In: Chemistry
What is the solubility ( in g/L) of lead(II) chromate, PbCrO4 in 0.13M potassium chromate, K2CrO4.
PbCrO4 --------> Pb+2 + CrO42-
0.13
s s
s 0.13+s
Ksp of PbCrO4 is 2.8*10-13
Ksp = [Pb+2][CrO42-]
2.8*10-13 = s * (0.13+s)
s = 2.15*10-12 mole/L
solubility of PbCrO4 = 2.15*10-12 moles/L
molar mass of PbCrO4 = 323g/mole
solubility of PbCrO4 = 2.15*10-12 moles/L * 323g/moles
= 6.94*10-10 g/L