Question

Write equations for the three-step ionization of phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) . Compare the degree of ionization for the three steps.

Short answer

1st ionization: \( \mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2 \mathrm{PO}_4^- \)2nd ionization: \( \mathrm{H}_2 \mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \)3rd ionization: \( \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \). The ionization degree decreases with each step.

Step-by-step solution

First Ionization

Write the equation for the first ionization of phosphoric acid, where one hydrogen ion is removed from \( \mathrm{H}_3 \mathrm{PO}_4 \): \[ \mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2 \mathrm{PO}_4^- \]

Second Ionization

Write the equation for the second ionization, where another hydrogen ion is removed from \( \mathrm{H}_2 \mathrm{PO}_4^- \): \[ \mathrm{H}_2 \mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \]

Third Ionization

Write the equation for the third ionization, where the final hydrogen ion is removed from \( \mathrm{HPO}_4^{2-} \): \[ \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \]

Compare Degrees of Ionization

Discuss the degree of ionization for each step. The first ionization of \( \mathrm{H}_3 \mathrm{PO}_4 \) is the strongest, followed by the second, and the third ionization is the weakest. This is because the removal of each successive hydrogen ion becomes more difficult as the molecule becomes progressively more negatively charged.

Key concepts

chemical equations

Understanding the ionization of phosphoric acid starts with writing the chemical equations for each step of the process. Phosphoric acid, \( \mathrm{H}_3 \mathrm{PO}_4 \), ionizes in three distinct steps, each releasing a hydrogen ion (H\(^+\)):

• The first ionization step shows \( \mathrm{H}_3 \mathrm{PO}_4 \) losing a hydrogen ion to form \( \mathrm{H}_2 \mathrm{PO}_4^- \): \[ \mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2 \mathrm{PO}_4^- \]
• The second ionization step shows \( \mathrm{H}_2 \mathrm{PO}_4^- \) losing another hydrogen ion to form \( \mathrm{HPO}_4^{2-} \): \[ \mathrm{H}_2 \mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \]
• The third ionization step shows \( \mathrm{HPO}_4^{2-} \) losing the final hydrogen ion to form \( \mathrm{PO}_4^{3-} \): \[ \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \]

These equations are crucial for understanding the processes involved in the ionization of phosphoric acid.
Writing chemical equations step-by-step helps visualize how a molecule breaks down and loses its hydrogen ions one at a time. It makes understanding the ionization process easier.

ionization steps

Phosphoric acid ionizes in three steps. Each step involves losing one hydrogen ion (H\(^+\)) and creating a new ion.

• In the first ionization step, \( \mathrm{H}_3 \mathrm{PO}_4 \) loses one H\(^+\): \[ \mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2 \mathrm{PO}_4^- \] This forms dihydrogen phosphate (\( \mathrm{H}_2 \mathrm{PO}_4^- \)).


• In the second ionization step, \( \mathrm{H}_2 \mathrm{PO}_4^- \) loses another H\(^+\): \[ \mathrm{H}_2 \mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \] This forms hydrogen phosphate (\( \mathrm{HPO}_4^{2-} \)).


• In the third ionization step, \( \mathrm{HPO}_4^{2-} \) loses the final H\(^+\): \[ \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \] This forms phosphate (\( \mathrm{PO}_4^{3-} \)).

Each step becomes more difficult because removing each successive hydrogen ion requires more energy. As the molecule loses a hydrogen ion, it becomes negatively charged, making it harder to lose the next hydrogen ion. This progressive change in charge affects the ease of ionization.

degree of ionization

The degree of ionization indicates how easily a molecule loses its hydrogen ions. In phosphoric acid, the degree of ionization decreases with each step.

• The first ionization has the highest degree of ionization because it is the easiest: \[ \mathrm{H}_3 \mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2 \mathrm{PO}_4^- \]
• The second ionization is more challenging, so its degree of ionization is lower: \[ \mathrm{H}_2 \mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \]
• The third ionization is the hardest, having the lowest degree of ionization: \[ \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \]

This trend is due to the increasing negative charge on the molecule.
As more hydrogen ions are removed, the molecule accumulates more negative charge.
This makes it more difficult to remove the next hydrogen ion.
The first ionization is easier because the molecule is neutral.
By the second and third steps, the molecule becomes more negatively charged, hence, more energy is needed to remove additional hydrogen ions.