Final answer:
To calculate the molarity of Ag+ in a saturated Ag3PO4 solution, we use the Ksp value of 1.8×10−18. We set up the equilibrium expression for the dissolution of Ag3PO4, solve for the molarity (x), and then multiply by 3 to find the molarity of Ag+ which results in 1.512×10−5M.
Step-by-step explanation:
The solubility constant (Ksp) for Ag3PO4 is given as 1.8×10−18. To find the molarity of Ag+ in a saturated solution of Ag3PO4, we need to set up the equilibrium expression based on the solubility product.
For the dissolution of Ag3PO4, the reaction is as follows:
Ag3PO4(s) ↔ 3Ag+(aq) + PO43−(aq)
Applying the solubility product constant expression, we have:
Ksp = [Ag+]3[PO43−]
If 'x' represents the molarity of Ag3PO4 that dissolves to form Ag+, then at equilibrium, [Ag+] = 3x and [PO43−] = x. Plugging these values back into the Ksp expression gives:
Ksp = (3x)3(x) = 27x4 = 1.8×10−18
Solving for x:
x4 = (1.8×10−18) / 27
x4 = 6.67×10−20
x = 5.04×10−6M
Therefore, [Ag+] = 3x = 3(5.04×10−6) = 1.512×10−5M, which is the molarity of Ag+ in a saturated solution of Ag3PO4.