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charle [14.2K]
3 years ago
13

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
Chemistry
2 answers:
Molodets [167]3 years ago
8 0
Molar solubility<span> 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:

</span>3Cu2+ + 2(AsO4)3-<span> = Cu3(AsO4)2
</span>
7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M
andriy [413]3 years ago
4 0

Answer:

The correct answer is option C.

Explanation:

Cu_3(AsO_4)_2\rightleftharpoons 3Cu^{2+}+2AsO_4^{3-}

                                     3S       2S

The solubility product of the Cu_3(AsO_4)_2 will be given by:

K_{sp}=(3S)^3\times (2S)^2=108S^5

7.6\times 10^{-36}=108S^5

0.0703\times 10^{-36}=S^5

S=3.71\times 10^{-8} M

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

Hence ,the correct answer is option C.

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<h3>Answer:</h3>

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