Question #43e92

1 Answer
Michael
May 26, 2017

(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).

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