In the following reaction, HS-(aq) + H2O(l) ⇌ S2-(aq) + H3O+(aq)
HS- is an acid and H2O is its conjugate base.
H3O+ is an acid and HS- is its conjugate base.
HS- is an acid and S2- is its conjugate base.
H2O is an acid and S2- is its conjugate base.
HS- is an acid and S2- is its conjugate base.
In the given reaction, HS- is an acid and H2O is its conjugate base. This is because in the forward reaction (HS-(aq) + H2O(l) ⇌ S2-(aq) + H3O+(aq)), HS- donates a proton (H+) to water (H2O), forming the hydrosulfide ion (S2-) and hydronium ion (H3O+). Therefore, HS- is acting as an acid by donating a proton.
H3O+ is an acid and HS- is its conjugate base. This is because in the reverse reaction (S2-(aq) + H3O+(aq) ⇌ HS-(aq) + H2O(l)), the hydrosulfide ion (HS-) accepts a proton (H+) from the hydronium ion (H3O+), forming HS- and water (H2O). Therefore, H3O+ is acting as an acid by donating a proton, and HS- is its conjugate base.
HS- is an acid and S2- is its conjugate base. As mentioned earlier, in the forward reaction, HS- donates a proton to water, forming S2- and H3O+. Consequently, HS- is acting as an acid and S2- is acting as its conjugate base.
H2O is not acting as an acid in this reaction. It is acting as a base by accepting a proton from HS- to form the hydrosulfide ion. Therefore, H2O is not an acid, and S2- is the conjugate base of HS-.