Right answer. Wrong reason. Entropy doesn't enter into it.
To do a VERY detailed answer:
HCl(aq) ==> H^+ + Cl^-
The excess H^+ from the HCl reacts with the OH^- from the ionization of Mg(OH)2 to form H2O. This reduces the concn of OH^- on the right and, by Le Chatelier's Principle, the Mg(OH)2 reaction shifts to the right to replace the OH^- lost by the reaction with HCl. So more of the solid (on the left side of the equation) dissolves to try to restore the OH^- that it had before the HCl messed things up. Remember the Ksp = (Mg^+2)(OH^-)^2 = constant.
Consider the following equilibrium:
Mg(OH)2 (s) <--> Mg 2+ (aq) + 2OH-(aq)
What happens to the amount of solid Mg(OH)2 when some HCl is added?
It decreases because it has least entropy?
So in the reaction, both the concentration of Mg(OH)2 and OH- decreases?
2 answers
Oh, so Mg(OH)2 only decreases on mole or solid. I got confused because OH- was on both sides. Thanks so much :)
3 answers