In a group 1 analysis, a student adds hydrochloric acid to the unknown solution to make [Cl-] = 0.15 M. Some PbCl2 precipitates. Calculate the concentration of Pb2+ remaining in solutions.

Answer 1

.............PbCl2 ==> Pb^2+ + 2Cl^-
I............solid......0......0.15
C............solid......x........x
E............solid......x......0.15+x

Ksp = (Pb2+)(Cl^-)^2
Ksp = (x)(x+0.15)
Solve for x.