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.