Question

It is found that up to 0.0110 g of SrF2 dissolves in 100 ml of aqueous solution at a certain temperature. Determine the value of ksp for SrF2.

Answer 1

The solubility product constant is ksp = 2725.47 * $10^{-12}$

The dissociation reaction of SrF2 is given below:

SrF2 ----------------> $Sr ^{2+}$ + 2$F^{-}$

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 = [ $Sr ^{2+}$] * [2$F^{-}$]^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 * $10^{-12}$

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