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

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