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 first need to understand the chemical equation for the dissolution of the compound, administratium fluoride (AdF3), in water. The general equation is as follows:
AdF3 (s) ↔ Ad3+ (aq) + 3F- (aq)
Based on the equation, we can see that 1 mole of AdF3 dissociates to produce 1 mole of Ad3+ ions and 3 moles of F- ions.
Given that the solubility of AdF3 in water is 3.091×10−4 M, we can determine the concentration of Ad3+ and F- ions in solution.
Concentration of Ad3+ = 3.091×10−4 M
Concentration of F- = 3 × (3.091×10−4 M) = 9.273×10−4 M
Now, we can substitute these concentrations into the Ksp expression:
Ksp = [Ad3+] × [F-]^3
Ksp = (3.091×10−4 M) × (9.273×10−4 M)^3
Calculating the value:
Ksp = (3.091×10−4 M) × (9.273×10−4 M)^3
= 2.200×10−16
Therefore, the value of the solubility product constant (Ksp) for administratium fluoride (AdF3) is 2.200×10−16.