Question

What is the water solubility of Hg2Br2 (Ksp = 6 x 10?23) in moles per liter?

(Answer shows to be 2 x 10-8 M. Please show all work to get this answer. Thanks!)

Answer 1

Step-by-step explanation:

Suppose the solubility of
`Hg_(2)Br_(2)` is x. Hence, upon dissociation equilibrium reaction for
`Hg_(2)Br_(2)` will be as follows.

`Hg_(2)Br_(2)(s) \rightleftharpoons Hg^(2+)_(2)(aq) + 2Br^(-)(aq)`

At equilibrium: x 2x

Therefore, equation for
`K_(sp)` will be as follows.

`K_(sp)` =
`[Hg^(2+)_(2)][Br^(-)]^(2)`

=
`(x) * (2x)^(2)`

=
`4x^(3)`

x =
`((K_(sp))/(4))^(1/3)`

x =
`((6 * 10^(-23))/(4))^(1/3)`

=
`2 * 10^(-8)`

Thus, we can conclude that the water solubility of
`Hg_(2)Br_(2)` is
`2 * 10^(-8)`.