Calculate the mass per liter of solid lead (ii) phosphate (ksp = 1.00 x 10-54) that should dissolve in 0.270 m lead (ii) nitrate solution.

According to the reaction equation:

Pb3 (PO4) 2 (s) ↔ 3Pb2 + (aq) + 2 PO4 3 - (aq)

when Ksp = [Pb2+]^3[PO43-]^2

when we have Ksp = 1 x 10^-54

and the initial [Pb2+] = 0.27 m

by substitution:

1 x 10^-54 = (0.27) ^3 * [PO43-]^2

∴ [PO43-] = 7 x 10^-27 M

∴ Mass of lead (II) phosphate = molarity * 1 mol Pb3 (PO4) 2 / 2 mol PO43-*molar mass

= 7 x 10^-27 * 1/2 * 811.5 g/mol

= 2.84 x 10^-24 g/L