The ksp of lead (ii) carbonate, pbco3, is 7.40 * 10-14. calculate the solubility of this compound in g/l.

The molar solubility is the Pb ions present. PbCO3 dissociate to form Pb ions and CO3 ions

Ksp 7.40x10^-14 is equal to (Pb ions) (CO3 ions)

Pb ions correspond to CO3 ions

(pb ions) (Pb ions) is equal to 7.40x10^-14

therefore Pb ions is square root of 7.4x10^14 which is 2.27x10^-7m

In g/l is (2.27x10^-7) x 267 (RFM of PbCO3) which is 6.06x10^-5 g/l