(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