Hydrogen selenide (H2Se) reacts with water according to the following equation.

Question

H2Se + H2O → SeH– + H3O+

Identify the acid, base, conjugate acid, and conjugate base in this reaction

(4 points)
H2Se is a(n)


H2O is a(n)

SeH– is a(n)

H3O+ is a(n)

Answer 1

In the reaction given:

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

we can identify the roles of each species using the Bronsted-Lowry acid-base theory, which defines acids as proton donors and bases as proton acceptors.

1. H2Se is a(n) acid because it donates a proton (H⁺) to water.

2. H2O is a(n) base because it accepts a proton (H⁺) from H2Se.

3. SeH⁻ is a(n) conjugate base of H2Se because it is what remains after H2Se donates a proton.

4. H3O⁺ is a(n) conjugate acid of H2O because it is formed when water accepts a proton.

So, summarizing:

• H2Se is an acid.
• H2O is a base.
• SeH⁻ is a conjugate base.
• H3O⁺ is a conjugate acid.