Question

Question #43e92

Answer 1

(a).

Explanation:

`sf(Mg(OH)_(2(s))` `sf(rightleftharpoons)` `sf(Mg(aq)^(2+)+2OH(aq)^-)`

`sf(K_(sp)=[Mg_((aq))^(2+)][OH_((aq))^-]^2=7.1xx10^(-12)color(white)(x)"mol"^3."l"^-3)`

If this is carried out in 0.50 M NaOH solution the position of equilibrium will be driven to the left thus supressing further solution of `sf(Mg^(2+))` ions.

To make things much easier for ourselves I am going to assume that the vast majority of the `sf(OH^-)` ions come from the NaOH.

`:.` `sf([Mg_((aq))^(2+)]=K_(sp)/[OH_((aq))^-]^2)`

`sf([Mg_((aq))^(2+)]=(7.1xx10^(-12))/(0.50^2)=2.8xx10^(-11)color(white)(x)"mol/l")`

This means the solubility of `sf(Mg_((aq))^(2+)=2.8xx10^(-11)color(white)(x)"mol/l")`.

This gives (a).