jika sebanyak 1,4 gram PbCl2 dapat larut dalam 200 ml larutan air, maka hitunglah Ksp PbCl2! ( Ar Pb = 207, Cl = 35,5)

m max= 1,4 g
v= 200 mL
Mr PbCl2 = Ar Pb + 2 Ar Cl
=207 + 2 (35,5)
= 207 + 71
= 278

s PbCl2 = (m max/Mr) x (1000/v)
= (1,4/278) x (1000/200)
= 0,005 x 5
= 2,5 x 10^-2

PbCl2 --> Pb2+      + 2 Cl-
  s              s                 2s

Ksp PbCl2 = [Pb2+] [Cl-]^2
= (s) (2s)^2
= 4 s^3
= 4 x (2,5 x 10^-2)^3
= 4 x 15,625 x 10^-6
= 62,5 x 10^-6
= 6,25 x 10^-5

semoga membantu