Respuesta :

superman1987
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
pstnonsonjoku

Using the Ksp of the solute and the idea of common ion effect, the mass of solute dissolved is 4.4 * 10^-30 g.

What is common ion effect?

Common ion effect refers to a situation in which a solvent already has an ion in common with the solute which prevents dissolution of the solute in the solvent.

The equation of the reaction is; Pb3(PO4)2(s) ==> 3Pb^2+(aq) + 2PO4^3-(aq)

Now we know that the Ksp of solid lead (ii) phosphate  is 1.00 x 10-54.

1.00 x 10-54 = [Pb^2+]^3 [PO4^3-]^2

[Pb^2+] = 0.270 M

[PO4^3-] = 7.12 * 10^-27 M

Then;

mass = 7.12 * 10^-27 M x 1 mol Pb3(PO4)2/2 mol PO43- x 811.5 g/mole = 4.4 * 10^-30 g

Learn more about molarity: https://brainly.com/question/11256472

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