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 solubility product constant (Ksp), we need to know the balanced chemical equation for the dissociation of administratium fluoride (AdF3) in water.
Let's assume the balanced equation for the dissociation is:
AdF3(s) ⇌ Ad3+(aq) + 3F-(aq)
The solubility of AdF3 in water is given as 3.091×10−4 M, which means that the concentration of Ad3+ ions in the saturated solution is 3.091×10−4 M, and the concentration of F- ions is 3 times higher, i.e., 3 × 3.091×10−4 M = 9.273×10−4 M.
The expression for the solubility product constant (Ksp) is the product of the concentrations of the products raised to the power of their stoichiometric coefficients.
So, in this case, the Ksp expression is:
Ksp = [Ad3+][F-]^3
Substituting the concentrations we obtained:
Ksp = (3.091×10−4)(9.273×10−4)^3
Evaluating this expression will give you the value of the solubility product constant (Ksp).