Question

Assume a sample is 0.01M in both Cu+ and Pb2+. If chloride is added, determine the precipitate that forms first (answer must be justified with a numerical answer). For separation to be considered good, determine if the concentration of the "first cation to precipitate" can be reduced to 5*10^-5M while the other remains in solution. Given: Ksp CuCl=1.9*10^-7, Ksp PbCl2=1.7*10^-5

Answer 1

i)

dissociation of CuCl2

CuCl(s) <–-------> Cu+(aq) + Cl-(aq)

Ksp = [Cu+] [Cl-]= 1.9×10^-7

[Cu+][Cl-] = 1.9×10^-7M²

0.01M [Cl-] = 1.9×10^-7M²

   [ Cl- ] = 1.9 × 10^-5M

So, For precipitation of CuCl minimum [Cl-] = 1.9×10^-5M

   Dissociation of PbCl2

   PbCl2(s) <------> Pb2+ (aq) + 2Cl-(aq)

   Ksp = [Pb2+] [ Cl-]²= 1.7×10^-5

[Pb2+] [Cl-]²= 1.7×10^-5M³

0.01M × [Cl-]²= 1.7×10^-5M³

   [Cl-]= 0.0412M

So, when precipitation of PbCl2 minimum [Cl-] is 0.0412M

Therefore,

CuCl will precipitate first

ii)

when PbCl2 start to precipitate

[Cl-] = 0.0412M

when [Cl-] = 0.0412M

[Cu+] × 0.0412M = 1.9×10^-7M²

[ Cu+] = 4.61×10^-6M

So, the first cation ( Cu+) can be reduced to 5×10^-5M while the other cation ( Pb2+) remains in solution