Question

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

Answer 1

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 2

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.