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.

Question

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^-?

DrBob222 · Answer

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.

DrBob222 · Answer

Check my thinking on this.

Sam · Answer

for a. i got Ag+ = 3.85x10^-11 b. Ag+ = 8x10^-9 still kind of lost on C and D