Hydrogen selenide (H2Se) reacts with water according the following equation H2Se + H2O SeH- + H3O+

1 answer

The reaction you provided describes the interaction of hydrogen selenide (H2Se) with water. In this reaction, hydrogen selenide acts as a weak acid and donates a proton (H⁺) to water, which results in the formation of selenide ions (SeH⁻) and hydronium ions (H3O⁺). The balanced equation can be summarized as follows:

\[ \text{H}_2\text{Se} + \text{H}_2\text{O} \rightleftharpoons \text{SeH}^- + \text{H}_3\text{O}^+ \]

In this equilibrium reaction:

  • H2Se is the weak acid.
  • H2O is acting as a base.
  • SeH⁻ is the conjugate base of H2Se.
  • H3O⁺ is the conjugate acid formed from the water molecule.

Since H2Se can donate a proton, it is classified as a Brønsted-Lowry acid in this context.