TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation: [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount Solution TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation: [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount.

1. TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1)
PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2)
In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M
When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at
another point, PbI2 begins to precipitate.
Find a point of TlI precipitation:
[Tl+][I-] = Ksp = 8.9E-8
(0.50)[I-] = 8.9E-8
[I-] = 1.78E-7 M
Find a point of PbI2 precipitation:
[Pb2+][I-]2 = 8.7E-9
(0.50)[I-]2 = 8.7E-9
[I-] = 1.32E-4 M
Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M)
When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just
begins to precipitates. At this point, we stop the ions separation.
Find remaining [Tl+]:
[Tl+][I-] = 8.9E-8
[Tl+](1.32E-4) = 8.9E-8
[Tl+] = 6.74E-4 M
The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13%
Tl+ is separated by 99.87% of original amount
Solution
TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1)
PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2)
In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M
When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at
another point, PbI2 begins to precipitate.
Find a point of TlI precipitation:
[Tl+][I-] = Ksp = 8.9E-8
(0.50)[I-] = 8.9E-8
[I-] = 1.78E-7 M
Find a point of PbI2 precipitation:

2. [Pb2+][I-]2 = 8.7E-9
(0.50)[I-]2 = 8.7E-9
[I-] = 1.32E-4 M
Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M)
When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just
begins to precipitates. At this point, we stop the ions separation.
Find remaining [Tl+]:
[Tl+][I-] = 8.9E-8
[Tl+](1.32E-4) = 8.9E-8
[Tl+] = 6.74E-4 M
The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13%
Tl+ is separated by 99.87% of original amount