the solubility of the fictitious compound administratium fluoride in water is 3.091e-4.calculate the value of solubility product ksp
See your PbCl2 problem.
To calculate the solubility product constant (Ksp), we need to know the molar concentration of the dissolved ions in a saturated solution. In this case, we only have the solubility value, which is given as 3.091e-4.
The solubility of a compound is usually given in moles per liter (M). However, we can use the given solubility to calculate the molar concentration.
Given:
Solubility (S) = 3.091e-4 M
Ksp is calculated by multiplying the concentration of the dissolved ions.
For the compound administratium fluoride (Af2):
Af2 (s) ⇌ A+ (aq) + 2F- (aq)
The equilibrium expression for the dissociation is:
Ksp = [A+][F-]^2
Since the compound dissociates into A+ and 2F-, the concentration of the A+ ion is the same as the solubility (S), and the concentration of the F- ion is double the solubility (2S).
Plugging these values into the Ksp expression:
Ksp = (S)(2S)^2
= 4S^3
Now, we can substitute the given solubility value:
Ksp = 4(3.091e-4)^3
Using a calculator, we can solve this equation to find the value of Ksp.
To calculate the solubility product constant (Ksp) for a compound, we need to know its molar solubility, which is the number of moles of the compound that dissolve in water per liter of solution.
The given information states that the solubility of administratium fluoride (AF) in water is 3.091e-4. This means that for every liter of water, 3.091e-4 moles of AF dissolve.
The solubility product constant (Ksp) can be calculated using the following equation:
Ksp = [A+]^m * [B-]^n
Where [A+] and [B-] are the concentrations of the cation (A+) and anion (B-) in the compound, respectively, and m and n are the coefficients of A+ and B- in the balanced equation.
In the case of administratium fluoride (AF), it can be assumed that AF dissociates completely into A+ and F- ions:
AF (s) ⇌ A+ (aq) + F- (aq)
Since the solubility of AF is given as 3.091e-4, the concentrations of A+ and F- ions can be assumed to be 3.091e-4 M based on stoichiometry. Therefore:
[A+] = 3.091e-4 M
[F-] = 3.091e-4 M
Now, substitute these values into the equation for Ksp:
Ksp = [A+]^m * [F-]^n = (3.091e-4)^1 * (3.091e-4)^1 = 9.559e-8
Therefore, the value of the solubility product constant (Ksp) for administratium fluoride is 9.559e-8.