In the reaction \( \text{H}_2\text{Se} + \text{H}_2\text{O} \rightarrow \text{SeH}^- + \text{H}_3\text{O}^+ \), hydrogen selenide (\( \text{H}_2\text{Se} \)) acts as the acid because it donates a proton (\( \text{H}^+ \)) to water. As a result, water (\( \text{H}_2\text{O} \)), which acts as a base, accepts the proton to form the hydronium ion (\( \text{H}_3\text{O}^+ \)). The conjugate base of the acid \( \text{H}_2\text{Se} \) is the selenide ion (\( \text{SeH}^- \)), which forms when \( \text{H}_2\text{Se} \) loses a proton. This behavior can be explained by the electronegativity of the elements involved: sulfur group elements, like selenium, have a tendency to donate protons, while water, being a polar molecule, readily accepts protons, enabling it to act as a base in this reaction.
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Use the periodic table to complete the activity.
An illustration shows the periodic table of elements with elements arranged in increasing order of atomic number in 18 columns (groups) and 7 rows (periods). The electronegativity of elements increases from left to right, across a period. The hydrogen on the left is included in group with medium electronegativity. Also, fluorine is shown as element with highest electronegativity.
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 the periodic table to explain why each substance acted as the acid or the base in this reaction.
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