consider equilibrium equation below
`Mg(NO3)2(aq) + 2Na(OH)(aq) <--> Mg(OH)2(s) + 2NaNO3 (aq)
If NH3 is added, according to Le Chatelier's Principle, which way would reaction shift?
I would think the rxn is amost at 100% now; however, adding more OH^- (from NH3 + H2O ==> NH4^+ + OH^-) would add more OH to the left, the rxn would need to try to reduce the OH^- and it can do that by forming more Mg(OH)2(s). So to the right.
To determine how the reaction would shift when NH3 is added, we need to consider Le Chatelier's principle. Le Chatelier's principle states that when a system at equilibrium is subjected to a change, it will undergo a shift to minimize the effect of that change and restore equilibrium.
In this reaction, NH3 (ammonia) is a weak base that can react with water to produce hydroxide ions (OH-) according to the following equation:
NH3(aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)
When NH3 is added, it increases the concentration of the hydroxide ions (OH-) in the reaction mixture. According to Le Chatelier's principle, the reaction will shift in the direction that consumes the excess OH- ions in order to restore equilibrium.
In the given equation, the hydroxide ions (OH-) are on the reactant side. If NH3 is added, it will react with the excess hydroxide ions (OH-) to form more water (H2O) and ammonium ions (NH4+). As a result, the concentration of hydroxide ions (OH-) will decrease, and the system will shift to the right to produce more hydroxide ions (OH-).
Therefore, when NH3 is added, the reaction will shift to the right to produce more Mg(OH)2(s) and NaNO3(aq) while consuming NH3 and OH- ions.