Chapter SH, Problem 22.9APP

To determine

To write: A balanced equation for ionization of nitric acid $\left({\text{HNO}}_3\right)$ in water

Explanation of Solution

Introduction:

  $\left({\text{HNO}}_3\right)$ ionizes in water to form a positive and a negative ion.

A balanced equation for ionization of ${\text{HNO}}_3$ , nitric acid in water can be written as:

  ${\text{HNO}}_3+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}+{\text{NO}}_3{}^{-}$

Here, nitric acid, ${\text{HNO}}_3$ ionizes in water, ${\text{H}}_2\text{O}$ to form hydronium ion, ${\text{H}}_3{\text{O}}^{+}$ and nitrate ion ${\text{NO}}_3{}^{-}$ .

Conclusion:

An equation for the given reaction is:

  ${\text{HNO}}_3+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}+{\text{NO}}_3{}^{-}$