Can Anyone Help Me With This Solubility Problem?
An aqueous solution of 1.90 M AgNO3 is slowly added from a buret to an aqueous solution of 0.0100 M Cl- and 0.250 M I-.
b) When the second ion begins to precipitate, what is the remaining concentration of the first ion?
Reposting this because i already know how to do the other ones. for part B, what is the setup i use to solve this? thx
Bob Pursley gave you a hint this morning. Here is another one.
Divide Ksp AgCl/Ksp AgI =
(Ag^+)(Cl^-)/(Ag^+)(I^&-) = (KspAgCl)/(Ksp AgI).
(Ag^+) cancels so you can get a ratio of
(Cl^-)/(I^-) = ??
Substitute numbers and solve for (Cl^-)
the ksp for AGCL is 1.8*10^-11, and the ksp for AGI came out to 8.5*10^-11. when i divided the numbers and set them equal to CL/0.250, the number came out really large? is that right?
Yes, the number is a large one. A couple of words of caution.
1. AGCL and AgCl are two different things. The same goes for AGI versus AgI. You need to use more caps to begin a sentence and fewer caps for symbols.
2. I don't believe either of your values for Ksp and that will affect the ratio you calculated (although, I agree it is a large number).
To solve part B of this solubility problem, you'll need to use the concept of the solubility product constant and the common ion effect.
The solubility product constant, Ksp, is an equilibrium constant that represents the equilibrium between a solid compound (precipitate) and its dissociated ions in a saturated solution. For the precipitation reaction between Ag+ and Cl-, the equilibrium expression would be:
Ag+(aq) + Cl-(aq) ⇌ AgCl(s)
Ksp = [Ag+][Cl-]
The common ion effect occurs when a solution containing a common ion (in this case, Cl-) reduces the solubility of a compound that contains the same ion (AgCl). When the second ion (I-) begins to precipitate, it means that the concentration of Cl- has reached a level where it is in equilibrium with AgCl.
To find the remaining concentration of Ag+ (the first ion), you need to calculate the concentration of Cl- when the second ion starts to precipitate. Since both AgCl and AgI are insoluble, we can assume that all the Ag+ ions are involved in the precipitation reaction. Therefore, the concentration of Ag+ at that point will be equal to the initial concentration of AgNO3 (1.90 M).
Now, we need to find the concentration of Cl- when the second ion starts to precipitate. To do this, we can use an ice table to set up an equilibrium expression and solve for the unknown concentration.
Let's assume that x M of AgCl precipitates. This means that x M of Ag+ and x M of Cl- are removed from the solution. At equilibrium, the concentration of Cl- will be (0.0100 M - x) since it started with 0.0100 M and lost x M due to the precipitation of AgCl.
From the stoichiometry of the reaction, we know that 1 mole of Ag+ reacts with 1 mole of Cl-. Therefore, the equilibrium concentration of Ag+ will also be (1.90 M - x).
Now, using the solubility product constant expression, we have:
Ksp = [Ag+][Cl-]
Ksp = (1.90 M - x) * (0.0100 M - x)
Since the Ksp value for AgCl is known, you can substitute it into the equation and solve for x. Once you've determined the value of x, you can subtract it from the initial concentration of Ag+ (1.90 M) to find the remaining concentration of the first ion.