Hydrogen selenide (H2Se) reacts with water according to the following equation.
H2Se + H2O → SeH– + H3O+
In three to five sentences, identify the acid, base, conjugate acid, and conjugate base in this reaction. Use patterns in the periodic table to explain why the substances you identified acted as the acid and the base in this reaction.
In this reaction, hydrogen selenide (H2Se) is the acid, as it donates a proton (H+) to water to form the hydronium ion (H3O+). Water is the base, as it accepts the proton to form the selenide ion (SeH-).
H2Se acts as the acid due to its ability to donate a proton, which is a characteristic of substances containing hydrogen that are located towards the right side of the periodic table. Water acts as the base due to its ability to accept a proton, a characteristic of substances containing oxygen that are located towards the left side of the periodic table.
The conjugate acid in this reaction is the hydronium ion (H3O+), which is formed when water accepts a proton. The conjugate base is the selenide ion (SeH-), which is formed when hydrogen selenide donates a proton. These species are formed as a result of the transfer of a proton in the reaction, leading to the formation of their respective conjugate pairs.