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

The correct answer is option C.

Step-by-step 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* (2S)^2=108S^5`

`7.6* 10^(-36)=108S^5`

`0.0703* 10^(-36)=S^5`

`S=3.71* 10^(-8) M`

The molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water is
`3.71* 10^(-8) M`.

Hence ,the correct answer is option C.