- #1
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
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