Determining Ksp from Solubility
PbCrO4 (s) Pb2+(aq) + CrO42-(aq)
Molar solubility of PbCrO4 = x x
5.8 x 10 -6g
100 mL
1000 ml
1 L
1mol PbCrO4
323.2 g
= 1.79 x 10 -7 M PbCrO4
1 Mole PbCrO4 = 1 mole Pb2+ and 1 mole CrO42-
Therefore [Pb2+] = [CrO42-] = 1.79 x 10-7 M
Ksp = [Pb2+] [CrO42-] = (1.79 x 10 -7 M)2 = 3.20 x 10-14
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