Calculate the solubility of PbCl2 if its solubility product is 3.2*10^-8 at 298K
PbCl2 (s) ⇌ Pb2 + (aq) + 2Cl- (aq)
Ksp = [Pb2 +] [Cl-]2
Let concentration of [Pb2 +] = x
Concentration of [Cl-] = x2
Ksp = 3.210-8
Now , xx2 = 3.210-8
x3 = 3.210-8
x = 3.1 10-2
Hence , solubility of lead (II) ions is 3.110-2 mol / L
Solubility of chloride (I) ions is 9.6110-4 mol / L
Ksp = [Pb2 +] [Cl-]2
Let concentration of [Pb2 +] = x
Concentration of [Cl-] = x2
Ksp = 3.210-8
Now , xx2 = 3.210-8
x3 = 3.210-8
x = 3.1 10-2
Hence , solubility of lead (II) ions is 3.110-2 mol / L
Solubility of chloride (I) ions is 9.6110-4 mol / L