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

100%sorry if i’m wronghope this good luck: )

The answer is the first option just took the test