The values of Ksp for AgBr and AgCl, are 7.7x10^-13 and 1.6 x 10^-10. A solution containing a mixture of 2.0x10^-2 M Br and 2.0 x 10^-2M Cl is a candidate for separating using selective precipitation. Solid AgNO3 is added, with a miniature spatula, and without changing the volume of solution. Silver nitrate is very solube.
b. At what concentration of Ag+ will AgCl first begin to precipitate?
I got 8.0x 10^-9 which is the correct answer, i just don't know how to get part c below
c.What will the residual concentration of Br- in solution be just as the condition in part b. reached?
thank you!
Ksp AgCl....(Ag^+)(Cl^-)...1.6E-10
-------- = ------------- = -------
Ksp AgBr....(Ag^+)(Br^-)...7.7E-13
(Cl^-)/(Br^-) = 207.79 (too many sig figures but you can round at the end).
Substitute the Cl^- value you have from part b and calculate Br^-
To find the concentration of Ag+ at which AgCl begins to precipitate, we can use the solubility product constant, Ksp, for AgCl.
The balanced equation for the dissociation of AgCl in water is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The Ksp expression for AgCl is given by:
Ksp = [Ag+][Cl-]
Now, let's assume that x is the concentration of Ag+ and Cl- ions that will be produced when AgCl precipitates. Therefore, the concentration of Ag+ will be x, and the concentration of Cl- will also be x.
Using the given Ksp value for AgCl (1.6 x 10^-10), we can write the equation:
Ksp = [Ag+][Cl-]
1.6 x 10^-10 = x * x
Taking the square root of both sides, we get:
√(1.6 x 10^-10) = x
x ≈ 4.0 x 10^-6 M
So the concentration of Ag+ at which AgCl first begins to precipitate is approximately 4.0 x 10^-6 M (which is the answer for part b).
Now, to find the residual concentration of Br- in solution when AgCl starts to precipitate, we need to determine the concentration of Cl- ions that combines with Ag+ ions to form AgCl. Since the concentration of Ag+ that causes precipitation is approximately 4.0 x 10^-6 M, the concentration of Cl- ions will also be 4.0 x 10^-6 M.
Given the initial concentration of Br- ions in the solution is 2.0 x 10^-2 M, and Cl- ions are formed from a mixture of Br- and Cl-, we can subtract the concentration of Cl- from the initial concentration of Br-:
Residual concentration of Br- = Initial concentration of Br- - Concentration of Cl-
Residual concentration of Br- = 2.0 x 10^-2 M - 4.0 x 10^-6 M
Residual concentration of Br- ≈ 1.9996 x 10^-2 M
So, the residual concentration of Br- in solution just as the condition in part b is reached is approximately 1.9996 x 10^-2 M (which is the answer for part c).
To find the concentration at which AgCl will first begin to precipitate (part b), you can use the solubility product constant (Ksp) of AgCl. The equation for the balanced chemical reaction is:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
The expression for the solubility product constant is:
Ksp = [Ag+][Cl-]
Since AgNO3 is added and it is very soluble, we can assume that [Ag+] comes from the dissociation of AgNO3. Therefore, [Ag+] = [AgNO3].
Let's assume that the concentration of AgNO3 added is x M. Then, using stoichiometry, the concentration of Ag+ is also x M, and the concentration of Cl- is 2.0 x 10^-2 M.
Plugging these values into the Ksp expression:
Ksp = (x)(2.0 x 10^-2)
Since the value of Ksp for AgCl is given as 1.6 x 10^-10, we can set up the equation:
1.6 x 10^-10 = (x)(2.0 x 10^-2)
Solving this equation for x, we get:
x = (1.6 x 10^-10) / (2.0 x 10^-2)
x ≈ 8.0 x 10^-9 M
Therefore, the concentration of Ag+ at which AgCl will first begin to precipitate is approximately 8.0 x 10^-9 M.
To find the residual concentration of Br- in solution at this point (part c), we can use the initial concentrations of Br- and Cl- and the stoichiometry of the reaction. Since AgCl is forming, the same amount of Ag+ is consumed, and 1 mole of Cl- is consumed for every 1 mole of AgCl formed. Therefore, the concentration of Cl- remaining is 2.0 x 10^-2 M - x.
Since 1 mole of Br- reacts with 1 mole of Ag+, and we assumed the same concentration of Ag+ as AgNO3, the concentration of Br- consumed is also x M.
The residual concentration of Br- in the solution will be the initial concentration of Br- minus the concentration of Br- consumed:
Residual concentration of Br- = 2.0 x 10^-2 M - x
Substituting the value of x we found earlier:
Residual concentration of Br- ≈ 2.0 x 10^-2 M - 8.0 x 10^-9 M
Residual concentration of Br- ≈ 2.0 x 10^-2 M (since 8.0 x 10^-9 M is negligible compared to 2.0 x 10^-2 M)
Therefore, the residual concentration of Br- in solution just as the condition in part b is reached is approximately 2.0 x 10^-2 M.