Question

If 550 mL of some Pb(NO3)2 solution is mixed with 400 mL of 6.70 x 10−2 M NaCl solution, what is the maximum concentration of the Pb(NO3)2 solution added if no solid PbCl2 forms? (Assume Ksp = 2.00 x 10−5 M at this temperature.) Enter the concentration in M.

Answer 1

concentration of [Cl-] = 400 x 6.70 x 10^-2 / (400 + 550)

                                  = 0.0282 M

PbCl2   ---------------> Pb+2   + 2Cl-

                                  Pb+2      0.0282

Ksp = [Pb+2][Cl-]^2

2.00 x 10^-5 = [Pb+2](0.0282)^2

[Pb+2] = 0.0251 M

concentration of Pb(NO3)2 = 0.0251 M