Ligand Exchange - are they reversible?

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jsmith613
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Are ALL ligand exchange reactions reversible?

Therefore, if the Kstab of one complex is LESS than that of the other complex, would it ONLY be possible to form the LESS stable complex by 'flooding' the mixture with the appropriate ion.

(these values are not real but they illustrate what I mean)
[Cu(H2O)6]2+ - log Kstab - +5
CuCl4- - log Kstab - +3

So to form the CuCl4 I would have to flood a solution of [Cu(H2O)6]2+ with chloride ions to sufficently move the position of eqm to the CuCl4 side

Is this correct
 
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jsmith613 said:
Is this correct

Generally speaking yes, it is just an equilibrium process. Sometimes reaction can be very slow, and even if the stability constants dictate that one ligand should be replaced with the other, it takes so long complex with lower stability constant may behave as a more stable one.