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Topic: Ag(I) halides and NH3  (Read 6551 times)

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Offline zeoblade

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Ag(I) halides and NH3
« on: March 05, 2010, 02:56:00 AM »
I noticed silver halides are very insoluble:

AgNO3(aq) + NaCl(s) --> AgCl(s) + NaNO3(aq)

The solubility product constant Ksp decreases down the halide group Cl > Br > I

However when NH3 is present, this ligand bites Ag and inhibits the silver halide precipitating:

AgNO3(aq) + 2NH3(l) --> [Ag(NH3)2]+(aq) + NO3-(aq)
[Ag(NH3)2]+(aq) + NaCl(s) --> No reaction

Except when KI is used, the silver diamine complex still can precipitate AgI! I can't explain why this can still occur because the NH3 ligands are supposed to be blocking precipitation with I, right? I noticed the Ksp of AgCl is 1.77x10-10 and AgI is 8.51x10-17, so could this be the reason why?

Offline Borek

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Re: Ag(I) halides and NH3
« Reply #1 on: March 05, 2010, 04:35:23 AM »
It is all in the equilibrium constants. Stability of the ammonia complex its too low to dissolve AgI. There are other ligands - like CN-, S2O32- - that make complexes strong enough to dissolve even silver iodide.
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Offline zeoblade

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Re: Ag(I) halides and NH3
« Reply #2 on: March 05, 2010, 04:58:43 AM »
I'm interpretaing what you're saying to be supportive of what I think is happening? the complex is not stable enough so AgI precipitates because AgI Ksp is so low but AgCl is low but not as low as AgI which is why AgCl is inhibited?

Offline Borek

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Re: Ag(I) halides and NH3
« Reply #3 on: March 05, 2010, 05:11:05 AM »
I'm interpretaing what you're saying to be supportive of what I think is happening? the complex is not stable enough so AgI precipitates because AgI Ksp is so low but AgCl is low but not as low as AgI which is why AgCl is inhibited?

Wow, I have spent five minutes untwisting what you wrote, but yes, I think we are talking about the same.
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