What pb2+ concentration should be maintained in PbNO3 to produce a solubility of 1.7×10−4 PbI2/L when PbI2 is added?
I found the ksp of PbI2 to be 7.1*10(-9), divided the ksp by the solubility, and took the square root, resulting in 6.5*10(-3). my answer is wrong :s wheres my mistake? :P
2 answers
nevermind got it. i was using the solubility for pb when it was for i :P
I would look at it this way.
PbI2 ==> Pb^2+ + 2I^-
1.7E-4....x......2*1.7E-4
7.1E-9 = (Pb^2+)(3.4E-4)^2
Solve for (Pb^2+).
0.0614M? which rounds to 0.061 to two s.f.
PbI2 ==> Pb^2+ + 2I^-
1.7E-4....x......2*1.7E-4
7.1E-9 = (Pb^2+)(3.4E-4)^2
Solve for (Pb^2+).
0.0614M? which rounds to 0.061 to two s.f.
1 answer