Molar mass of PbCrO4,
MM = 1*MM(Pb) + 1*MM(Cr) + 4*MM(O)
= 1*207.2 + 1*52.0 + 4*16.0
= 323.2 g/mol
Molar mass of PbCrO4= 323.2 g/mol
s = 4*10^-5 g/L
To covert it to mol/L, divide it by molar mass
s = 4*10^-5 g/L / 323.2 g/mol
s = 1.238*10^-7 mol/L
At equilibrium:
PbCrO4 <----> Pb2+ + CrO42-
s s
Ksp = [Pb2+][CrO42-]
Ksp = (s)*(s)
Ksp = 1(s)^2
Ksp = 1(1.238*10^-7)^2
Ksp = 1.532*10^-14
Answer: 1.5*10^-14
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