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Chapter 10: Problem 331

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

Expert verified
The stepwise dissociation of pyrophosphoric acid (H鈧凱鈧侽鈧�) involves four reactions and four conjugate acid-base pairs: 1. \( \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \) (H鈧凱鈧侽鈧� and H鈧働鈧侽鈧団伝) 2. \( \mathrm{H}_{3} \mathrm{P}_{2} \mathrm{O}_{7}^- (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \) (H鈧働鈧侽鈧団伝 and H鈧侾鈧侽鈧嚶测伝) 3. \( \mathrm{H}_{2} \mathrm{P}_{2} \mathrm{O}_{7}^{2-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \) (H鈧侾鈧侽鈧嚶测伝 and HP鈧侽鈧嚶斥伝) 4. \( \mathrm{H}\mathrm{P}_{2} \mathrm{O}_{7}^{3-} (\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+} (\mathrm{aq}) + \mathrm{P}_{2} \mathrm{O}_{7}^{4-} (\mathrm{aq}) \) (HP鈧侽鈧嚶斥伝 and P鈧侽鈧団伌鈦�)

Step by step solution

01

First dissociation

In the first step, pyrophosphoric acid (H鈧凱鈧侽鈧�) releases one hydrogen ion (H鈦�) and becomes an anion with the formula H鈧働鈧侽鈧団伝. 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: H鈧凱鈧侽鈧� (acid) and H鈧働鈧侽鈧団伝 (base)
02

Second dissociation

In the second step, H鈧働鈧侽鈧団伝 releases one hydrogen ion (H鈦�) and becomes an anion with the formula H鈧侾鈧侽鈧嚶测伝. 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: H鈧働鈧侽鈧団伝 (acid) and H鈧侾鈧侽鈧嚶测伝 (base)
03

Third dissociation

In the third step, H鈧侾鈧侽鈧嚶测伝 releases one hydrogen ion (H鈦�) and becomes an anion with the formula HP鈧侽鈧嚶斥伝. 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: H鈧侾鈧侽鈧嚶测伝 (acid) and HP鈧侽鈧嚶斥伝 (base)
04

Fourth dissociation

In the fourth step, HP鈧侽鈧嚶斥伝 releases one hydrogen ion (H鈦�) and becomes an anion with the formula P鈧侽鈧団伌鈦�. 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: HP鈧侽鈧嚶斥伝 (acid) and P鈧侽鈧団伌鈦� (base) In conclusion, the stepwise dissociation of pyrophosphoric acid (H鈧凱鈧侽鈧�) involves four reactions, and there are four conjugate acid-base pairs, as shown above.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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