In a saturated solution of silver phosphate, the concentration of silver ion is 4.5 x 10^-4 mol/L. The Ksp of silver phosphate would be which of the following?

6.8  10-8 1.0  10-11
1.4  10-14 none of the above
1.5  10-15

To determine the Ksp (solubility product constant) of silver phosphate, we need to write the balanced chemical equation and set up the expression for the equilibrium constant.

The balanced chemical equation for the dissociation of silver phosphate (Ag3PO4) is as follows:

Ag3PO4(s) ⇌ 3Ag+(aq) + PO4^3-(aq)

The solubility product constant (Ksp) expression for the dissociation of silver phosphate is:

Ksp = [Ag+]^3 [PO4^3-]

Given that the concentration of silver ion (Ag+) is 4.5 x 10^-4 mol/L, we can substitute this value into the Ksp expression:

Ksp = (4.5 x 10^-4)^3 [PO4^3-]

Now we need to consider the concentration of phosphate ion (PO4^3-) in the saturated solution. The concentration of phosphate ion can be determined by using the stoichiometry of the balanced equation. Since the stoichiometric coefficient of phosphate ion is 1, the concentration of phosphate ion will be equal to the concentration of silver ion:

[PO4^3-] = 4.5 x 10^-4 mol/L

Substituting this value into the Ksp expression:

Ksp = (4.5 x 10^-4)^3 (4.5 x 10^-4)

Simplifying the calculation:

Ksp = 1.5 x 10^-15

Therefore, the Ksp of silver phosphate is 1.5 x 10^-15.

The correct option from the given choices is:
1.5  10^-15

See above.