The fictitious compound, pandemonium fluoride (PnF2) has a Ksp value in water of 3.091×10−9M3 at room temperature. Calculate the solubility of PnF2 in water. Express your answer in units of molarity.

To calculate the solubility of PnF2 in water, we need to use the Ksp value. Ksp stands for the solubility product constant and represents the equilibrium between the dissolved ions in a saturated solution.

The general equation for the solubility product expression of a compound is:
Ksp = [A]^x[B]^y

Where [A] and [B] are the molar concentrations of the ions A and B in the compound, and x and y are their respective stoichiometric coefficients in the balanced chemical equation.

In the case of PnF2, the compound dissociates into two fluoride ions (F-) for every one PnF2 molecule, so the equation becomes:
Ksp = [F-]^2

We are given the Ksp value of PnF2, which is 3.091×10−9 M^3 at room temperature. Since the stoichiometric coefficient for fluoride ions is 2, we can rewrite the equation as:
3.091×10−9 M^3 = [F-]^2

To find the solubility of PnF2 in water (the concentration of fluoride ions), we need to take the square root of the Ksp value:
[F-] = √(3.091×10−9 M^3)

Calculating this expression, we find that the solubility of PnF2 in water, represented by the concentration of fluoride ions, is approximately 5.56×10^-5 M (molarity).