Chapter 22, Problem 12STP

To determine

To Write: The chemical equation for the ionization reaction of hydrogen sulfide in water.

Explanation of Solution

Introduction:Hydrogen sulfide forms a weak acid when dissolved in water.

The chemical formula of thehydrogen sulfide is ${\text{H}}_2\text{S}$ .

When dissolved in water, one molecule of ${\text{H}}_2\text{S}$ reacts with one molecule of water to give a ${\text{HS}}^{-}$ ion and a ${\text{H}}_3{\text{O}}^{+}$ ion. The ${\text{HS}}^{-}$ ion further reacts with another water molecule to produce a ${\text{S}}^{-}$ ion and a ${\text{H}}_3{\text{O}}^{+}$ ion.

The reaction can be written as the following chemical equation:

  $\begin{array}{l}{\text{H}}_2{\text{S+H}}_2\text{O}\to {\text{HS}}^{-}+{\text{H}}_3{\text{O}}^{+}\\ {\text{HS}}^{-}{\text{+H}}_2\text{O}\to {\text{ S}}^{-}+{\text{H}}_3{\text{O}}^{+}\end{array}$

Conclusion: The ionization reaction of hydrogen sulfide in water produces a ${\text{S}}^{-}$ ion and two ${\text{H}}_3{\text{O}}^{+}$ ions.