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tkahn6
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Effects of Adding Water to Cobalt Chloride Equilibrium
[tex]Co(H_{2}O)_{6} ^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow CoCl_{4} ^{2-} (aq) + 6H_{2}O (l)[/tex]
[tex] Pink \Leftrightarrow Blue [/tex]
We're learning about Le Chatelier's Principal.
My question is:
Why does the reaction shift the left when adding [tex] H_{2}O [/tex]?
My first thought was that stress is being added to the right side of the equation and so equilibrium shifts left. This is incorrect because [tex] H_{2}O [/tex] is a liquid and therefore is not part of the equilibrium reaction (it has no concentration).
The addition of water does however dilute both sides of the equation. This is the explanation my teacher gave but going back over it, diluting the concentrations of the products and reactants would give you a larger [tex]K_{c}[/tex] which would shift it to the right.
Thank you.
Homework Statement
[tex]Co(H_{2}O)_{6} ^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow CoCl_{4} ^{2-} (aq) + 6H_{2}O (l)[/tex]
[tex] Pink \Leftrightarrow Blue [/tex]
We're learning about Le Chatelier's Principal.
My question is:
Why does the reaction shift the left when adding [tex] H_{2}O [/tex]?
Homework Equations
The Attempt at a Solution
My first thought was that stress is being added to the right side of the equation and so equilibrium shifts left. This is incorrect because [tex] H_{2}O [/tex] is a liquid and therefore is not part of the equilibrium reaction (it has no concentration).
The addition of water does however dilute both sides of the equation. This is the explanation my teacher gave but going back over it, diluting the concentrations of the products and reactants would give you a larger [tex]K_{c}[/tex] which would shift it to the right.
Thank you.
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