Respuesta :
The solubility product constant is ksp = 2725.47 * 10⁻¹²
The dissociation reaction of SrF2 is given below:
SrF2 ----------------> Sr⁺² + 2F⁻
Let us make an ICE table for the reactant and product side:
I 1 0 0
C -s s 2s
E 1 - s s 2s
where,
I denote initial.
C denotes change.
E denotes equilibrium.
ksp is called the solubility product constant.
It is the equilibrium constant for the dissolution of solid substances into an aqueous solution.
A solubility product is the product of the concentration of ions that are present in a saturated solution of an ionic compound.
The ksp can be calculated as follows:
ksp = [ Sr⁺²] * [2F⁻]^2
= (s) * (2s)^2
where s is the solubility.
m = 0.011 g
n = 0.011 / 125.62
where n is the number of moles.
molar mass of SrF2 = 125.62 g / mol
n = 8.8 * 10⁻⁵
This is for one mole.
For 2 moles it will be
2s = 17.6 * 10⁻⁵ / 0.1 mol / L
= 17.6 * 10⁻⁴ mol / L
100 ml = 100 * 10⁻³ = 0.1 L
Therefore,
s = 8.8 * 10⁻⁵ / 0.1
= 8.8 * 10⁻⁴ mol / L
ksp = (8.8 * 10⁻⁴ ) * (17.6 *10⁻⁴ )^2
ksp = 2725.47 * 10⁻¹²
The solubility product constant is ksp = 2725.47 *10⁻¹²
For more information on Solubility constant click on the link below:
brainly.com/question/1419865
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