Zn^2+(aq) + 2OH^-(aq) ----> Zn(OH)2(s)
a) how the addition of HCl affects the equilibrium? my answer is it will shift to the right
b) how the addition of a small amount of NaOh affects the equilibrium?my answer is it will shift to the left
c) how the addition of a large amount of NaOH affects the equilibrium? my answer is it will shift to the left
d) rewrite the net equation for this equilibrium and indicate if this is an exothermic or endothermic reaction? my ansewr is endothermic reaction
You need to re-read your text about Le Chatelier's Principle. That principle says, in simple terms, that a system in equilibrium will try to undo what we do to it.
Zn^2+(aq) + 2OH^-(aq) ==> Zn(OH)2(s)
a. If HCl is added, it reacts with the OH (H^+ + OH^- ==> H2O) which reduces the OH^-. So the system will shift to increased the OH. The ONLY way it can do that (three choices are: shift right, shift left, doesn't shift) is to shift to the left.
b. If you add a small amount of OH^- you increase OH so the system will try to reduce that and the only way it can do it is to make the reaction go forward (shift to the right).
c. They tricked you on this one. A LARGE amount of NaOH makes this reaction.
Zn(OH)2(s) + 2OH^-(aq) ==> Zn(OH)4^2-(aq)
So addition of a large amount of NaOH dissolves Zn(OH)2^2- so the reaction goes to the right.
thanks for the explanation pls is it endothermic or exothermic reaction i'm confused with this
a) The addition of HCl will shift the equilibrium to the LEFT. This is because HCl is a strong acid that will react with the OH- ions, reducing their concentration. According to Le Chatelier's principle, when the concentration of one of the reactants (OH-) decreases, the system will shift in the direction that produces more of that reactant. Therefore, the equilibrium will shift to the left to regenerate the OH- ions that were consumed by the reaction with HCl.
b) The addition of a small amount of NaOH will shift the equilibrium to the RIGHT. This is because NaOH is a strong base that will react with the Zn2+ ions, reducing their concentration. According to Le Chatelier's principle, when the concentration of one of the reactants (Zn2+) decreases, the system will shift in the direction that produces more of that reactant. Therefore, the equilibrium will shift to the right to regenerate the Zn2+ ions that were consumed by the reaction with NaOH.
c) The addition of a large amount of NaOH will shift the equilibrium to the LEFT. This is because the large amount of NaOH will react with almost all of the Zn2+ ions to form Zn(OH)2 precipitate. According to Le Chatelier's principle, when one of the reactants (Zn2+) is consumed almost completely, the system will shift in the direction that produces more of that reactant. Therefore, the equilibrium will shift to the left to regenerate the Zn2+ ions that were consumed by the reaction with NaOH.
d) The net equation for this equilibrium is Zn^2+(aq) + 2OH^-(aq) ↔ Zn(OH)2(s).
This reaction is an exothermic reaction.
a) Adding HCl to the system will increase the concentration of H+ ions in solution. According to Le Chatelier's principle, the system will shift to counteract this increase by favoring the reaction that produces OH- ions, which in this case is the forward reaction. Hence, the equilibrium will shift to the right.
b) Adding a small amount of NaOH will increase the concentration of OH- ions in solution. According to Le Chatelier's principle, the system will shift to counteract this increase by favoring the reaction that consumes OH- ions, which in this case is the reverse reaction. Hence, the equilibrium will shift to the left.
c) Adding a large amount of NaOH will significantly increase the concentration of OH- ions in solution. According to Le Chatelier's principle, the system will try to counteract this increase by favoring the reaction that consumes OH- ions, which in this case is the reverse reaction. Therefore, the equilibrium will shift to the left.
d) The net equation for the equilibrium is:
Zn^2+(aq) + 2OH^-(aq) ↔ Zn(OH)2(s)
This reaction involves the formation of a precipitate (Zn(OH)2), which suggests that it is an endothermic reaction. In an endothermic reaction, heat is absorbed or gained from the surroundings.