Use the periodic table to complete the activity.

Question

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

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 following:

1. H2Se is a(n) acid. It donates a proton (H⁺) to water in this reaction, forming the conjugate base.

2. H2O is a(n) base. It accepts a proton from hydrogen selenide, leading to the formation of the hydronium ion (H3O⁺).

3. SeH⁻ is a(n) conjugate base. It is the species formed after H2Se donates a proton.

4. H3O⁺ is a(n) conjugate acid. It is the species formed when water (acting as a base) accepts a proton.

To summarize:

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