc? What is the sign of delta-H?) Increasing T shifted the equilibrium to the right (in the direction I wrote the original equation in part a above).
Kc = ([CoCl^-4)]*[H2O]^6)/([Cl^-]^4*[Co(H2O)6^+2]) = ?? So if Kc increases, then you know the products have become larger in number and the reactants have become smaller in number. That means heating it causes Kc to become larger. With heat included, I would write the reaction like this:
Co(H2O)6^+2 + 4Cl^- + heat ==> CoCl4^- + 6H2O. Is that not right. The heat on the left makes it go to the right. So the sign of delta H will be _______??
The color change aspect of the reaction is simply to show you that something is going on. When it is pink you know you have Co(H2O)6^+2 and when it is blue you have CoCl4^-2. Review the text on Le Chatelier's Principle. This principle is VERY important and it is important that you understand how a reaction shifts from one side to another.
Check my thinking. Check for any typos. I hope I bolded only my answers and not any of the questions. Repost with specific questions if this isn't clear.

(I could easily answer the question in parentheses if this was dealing with a precipitate-reaction. But I don't really get the color change aspect of this)



Thank you! That confirmed some of my thoughts. I think, for some of the questions, I was trying to look deeper into it that necessary. Thank you for your help and clarification.

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