Question #43e92

1 Answer
May 26, 2017

Answer:

(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).