Ask A Scientist Chemistry Archive

name         Nizar M.
status       educator
age          30s

Question -   My question is about selective precipitation.
Consider a solution of 0.1M Chloride ions and 0.01M chromate ions.By
adding silver nitrate solution, chloride ions precipitate first as silver
chloride;however, even when addition of silver nitrate continues such that
silver chromate starts forming some of chloride ions remain in
solution.What is the chemistry behind the remaining unprecipitated
chloride ions?Ksp(AgCl)=2.8Exp-10
                           Ksp(Ag2CrO4)=1.9Exp-12
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Let us walk through this step at a time: Starting with 0.1 m chloride
(assume sodium salt for all added salts). At the addition of the first drop
of Ag+(NO3-) we have the equilibrium: [Ag+][Cl-]=Ksp(1)=1.76x10^-10 and
[Cl-]=0.1, so the initial concentration of Ag+ will be very small,
surpressed by the common ion effect:
  [Ag+]=1.76x10^-10/0.1=1.76x10^-11.

When the chloride has been quantitatively titrated (ignore for the moment
how we might know that) then the situation is: AgCl(solid) = (Ag+) + (Cl-)
and the concentration of (Ag+)=(Cl-): [Ag+][Cl-]=Ksp(1)=1.76x10^-10 or
[Ag+]=[Cl-]=1.33x10^-5. The concentration of (Ag+) increases from its
initial value of
1.76x10^-11 to 1.33x10^-5, while the concentration of (Cl-) decreases from
its initial
value of 0.1 to a value of 1.33x10^-5.

Now the addition of 0.01 molar (CrO4=). The equilibrium here is:
Ag2CrO4(solid)= 2(Ag+) +(CrO4=) and Ksp(2)=[Ag+]^2[CrO4=] = 1.12x10^-12
The equilibrium for this reaction alone is: [2X]^2[X]=4X^3=1.12x10^-12 or:
[CrO4=] = X = 6.54x10^-5 and [Ag+] = 2X = 1.30x10^-4 all in mols / l.

The principle here is that EVERY equilibrium must be independently
satisfied, so if the chromate equilibrium dictates that the concentration of
[Ag+] = 1.30x10^-4 then the silver chloride equilibrium must also be
satisfied with a concentration of
[Ag+]=1.30x10^-4. That is [Ag+][Cl-]=1.76x10^-10 or
[Cl-]=1.76x10^-10/1.30x10^-4
That is: [Cl-] = 1.35x10-6.

The net result is a further surpression of [Cl-] from the value it has for
the solubility of
AgCl alone. This does not always occur. For example, AgCl is insoluble, but
the addition of ammonia forms the silver ammonia complex, which is more
stable than AgCl and the addition of sufficient ammonia causes AgCl to
redissolve.

Vince Calder
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When you say "some" of the chloride ions remain in solution, what
concentration do you mean?  Solubility products specify maximum
concentrations of components in solution at equilibrium.  For ease of
notation, let us call KSp of AgCl Ksp1 and Ksp of Ag2CrO4 Ksp2.  This gives
us two inequalities:

   [Ag+][Cl-] < Ksp1
   [Ag+]^2[CrO4=] < Ksp2

The second equation means that, if you add silver ions to a solution
containing chromate ions, silver chromate will begin to precipitate out when
[Ag+]^2[CrO4=] exceeds Ksp2.  The level of [Ag+] required for this to happen
can be found by

   [Ag+]^2[CrO4=] > Ksp2
   [Ag+]^2 > Ksp2/[CrO4=]
   [Ag+] > sqrt(Ksp2/[CrO4=])

Given that the initial concentration of chromate ions, [CrO4=], is 0.01 M,
this gives

   [Ag+] > sqrt(1.9e-12/0.01)
   [Ag+] > sqrt(1.9e-10)
   [Ag+] > 1.38e-5

So, silver chromate will begin to precipitate out when the free silver ion
concentration exceeds 1.38e-5 M.  What is the chloride concentration then?
The first equation tells us the maximum concentration of chloride in
solution when silver is present:

   [Ag+][Cl-] < Ksp1
   [Cl-] < Ksp1/[Ag+]
   [Cl-] < 2.8e-10/1.38e-5
   [Cl-] < 2.08e-5

According to these two solubility products, there can be as much as 2e-5
molar chloride still present when silver chromate begins to precipitate out.

This would not explain if the concentration of chloride when silver chromate
begins to precipitate from your system is higher than this.  If that is the
case, there are a few other things that might be going on.  One is that the
mixing in your system may not be completely efficient, so that different
high-silver and high-chloride zones are present.  Then, silver chromate will
precipitate in high-silver zones while chloride still persists in solution
in regions where the silver concentration is lower.

Another effect to consider is that precipitation is not always an
instantaneous process.  It is possible that one solid may precipitate sooner
than another, even if its solubility product is higher.  I doubt that this
would be the case here, because silver halide salts precipitate quite
readily.

Another possibility is that some of the chromate, by forming a
chlorochromate complex with the chloride, sequesters some of the chloride
from solution so that it is not available for incorporation into the silver
chloride precipitate.

   CrO4(2-) + Cl- P 2H+ --> CrO3Cl- + H2O

The problem with this explanation is that chlorochromate is a weakly-bound
complex that only forms in the presence of excess chloride, which cannot be
the case here.  Are other ions present in your system?  A soluble complex
containing chloride may be forming, since free chloride will, as you know,
be efficiently scavenged by the silver ions.

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois
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