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.
0.917655e-3
To calculate the solubility of PnF2 in water, we need to use the Ksp value given. The Ksp value represents the solubility product constant and indicates the extent to which a compound dissociates or dissolves in water.
The formula for the solubility product constant is as follows:
Ksp = [Pn2+][F-]^2
Where [Pn2+] represents the concentration of PnF2 dissolved in water and [F-] represents the concentration of fluoride ions in water.
Since the stoichiometry of the compound is 1:2 (1 Pn2+ ion for 2 F- ions), we can assume that the concentration of Pn2+ is equivalent to half of the fluoride ion concentration. Therefore, we'll represent the fluoride ion concentration as 'x' and the Pn2+ concentration as '0.5x'.
Now, we can substitute these values into the Ksp expression:
3.091×10^−9 = (0.5x)(x)^2
Rearranging the equation, we get:
3.091×10^−9 = 0.5x^3
Simplifying further:
x^3 = (3.091×10^−9) / 0.5
x^3 = 6.182×10^−9
To find the solubility of PnF2 in water, we need to take the cube root of both sides:
x ≈ (6.182×10^−9)^(1/3)
Using a scientific calculator or an online calculator for the cube root, we can find the approximate value of x, which is the concentration of fluoride ions in water.
Converting the concentration to molarity, we divide by the volume of the solution in liters. Assuming a volume of 1 liter, the solubility of PnF2 in water is approximately:
Solubility ≈ x / 1
Remember to use the appropriate units and significant figures when expressing your answer.