The values of Ksp for silver bromide, AgBr is 7.7x10^-13, and silver chloride, AgCl is 1.6x10^-10. A solution containing a mixture of 2.0x10^-2 M Br^- and 2.0x10^-2 M Cl^- is a candidate for separation using selective precipitation. Solid AgNO3 is added without changing the volume of solution. Silver nitrate is very soluble.
a. At what concentration of Ag^+ will silver bromide first begin to precipitate?
b. At what concentration of Ag^+ will silver chloride first begin to precipitate?
c. What will the residual concentration of Br^- in solution be just as the condition in part b is reached?
d. AgI has Ksp=8.3x10^-17. Using the same method, would I^- be easier, or harder to separate from Cl^- compared to separating Br^- from Cl^-?
AgCl --> Ag^+ + Cl^-
AgBr --> Ag^+ + Br^-
a. Obviously, AgBr will ppt first because it has the smaller Ksp.
(Ag^+)(Br^-) = 7.7 x 10^-13
You know Br^-, you can calculate Ag^+ needed to ppt the first molecule of AgBr.
b. Do the same for AgCl to determine the Ag^+ needed to ppt AgCl with 0.02 M Cl^-.
c.
Ksp AgCl...(Ag^+)(Cl^-)... 1.6 x 10^-10
-------- = ------------ = ------------
Ksp AgBr...(Ag^+)(Br^-)... 7.7 x 10^-13
Note Ag^+ cancels, you can calculate the ratio of (Cl^-)/(Br^-), substitute numbers and calculate (Br^-).
I will d to you.
Check my thinking on this.
for a. i got Ag+ = 3.85x10^-11
b. Ag+ = 8x10^-9
still kind of lost on C and D
To determine the answers to these questions, we can use the principles of solubility and the concept of Ksp (solubility product constant). Ksp is the product of the concentrations of the ions in a saturated solution.
First, let's break down the problem and tackle each part step by step:
a. To find the concentration of Ag+ at which silver bromide (AgBr) will begin to precipitate, we need to set up the solubility equilibrium expression for AgBr and compare it to the given Ksp value.
The solubility equilibrium expression for AgBr is:
AgBr ⇌ Ag+ + Br-
The equilibrium constant expression, Ksp, can be written as:
Ksp = [Ag+][Br-]
Given that the Ksp of AgBr is 7.7x10^-13, we can substitute this value into the Ksp expression:
7.7x10^-13 = [Ag+][Br-]
Since the concentration of Br- is given as 2.0x10^-2 M, we can rearrange the equation to solve for [Ag+]:
[Ag+] = Ksp / [Br-]
[Ag+] = (7.7x10^-13) / (2.0x10^-2)
Evaluating this expression will give us the concentration of Ag+ at which silver bromide will start to precipitate.
b. Similarly, to find the concentration of Ag+ at which silver chloride (AgCl) will begin to precipitate, we need to set up the solubility equilibrium expression for AgCl and compare it to the given Ksp value.
The solubility equilibrium expression for AgCl is:
AgCl ⇌ Ag+ + Cl-
The equilibrium constant expression, Ksp, can be written as:
Ksp = [Ag+][Cl-]
Given that the Ksp of AgCl is 1.6x10^-10, we can substitute this value into the Ksp expression:
1.6x10^-10 = [Ag+][Cl-]
Since the concentration of Cl- is given as 2.0x10^-2 M, we can rearrange the equation to solve for [Ag+]:
[Ag+] = Ksp / [Cl-]
[Ag+] = (1.6x10^-10) / (2.0x10^-2)
Again, evaluating this expression will give us the concentration of Ag+ at which silver chloride will start to precipitate.
c. To determine the residual concentration of Br- at the point when silver chloride starts to precipitate, we can assume that the concentration of Ag+ reaches the value found in part b.
So, using the concentration of Ag+ found in part b, we can substitute it back into the equilibrium expression for AgBr to find the concentration of Br-:
[Ag+] = [Br-] (since the stoichiometry of the reaction is 1:1)
[Br-] = [Ag+] = (1.6x10^-10) / (2.0x10^-2)
d. Finally, to evaluate whether I- would be easier or harder to separate from Cl- compared to Br-, we can compare their respective Ksp values.
Given that the Ksp of AgI is 8.3x10^-17, we can compare it to the Ksp values of AgBr (7.7x10^-13) and AgCl (1.6x10^-10).
The lower the Ksp value, the less soluble the compound is. Therefore, since the Ksp value for AgI is lower than both AgBr and AgCl, it suggests that I- would be even easier to separate from Cl- compared to Br-.