hello,
i have some problem with solubility in a solution of Cl- and I-
if i introduce Ag+ i gona form in first AgI because is less solubility
but i gona form all AgI and after all AgCl ?
if the Ks of AgI and AgCl was different (impossible) AgCl can be form in first ?
thx for helpTexium - 9-5-2015 at 06:22
AgI will precipitate first, and nearly all of it will precipitate before any AgCl will begin to precipitate. The amount of iodide that remains in
solution when AgCl begins to precipitate will be equal to the Ksp of AgI divided by [Ag<sup>+</sup>] at the point when all of
the AgI has precipitated, because it is at equilibrium concentration when it is at that (very, very small) value.
Ksp = [Ag<sup>+</sup>] [I<sup>-</sup>]Boffis - 9-5-2015 at 06:55
What is it you are trying to do? A homework exercise?
By "ks" are you referring to solubility products? As far as I recall the solubility products of AgCl and AgI are about 10 and 16.
Since both are deemed to dissolve via AgHal => Ag+ + Hal- the solubility products are = -Log[(Mol conc Ag+) x (Mol Conc Hal-)
Therefore the concentration of silver in its chloride solution will be 10-5 Molar and AgI will be 10-8 M. There is therefore no simple method of
precipitating the iodide first. However, the ammino complex of silver gives a concentration between the two so if you carry out the precipitation in
ammonia solution or with amoniacal silver solution only the iodide will precipitate. Alternatively you can precipitate the lot and then leach the
chloride with ammonia solution.
The only way I can think of precipitating the Cl- first would be to oxide the iodide to iodate (don´t know how you would do this) silver metaiodate
has a solubility product of about 8 so the chloride would ppt first but the difference is small so seperation may not be clean. You could then reduce
the iodate back to iodide.
If you are trying to seperate iodide from chloride the easiest method (and the one I use) is to oxidize the solution with hypochlorite bleach and
extract the iodine with dichloromethane or chloroform.blogfast25 - 9-5-2015 at 07:36
The solubility product for a silver halide AgX is Ks = [Ag<sup>+</sup>] [X<sup>-</sup>]. For AgI it’s 8.5 x
10<sup>-17</sup>, for AgCl it’s 1.8 x 10<sup>-10</sup> (http://bilbo.chm.uri.edu/CHM112/tables/KspTable.htm)
AgI is thus the least soluble one and the difference in Ks is large enough so that the AgI precipitates first and completely.
The Ks and solubility S are related as follows. If the solubility of AgCl is S mol/L (M), then [Ag<sup>+</sup>] =
[Cl<sup>-</sup>] = S. So:
Ks = S x S = S<sup>2</sup> and S = √S = 0.0000134 M
By "ks" are you referring to solubility products? As far as I recall the solubility products of AgCl and AgI are about 10 and 16.
Since both are deemed to dissolve via AgHal => Ag+ + Hal- the solubility products are = -Log[(Mol conc Ag+) x (Mol Conc Hal-)
Those are really the pKs = - logKs
[Edited on 9-5-2015 by blogfast25]highuse - 10-5-2015 at 00:32
thanks for help !
i gonna have my final exam of chimistry of solution, and i don't understand what decide who precipitade first, the solubility "s"
or the solubility product "Ks" / "pKs" ?
for example : Zn3(PO4)2 have a pKs= 32 so Ks= 10^-32 and a solubility = (Ks/108)^1/5 = 1.6 x 10^-7 mol/L
and Cr(OH)3 pKs = 31 and s= 7.8 x 10^-9
i gonna have my final exam of chimistry of solution, and i don't understand what decide who precipitade first, the solubility "s"
or the solubility product "Ks" / "pKs" ?
If either is exceeded the compound will precipitate. Usually we use the Ks/pKs because it's easier.
