What would happen if #"HCl"# is added to a solution of cobalt (II) ions?

1 Answer
Apr 30, 2017

Answer:

The green complex #sf([CuCl_4]^(2-))# is formed.

Explanation:

I am assuming we are in aqueous conditions.

The aqueous copper(II) ion consists of a central #sf(Cu^(2+))# ion surrounded by 6 #sf(H_2O)# ligands:

It has the formula #sf([Cu(H_2O)_6]^(2+))# and is blue in colour. In solution it looks like this:

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If a large excess of chloride ions is added the water ligands are displaced as the following equilibrium is established:

#sf([Cu(H_2O)_6]^(2+)+4Cl^(-)rightleftharpoonsCuCl_4^(2-)+6H_2O)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(rarr)#

#sf(" "color(blue)(blue)" "color(green)(green))#

Concentrated hydrochloric acid contains a large amount of chloride ions so Le Chatelier's Principle tells us that adding this will cause the position of equilibrium to shift to the right producing green #sf(CuCl_4^(2-))# ions.

They have a tetrahedral structure:

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The solution looks like this:

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If excess water is now added the position of equilibrium is driven back to the left and the blue colour returns.

#sf([Cu(H_2O)_6]^(2+)+4Cl^(-)rightleftharpoonsCuCl_4^(2-)+6H_2O)#

#stackrel(color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx))(color(blue)(larr)#

#sf(" "color(blue)(blue)" "color(green)(green))#