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?

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.

Oh, so Mg(OH)2 only decreases on mole or solid. I got confused because OH- was on both sides. Thanks so much :)

When some HCl is added to the equilibrium between solid Mg(OH)2 and its ions Mg2+ and OH-, the addition of HCl causes a shift in the equilibrium position.

HCl is a strong acid and dissociates completely in water, producing H+ ions. The H+ ions react with the OH- ions from the equilibrium to form water, according to the reaction: H+ (aq) + OH- (aq) --> H2O (l). This reaction consumes the OH- ions and shifts the equilibrium in the reverse direction to replace the OH- ions being consumed.

Since the equilibrium is being shifted in the reverse direction, the concentration of OH- ions decreases. However, as the OH- ions are consumed, the equilibrium tries to compensate by dissociating some of the solid Mg(OH)2 to produce more OH- ions. Therefore, the amount of solid Mg(OH)2 also decreases.

In summary, when HCl is added, both the concentration of OH- ions and the amount of solid Mg(OH)2 decrease.

When HCl is added to the system in equilibrium, it reacts with the OH- ions, forming water. The reaction is as follows:

HCl (aq) + OH- (aq) --> H2O (l) + Cl- (aq)

In this reaction, HCl donates a proton (H+) to the OH- ion, forming water (H2O) and leaving behind a chloride ion (Cl-). As a result, the concentration of OH- ions decreases.

Now, according to Le Chatelier's principle, when the concentration of one of the reactants in an equilibrium system decreases, the equilibrium shifts in such a way to counteract this change. In this case, since the concentration of OH- ions decreases, the equilibrium shifts to the left to produce more OH- ions. This means that some of the solid Mg(OH)2 will dissociate to replenish the OH- ions in the solution.

So, in summary, when HCl is added, the amount of solid Mg(OH)2 decreases because it dissociates to replenish the OH- ions and maintain the equilibrium. However, it's important to note that the decrease in the amount of solid Mg(OH)2 is primarily due to the reaction with HCl, rather than a decrease in entropy.