A student attempts to separate Cl^- and I^- from a solution that is 0.05 M Cl^-

Question

A student attempts to separate Cl^- and I^- from a solution that is 0.05 M Cl^- and 0.10 M I^- with Pb(NO_3)_2. As Pb(NO_3)_2 is added, which anion will precipitate first and what will the concentration of the anion be when the second anion begins to precipitate? K_sp = 1.7 times 10^-5 PbCl_2 and K_sp = 9.8 times 10^-9 for Pbl_2. iodide precipitates first; 0.05 M iodide iodide precipitates first; 0.0012 M iodide chloride precipitates first; 0.0012 chloride chloride precipitates first; 0.10 M chloride iodide precipitates first; 0 0068 M iodide

Explanation / Answer

The one with lower Ksp value will precipitate 1st
So, PbI2 will precipitate first

for PbCl2 to precipitate
PbCl2 <------> Pb2+ + 2Cl-
Ksp = [Pb2+] [Cl-]^2
1.7*10^-5 = [Pb2+] * (0.05)^2
[Pb2+] = 6.8*10^-3 M

at this time for PbI2:
PbI2 <------> Pb2+ + 2I-
Ksp = [Pb2+] [I-]^2
1.7*10^-5 = (6.8*10^-3) [I-]^2
[I-]^2 = 2.5*10^-3
[I-] = 0.05 M

Answer:
Iodide precipitates first; 0.05 M iodide

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