203k views
3 votes
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

1 Answer

2 votes

Answer:

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.

User Dylan Gattey
by
8.4k points