Precipitates ```Name: Nizar M. Status: educator Age: 30s Location: N/A Country: N/A Date: 2001 - 2002 ``` 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 Replies: 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 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 Click here to return to the Chemistry Archives

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