Final answer:
The solubility of lead(II) carbonate (PbCO3) in g/L can be calculated using the solubility product constant (Ksp) of the compound. The solubility is approximately 2.30 × 10-5 g/L.
Step-by-step explanation:
To calculate the solubility of lead(II) carbonate (PbCO3) in g/L, you need to use the solubility product constant (Ksp) of the compound. The Ksp of PbCO3 is given as 7.40 × 10-14. The solubility of a compound can be calculated from its Ksp using the formula:
Solubility = sqrt(Ksp)
Therefore, the solubility of lead(II) carbonate is:
Solubility = sqrt(7.40 × 10-14) ≈ 8.60 × 10-8 M
Since 1 mole of PbCO3 weighs 267.22 g, you can convert the solubility from molar concentration to mass concentration:
Solubility (g/L) = Solubility (M) * Molecular weight of PbCO3
Solubility (g/L) = 8.60 × 10-8 * 267.22 g/mol ≈ 2.30 × 10-5 g/L