The Ksp for silver(I) phosphate is 1.8 × 10^–18. Determine the silver ion concentration in a saturated solution of silver(I) phosphate

To determine the silver ion concentration in a saturated solution of silver(I) phosphate, we need to use the solubility product constant (Ksp) and the stoichiometry of the balanced equation.

The balanced equation for the dissociation of silver(I) phosphate (Ag3PO4) into silver ions (Ag+) and phosphate ions (PO43-) is:

Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq)

According to the Ksp expression for silver(I) phosphate, we have:

Ksp = [Ag+]^3 [PO43-]

Since the stoichiometric coefficient in front of Ag+ is 3, we can assume that the concentration of Ag+ in the saturated solution will be 3 times the concentration of PO43-. Therefore, we can rewrite the Ksp expression as:

Ksp = (3[Ag+])^3 [PO43-]

Given that Ksp = 1.8 × 10^–18, we can substitute this value into the equation:

1.8 × 10^–18 = (3[Ag+])^3 [PO43-]

Now, we need to solve for [Ag+]. To do this, we need to know the concentration of the phosphate ions ([PO43-]) in the saturated solution. Without this information, we cannot directly calculate the silver ion concentration.

If you have the concentration of phosphate ions in the saturated solution, we can proceed with the calculations.

To determine the silver ion concentration in a saturated solution of silver(I) phosphate, we can use the solubility product constant (Ksp) of silver(I) phosphate.

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

Ag3PO4(s) ⇌ 3 Ag+(aq) + PO4^3-(aq)

According to the equation, every 1 mole of silver(I) phosphate dissociates to form 3 moles of Ag+ ions.

Since the Ksp value is given as 1.8 × 10^–18, this means that at equilibrium, the product of the concentrations of the silver ion (Ag+) and phosphate ion (PO4^3-) should be equal to Ksp.

Therefore, let's assume the concentration of the silver ion in the saturated solution of silver(I) phosphate is represented by [Ag+]. We can consider the concentration of the phosphate ion to be 1 because the solubility constant is based on the saturation of the silver(I) phosphate.

Using the equation for the solubility product constant, we can write:

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

Substituting the values, we get:

1.8 × 10^–18 = [Ag+]^3 * 1

Simplifying further, we find:

[Ag+]^3 = 1.8 × 10^–18

To solve for [Ag+], we take the cube root:

[Ag+] = (1.8 × 10^–18)^(1/3)

Calculating this, we find:

[Ag+] ≈ 5.49 × 10^–6 M

Therefore, the silver ion concentration in a saturated solution of silver(I) phosphate is approximately 5.49 × 10^–6 M.

..................Ag3PO4 ==> 3Ag^+ + [PO4]^3-

I.................solid.................0...............0
C................solid.................3x.............x
E................solid..................3x.............x

Ksp = (Ag^+)^3[PO4]^3-
Plug the E line into the Ksp expression above and solve for x = solubility Ag3PO4 in mols/L
Post your work if you get stuck.