The solubility of the fictitious compound, administratium fluoride (AdF3) in water is 3.091×10−4 M. Calculate the value of the solubility product Ksp.
To calculate the value of the solubility product constant (Ksp), we need to know the equation for the dissociation of administratium fluoride (AdF3) in water. For a compound that is not listed in any source, it is difficult to find its exact dissociation equation. However, we can assume that it will dissociate into Ad3+ and F- ions, as most ionic compounds do.
The dissociation equation can be written as follows:
AdF3 (s) ⇌ Ad3+ (aq) + 3F- (aq)
The solubility of AdF3 in water is given as 3.091×10−4 M. This means that at equilibrium, the concentration of Ad3+ ions and F- ions in the solution is 3.091×10−4 M.
From the equation, we can see that the molar ratio between AdF3 and Ad3+ ions is 1:1. Therefore, the concentration of Ad3+ ions is also 3.091×10−4 M.
Since AdF3 dissociates into 3F- ions, the concentration of F- ions will be three times the concentration of Ad3+ ions. Therefore, the concentration of F- ions is 3 × 3.091×10−4 M = 9.273×10−4 M.
Now, we have the concentrations of Ad3+ and F- ions in the solution. To calculate the solubility product constant (Ksp), we multiply these concentrations together.
Ksp = [Ad3+] × [F-]^3
= (3.091×10−4 M) × (9.273×10−4 M)^3
Calculating this expression will give you the value of the solubility product constant (Ksp) for administratium fluoride (AdF3) in water.