Question
Hydrogen selenide (H2Se) reacts with water according the following equation H2Se + H2O SeH- + H3O+
Answers
GPT-4o mini
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.
\[
\text{H}_2\text{Se} + \text{H}_2\text{O} \rightleftharpoons \text{SeH}^- + \text{H}_3\text{O}^+
\]
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.