Complete these bronsted lowry reactions
HS^- + H^+ --->
HS^- + OH^- ---->
HS^- + H^+ ---> H2S
HS^- + OH^- ----> H2O + S^-2
HS^- + H^+ ---> H2S (This reaction is a proton transfer reaction where the HS^- accepts a proton from H^+ to form H2S)
HS^- + OH^- ---> S^2- + H2O (This reaction is also a proton transfer reaction where the HS^- donates a proton to OH^- to form H2O and S^2-)
To complete the Bronsted-Lowry reactions, we need to determine the products of the reactions.
Reaction 1:
HS^- + H^+ --> H2S
The reaction between the hydrosulfide ion (HS^-) and the hydrogen ion (H^+) will result in the formation of hydrogen sulfide (H2S).
Reaction 2:
HS^- + OH^- --> S^2- + H2O
The reaction between the hydrosulfide ion (HS^-) and the hydroxide ion (OH^-) will result in the formation of sulfide ion (S^2-) and water (H2O).
So, the completed reactions are:
HS^- + H^+ --> H2S
HS^- + OH^- --> S^2- + H2O
To complete the Bronsted-Lowry reactions, we need to identify the acid and base on each side of the equation.
1. HS^- + H^+ -->
In this reaction, HS^- (hydrogen sulfide ion) can act as a base by accepting a proton (H^+), and H^+ is the acidic proton that can be donated. Therefore, the reaction can be completed as:
HS^- (base) + H^+ (acid) --> H2S (hydrosulfuric acid)
The reaction can be written as:
HS^- + H^+ --> H2S
2. HS^- + OH^- -->
In this reaction, HS^- (hydrogen sulfide ion) can again act as a base by accepting a proton (H^+), and OH^- (hydroxide ion) can act as the acid by donating the proton. Therefore, the reaction can be completed as:
HS^- (base) + OH^- (acid) --> S^2- (sulfide ion) + H2O (water)
The reaction can be written as:
HS^- + OH^- --> S^2- + H2O
In summary:
1. HS^- + H^+ --> H2S
2. HS^- + OH^- --> S^2- + H2O