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Question about Le Chatelier's Principle (1 Viewer)

Tumble

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Hi guys, my first post here :smile:

In the Jacaranda textbook (Pg 141-142) the reversible reaction Cu(H2O)42+ + 4Cl- ⇌ CuCl42– + 4H2O(l) is given.

The textbook writes: "When water is added to the green mixture, the disturbance causes the equilibrium to shift to counteract the change. The equilibrium shifts to the left to use up some of the added water. Thus, the solution becomes bluer and less green."

However adding water won't change the concentration of water, so why does the system shift to the left?
 

ichila101

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Hi guys, my first post here [emoji2]

In the Jacaranda textbook (Pg 141-142) the reversible reaction Cu(H2O)42+ + 4Cl- ⇌ CuCl42– + 4H2O(l) is given.

The textbook writes: "When water is added to the green mixture, the disturbance causes the equilibrium to shift to counteract the change. The equilibrium shifts to the left to use up some of the added water. Thus, the solution becomes bluer and less green."

However adding water won't change the concentration of water, so why does the system shift to the left?
Does the book read that adding water won't change the concentration of water cause an increase of water will increase its concentration

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ichila101

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Isn't there a general rule that adding a pure solid/liquid to a system doesn't affect the equilibrium?
It's talked about here: http://digipac.ca/chemical/mtom/contents/chapter3/chap3_7_3.htm
And a little bit here: https://web.mst.edu/~gbert/LeChatelier/LeChatelier.html
Yes but it is written in the links that you showed, if it were pure water then adding water would not increase the concentration of water. However the above reaction you wrote is not pure water but is a solution so the addition of water will increase the concentration of water in that solution.
 

Tumble

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Ahh I see what you're getting at, thanks.
I was under the impression that concentration meant the amount of substance dissolved in water, and hence water's concentration was always constant.
 

ciaodon

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To suit the question, when there is an increase of H_2 O in the green solution, the equilibrium is said to shift left (given), meaning it will favour the reverse reaction which decreases the concentration of H_2 O present in solution- resulting in a colour change.

Cu(H_2 O)_4 2+ is not pure H_2 O as you assume the Cu ion has bonded with the H_2 O molecule to form an aqueous solution of Cu(H_2 O)_4 2+, meaning that a judgement on the change in concentration can be made (as moles determine concentration).

*I hope this helped & pls tell me if it didn't make sense :angel:
 

ciaodon

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To suit the question, when there is an increase of H_2 O in the green solution, the equilibrium is said to shift left (given), meaning it will favour the reverse reaction which decreases the concentration of H_2 O present in solution- resulting in a colour change.

Cu(H_2 O)_4 2+ is not pure H_2 O as you assume the Cu ion has bonded with the H_2 O molecule to form an aqueous solution of Cu(H_2 O)_4 2+, meaning that a judgement on the change in concentration can be made (as moles determine concentration).

*I hope this helped & pls tell me if it didn't make sense :angel:
*Just by extension, because the reverse reaction is being favoured it means that more of the H_2 O and CuCl_4 2- is consumed in order to obtain a greater yield of Cu(H_2 O)_4 2+ and Cl 1- in this instance. This could possibly mean that the solution of Cu(H_2 O)_4 2+ has colour-altering potential as more Cu(H_2 O)_4 2+ could make green turn to blue// this judgment could also be applied to ions of Cl 1-
 
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