CORRECT


(4.a) 1.8 x 10-5 M



PbCl2(s) <---> Pb+2(aq) + Cl-(aq)
Initial (I) 1.7 x 10-3M 3.4 x 10-3M
Change(C) +0.10M


_____________________________________________________________

Equilibrium (E) 1.7 x 10-3M 0.1034M


+0.10 M This is what comes from the common ion from the NaCl

3.4 x 10-3 M This is what comes from dissolving PbCl2(s)

0.10 M + 3.4 x 10-3 M = 0.1034 M



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

(1.7 x 10-3)(0.1034)2 = 1.8 x 10-5

Return to Problem Set