Question

Calculate the mass per liter of solid lead (II) phosphate (Ksp = 1.00 x 10-54) that should dissolve in 0.320 M lead (II) nitrate solution.

Answer 1

Pb(N03)2 --> Pb+2 + 2 NO3-

since lead nitrate is very soluble in water

all of it will split into respective ions

now from the reaction

[Pb+2] = [Pb(NO3)2] taken = 0.32

[Pb+2] = 0.32 M

now

Pb3(PO4)2 --> 3 Pb+2 + 2 PO43-

now

Ksp = [Pb+2]^3 [PO43-]^2

1 x 10-54 = [0.32]^3 [PO43-]^2

[PO43-] = 5.524 x 10-27 M

so

now we know that

concentration of Pb2(PO4)3 = (concentration of PO43- ) / 3

concentration of Pb2(PO4)3 = 5.524 10-27 / 3

concentration of Pb2(PO4)3 = 1.84 x 10-27 mol / L

now

concentration in mass / L = molar mass x conc in mol/L

= (811.5427 g/mol) x 1.84 x 10-27 mol/L

= 1.494 x 10-24 g/L

so

1.494 x 10^-24 g / L of solid lead (ll) phosphate should be dissolved