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Calculate the molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water. Use 7.6 x 10^-36 as the solubility product constant of Cu3(AsO4)2.
9.1 x 10^-4 M
3.4 x 10^-2 M
3.7 x 10^-8 M
8.7 x 10^-2 M

Respuesta :

MissPhiladelphia
Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

3Cu2+ + 2(AsO4)3- = Cu3(AsO4)2

7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M

Answer:

The correct answer is option C.

Explanation:

[tex]Cu_3(AsO_4)_2\rightleftharpoons 3Cu^{2+}+2AsO_4^{3-}[/tex]

                                     3S       2S

The solubility product of the [tex]Cu_3(AsO_4)_2[/tex] will be given by:

[tex]K_{sp}=(3S)^3\times (2S)^2=108S^5[/tex]

[tex]7.6\times 10^{-36}=108S^5[/tex]

[tex]0.0703\times 10^{-36}=S^5[/tex]

[tex] S=3.71\times 10^{-8} M[/tex]

The molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water is [tex]3.71\times 10^{-8} M[/tex].

Hence ,the correct answer is option C.

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