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.
........AdF3 ==> Ad^3+ + 3F^-
I.......solid.....0.......0
C.....x dissolves..x......3x
E.......solid......x......3x
Ksp = (Ad^3+)(F^-)^3
Ksp = (x)(3x)^3
You know x = 3.091E-4. Substitute and solve for Ksp.
2.46x10^-13
To calculate the solubility product constant (Ksp) for administratium fluoride (AdF3) in water, we first need to write the balanced chemical equation for the dissociation of AdF3 in water and determine the stoichiometry.
The formula for administratium fluoride (AdF3) suggests that one molecule of AdF3 dissociates into one administratium ion (Ad3+) and three fluoride ions (F-). Therefore, we can write the balanced equation as follows:
AdF3 ⇌ Ad3+ + 3F-
According to the solubility information provided, the concentration of the administratium ion (Ad3+) is negligible since it is a very insoluble compound. Hence, we can assume that the concentration of Ad3+ is close to zero.
Now, let's define the solubility of AdF3 in water as "s" and write the solubility product expression:
Ksp = [Ad3+][F-]^3
Since the concentration of Ad3+ is negligible, we can assume that [Ad3+] ≈ 0.
Therefore, the solubility product expression simplifies to:
Ksp = [F-]^3
Now we can substitute the given solubility of AdF3 in water (3.091×10−4 M) into the solubility product expression:
Ksp = (3.091×10−4)^3
Calculating this expression:
Ksp = 3.091×10−4 × 3.091×10−4 × 3.091×10−4
Ksp ≈ 2.89×10−11
Thus, the value of the solubility product constant (Ksp) for AdF3 in water is approximately 2.89×10−11.