The solubility of the fictitious compound, administratium fluoride ( AdF3 ) in water is 3.091x10^-4M . Calculate the value of the solubility product Ksp .

Ksp = 2.46e-13

Thank you

To calculate the value of the solubility product (Ksp), you need to use the information provided about the solubility of the compound in water.

First, let's write the equation for the dissolution of administratium fluoride (AdF3) in water:

AdF3(s) ⇌ Ad3+(aq) + 3F-(aq)

The solubility of AdF3 is given as 3.091x10^-4 M, which corresponds to the concentration of the Ad3+ ions. Since AdF3 dissolves in a 1:3 ratio, the concentration of fluoride ions (F-) will be three times the concentration of Ad3+:

[Ad3+] = 3.091x10^-4 M
[F-] = 3 × 3.091x10^-4 M = 9.273x10^-4 M

Now, we can write the expression for the solubility product (Ksp) using the concentrations of the ions:

Ksp = [Ad3+][F-]^3

Plugging in the values, we get:

Ksp = (3.091x10^-4 M)(9.273x10^-4 M)^3
= 2.478x10^-17 M^4

Therefore, the value of the solubility product (Ksp) for administratium fluoride (AdF3) is 2.478x10^-17 M^4.