Sciencemadness Discussion Board

Does chromic acid react with AgI?

silvergrahm - 6-4-2013 at 23:57

Not a school question, believe it or not...

(H2O)CrO3 + AgI --> ?

Thanks!

[Edited on 7-4-2013 by silvergrahm]

Finnnicus - 7-4-2013 at 00:43

H2CrO4 + 2AgI --> Ag2CrO4 + 2HI
I Think this is a equilibrium reaction

weiming1998 - 7-4-2013 at 00:59

Quote: Originally posted by Finnnicus  
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

Quote: Originally posted by weiming1998  
[...]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

Quote: Originally posted by Finnnicus  
H2CrO4 + 2AgI --> Ag2CrO4 + 2HI
I Think this is a equilibrium reaction

I do not. While a surprising number of reactions are equilibrium reactions, this is not one of them. any HI that formed would immediately be oxidized


[Edited on 7-4-2013 by AndersHoveland]

Finnnicus - 7-4-2013 at 17:33

Now I feel dumber than silvergraham... :(

silvergrahm - 7-4-2013 at 17:37

Quote: Originally posted by Finnnicus  
Now I feel dumber than silvergraham... :(


WTF BRO.

Finnnicus - 7-4-2013 at 17:40

Calm down, just teasing. Back to the science now.

silvergrahm - 7-4-2013 at 17:50

Quote: Originally posted by Finnnicus  
Calm down, just teasing. Back to the science now.


No, I know nothing of chemistry.

AJKOER - 8-4-2013 at 18:38

Quote: Originally posted by Finnnicus  
H2CrO4 + 2AgI --> Ag2CrO4 + 2HI
I Think this is a equilibrium reaction


Per this source (http://chromium.atomistry.com/silver_chromate.html ) to quote:

"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.
---------------------------------------------

In the presence of sunlight (AgI is one of the more photo-sensitive Silver salts), one could have (in the proper medium):

2 AgI --uv--> 2 Ag + I2

and upon adding H2CrO4 may also be successful.