Scandium fluoride is 1.56 * 10¹¹ times more soluble in pure water than it is in a solution that is 0.00837 M in potassium fluoride.
The molar solubilty of scandium fluoride in pure water and in 0.00837 M potassium fluoride is calculated from the Ksp value of Scandium fluoride.
Equation for the dissolution of scandium fluoride in water is given below:
Molar solubilty in pure water is calculated thus:
[tex][Sc^{3+}] = x; [F^{-}] = x[/tex]
[tex]K_{sp} = [Sc^{3+}] * [F^{-}]^{3}[/tex]
[tex]5.81 *10^{-24} = x * x^{3} = x^{4}[/tex]
x = 1.55 * 10⁻⁶ M
Molar solubilty in 0.00837 M potassium fluoride is calculated thus:
[tex][Sc^{3+}] = x; [F^{-}] = (0.00837 + x)[/tex]
[tex]K_{sp} = [Sc^{3+}] * [F^{-}]^{3}[/tex]
[tex]5.81 *10^{-24} = x * (0.00837 + x)^{3}[/tex]
Since the Ksp value is very small, we assume that x <<< 0.008367, therefore;
[tex]5.81 *10^{-24} = x * (0.00837)^{3}[/tex]
x = 9.91 * 10⁻¹⁸ M
Comapring the molar solubilties in water and in 0.00837 M in potassium fluoride:
1.55 * 10⁻⁶ M/9.91 * 10⁻¹⁸ M = 1.56 * 10¹¹
Therefore, scandium fluoride is 1.56 * 10¹¹ times more soluble in pure water than it is in a solution that is 0.00837 M in potassium fluoride.
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