abeosehbreannah
Answered

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

Verified Answer :

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:

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

                                     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.