Question

Write the equations for the stepwise dissociation of pyrophosphoric acid, \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7} .\) Identify all conjugate acidbase pairs.

Short answer

The stepwise dissociation of pyrophosphoric acid (H₄P₂O₇) involves four reactions and four conjugate acid-base pairs: 1. \( \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \) (H₄P₂O₇ and H₃P₂O₇⁻) 2. \( \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \) (H₃P₂O₇⁻ and H₂P₂O₇²⁻) 3. \( \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \) (H₂P₂O₇²⁻ and HP₂O₇³⁻) 4. \( \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{P}_{2} \mathrm{O}_{7}^{4-} (\mathrm{aq}) \) (HP₂O₇³⁻ and P₂O₇⁴⁻)

Step-by-step solution

First dissociation

In the first step, pyrophosphoric acid (H₄P₂O₇) releases one hydrogen ion (H⁺) and becomes an anion with the formula H₃P₂O₇⁻. The reaction can be written as: \[ \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \] Conjugate acid-base pair 1: H₄P₂O₇ (acid) and H₃P₂O₇⁻ (base)

Second dissociation

In the second step, H₃P₂O₇⁻ releases one hydrogen ion (H⁺) and becomes an anion with the formula H₂P₂O₇²⁻. The reaction can be written as: \[ \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \] Conjugate acid-base pair 2: H₃P₂O₇⁻ (acid) and H₂P₂O₇²⁻ (base)

Third dissociation

In the third step, H₂P₂O₇²⁻ releases one hydrogen ion (H⁺) and becomes an anion with the formula HP₂O₇³⁻. The reaction can be written as: \[ \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \] Conjugate acid-base pair 3: H₂P₂O₇²⁻ (acid) and HP₂O₇³⁻ (base)

Fourth dissociation

In the fourth step, HP₂O₇³⁻ releases one hydrogen ion (H⁺) and becomes an anion with the formula P₂O₇⁴⁻. The reaction can be written as: \[ \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{P}_{2} \mathrm{O}_{7}^{4-} (\mathrm{aq}) \] Conjugate acid-base pair 4: HP₂O₇³⁻ (acid) and P₂O₇⁴⁻ (base) In conclusion, the stepwise dissociation of pyrophosphoric acid (H₄P₂O₇) involves four reactions, and there are four conjugate acid-base pairs, as shown above.

Key concepts

Pyrophosphoric Acid

Pyrophosphoric acid, chemically denoted as \( \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7 \), is an inorganic compound that consists of four hydrogen atoms, two phosphorus atoms, and seven oxygen atoms. This unique composition makes it a polyprotic acid.
Polyprotic acids are capable of losing multiple protons, with each loss occurring one step at a time. In the context of pyrophosphoric acid, it can lose up to four protons in total through stepwise dissociation.
In aqueous solutions, pyrophosphoric acid plays an important role not only due to its acidic nature but also because it can form phosphate-based compounds. Understanding each dissociation step provides insights into its reactivity and potential applications.

• First dissociation: It starts as \( \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7 \) and releases a proton to become \( \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \).
• Second dissociation: This leads to \( \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \).
• Third dissociation: Next, it becomes \( \mathrm{HP}_2 \mathrm{O}_7^{3-} \).
• Fourth dissociation: Finally, it results in \( \mathrm{P}_2 \mathrm{O}_7^{4-} \).

This cascading transformation highlights its role as a multi-step acid, crucial for various chemical processes.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs form the cornerstone of the Bronsted-Lowry acid-base theory. In the dissociation reactions of pyrophosphoric acid, each step reveals a new conjugate acid-base pair. This concept helps identify the proton donor (acid) and proton acceptor (base) in each reaction.
Let's look at how this occurs for pyrophosphoric acid:

• First pair: \( \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7 \) (acid) and \( \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \) (base).
• Second pair: \( \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \) (acid) and \( \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \) (base).
• Third pair: \( \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \) (acid) and \( \mathrm{HP}_2 \mathrm{O}_7^{3-} \) (base).
• Fourth pair: \( \mathrm{HP}_2 \mathrm{O}_7^{3-} \) (acid) and \( \mathrm{P}_2 \mathrm{O}_7^{4-} \) (base).

In each pair, the acid donates a proton to become its conjugate base, while the base accepts a proton to become its conjugate acid. Each new species can undergo further dissociation, demonstrating the dynamic nature of these equilibria.

Chemical Equations

Chemical equations are vital tools for understanding chemical reactions. They represent reactants and products and help us visualize the transformation from one to another.
In the case of pyrophosphoric acid dissociation, four chemical equations illustrate each step of the process. These equations not only depict the separation of hydrogen ions but also reveal the formation of new anions at each stage.
Each dissociation step has its specific equation:

• First: \( \mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7 \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \)
• Second: \( \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \)
• Third: \( \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{HP}_2 \mathrm{O}_7^{3-} \)
• Fourth: \( \mathrm{HP}_2 \mathrm{O}_7^{3-} \rightleftharpoons \mathrm{H}^+ + \mathrm{P}_2 \mathrm{O}_7^{4-} \)

These equations help express the conservation of mass and charge, and they balance the number of atoms on each side. Understanding these helps in gauging the extent and direction of the reactions as well.

Anion Formation

Anion formation occurs during the dissociation of pyrophosphoric acid. As hydrogen ions are released, the remaining part of the molecule becomes negatively charged, forming an anion.
During each step of pyrophosphoric acid dissociation:

First step: The anion formed is \( \mathrm{H}_3 \mathrm{P}_2 \mathrm{O}_7^- \).

Second step: The anion becomes \( \mathrm{H}_2 \mathrm{P}_2 \mathrm{O}_7^{2-} \).

Third step: It further transforms into \( \mathrm{HP}_2 \mathrm{O}_7^{3-} \).

Fourth step: Finally, it results in \( \mathrm{P}_2 \mathrm{O}_7^{4-} \).

Anion formation is key to understanding the behavior of a molecule in solution, affecting its reactivity and interactions with other species. These anions can participate in further reactions, forming new compounds or equilibria in chemical processes.