H2CrO4 + 2AgI --> Ag2CrO4 + 2HI
I Think this is a equilibrium reaction
The Ksp of silver iodide is 8.52×10^-17, while the Ksp of silver chromate is 1.12×10^-12 (http://www.ktf-split.hr/periodni/en/abc/kpt.html). Thus, the equilibrium would be strongly shifted to the left, so such a reaction would not be
favourable at all (especially considering that HI is a much stronger acid than chromic acid). However, chromic acid (especially when heated) will most
likely oxidise AgI into elemental iodine, leaving a bunch of silver ions that will form silver chromate. Some Cr3+ ions will also be formed, from the
reduction of chromic acid.
So in conclusion, a messy slurry of silver chromate, iodine and unreacted silver iodide will be the product.
[Edited on 7-4-2013 by weiming1998]Fantasma4500 - 7-4-2013 at 05:52
you should perhaps also consider that HI goes back and forth in between H2 and I2 to HI
so even if you see iodine there should be some HI present, also..woelen - 7-4-2013 at 06:50
[...]The Ksp of silver iodide is 8.52×10^-17, while the Ksp of silver chromate is 1.12×10^-12 (http://www.ktf-split.hr/periodni/en/abc/kpt.html). Thus, the equilibrium would be strongly shifted to the left, so such a reaction would not be
favourable at all [...]
In principle I agree with you, but I want to point out a subtlety, which is important to know!
The Ksp of (nearly) insoluble salts is the product of all concentrations of the ions, raised to the power of theur ratio in their formula, e.g.
Ksp of Ag2CrO4 is [Ag(+)]^2 * [CrO4(2-)] while Ksp of AgI is [Ag(+)]*[I(-)]. The left one is a third power of concentrations, while the second one is
a second power of concentrations. So, a Ksp of 10^(-15) of a compound A2B can result in a higher concentration than a Ksp of 10^(-12) of a compound
AB.
[Edited on 7-4-13 by woelen]AndersHoveland - 7-4-2013 at 07:16
"Silver Chromate, Ag2CrO4, is obtained as a reddish-brown precipitate on mixing solutions of soluble silver salts with those of chromate or dichromate
of potassium - preferably the former."
So, I would try dissolving AgI in alcohol (or, dilute acetic acid) and adding H2CrO4.
H2CrO4 + 2AgI <--> Ag2CrO4 + 2HI
Heating (in an open container) may allow HI to escape moving the equilibrium to the right.
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In the presence of sunlight (AgI is one of the more photo-sensitive Silver salts), one could have (in the proper medium):