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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Bob Sacamano on October 27, 2010, 10:34:44 PM

Title: Reaction of [Co(NH3)5(H20)]3+ with HCl
Post by: Bob Sacamano on October 27, 2010, 10:34:44 PM
When [Co(NH3)5(H20)]3+ is treated with HCl, [Co(NH3)5Cl]Cl2 is produced. Is a strong acid required to carry out this reaction or would a chloride salt suffice to produce the desired product?

I don't really know where to go with this, is the proton somehow removing the water from the first coordination sphere allowing Cl to bind?
Title: Re: Reaction of [Co(NH3)5(H20)]3+ with HCl
Post by: majorjp on November 04, 2010, 06:13:41 AM
The water molecule is not a particularly strong ligand so it can be easily replace just by a Cl- anion, therefore any chloride salt should suffice as long as that salt does not contain NH4+ (because the ammonia cation can form NH3)or NO2-. The reason being that these two molecule will compete with the chloride anion to replace the water ligand.
Title: Re: Reaction of [Co(NH3)5(H20)]3+ with HCl
Post by: rackye on November 05, 2010, 08:46:13 PM
I'm not quite sure about it because chlorine anion by itself can not replace H2O because it's field strength is weaker
Title: Re: Reaction of [Co(NH3)5(H20)]3+ with HCl
Post by: majorjp on November 06, 2010, 08:10:31 AM
True, that is why the concentration of the Cl- anions is the true determining factor. Strong acids are used because they normally result in a high concentration of Cl- anions. There is a limit, due to solubility, of the Cl- anions from that can result from a chloride salt. Being as this reaction is an equilibrium reaction that depends on concentration, water can be replaced by adjusting the other reagents concentrations. The acid is typically the more useful source of the Chloride anions though in general reactions.