Problem 50 - Chemistry The Molecular Science

Question

Write balanced chemical equations that show phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), ionizing stepwise as a polyprotic acid.

Step-by-Step Solution

Answer

Phosphoric acid ionizes: \( \mathrm{H}_3\mathrm{PO}_4 \rightarrow \mathrm{H}^+ + \mathrm{H}_2\mathrm{PO}_4^- \); \( \mathrm{H}_2\mathrm{PO}_4^- \rightarrow \mathrm{H}^+ + \mathrm{HPO}_4^{2-} \); \( \mathrm{HPO}_4^{2-} \rightarrow \mathrm{H}^+ + \mathrm{PO}_4^{3-} \).

Step 1: Understand Phosphoric Acid Ionization

Phosphoric acid, \( \mathrm{H}_3\mathrm{PO}_4 \), is a triprotic acid, meaning it can donate three protons \( \mathrm{H}^+ \) ions stepwise. Each step represents the loss of a single proton.

Step 2: First Ionization Step

In the first ionization step, phosphoric acid donates one proton to form the dihydrogen phosphate ion. The chemical equation for this step is: \[ \mathrm{H}_3\mathrm{PO}_4 (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{H}_2\mathrm{PO}_4^- (aq) \] Here, \( \mathrm{H}_3\mathrm{PO}_4 \) loses one \( \mathrm{H}^+ \) to become \( \mathrm{H}_2\mathrm{PO}_4^- \).

Step 3: Second Ionization Step

In the second ionization step, the \( \mathrm{H}_2\mathrm{PO}_4^- \) ion loses another proton to form the hydrogen phosphate ion. The equation for this step is: \[ \mathrm{H}_2\mathrm{PO}_4^- (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{HPO}_4^{2-} (aq) \] Here, \( \mathrm{H}_2\mathrm{PO}_4^- \) donates one more \( \mathrm{H}^+ \) ion.

Step 4: Third Ionization Step

In the final ionization step, the hydrogen phosphate ion loses its last proton, forming the phosphate ion. The balanced equation is: \[ \mathrm{HPO}_4^{2-} (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{PO}_4^{3-} (aq) \] At this stage, \( \mathrm{HPO}_4^{2-} \) loses its last \( \mathrm{H}^+ \).

Key Concepts

Chemical Equations

Chemical equations are essential tools that chemists use to succinctly represent chemical reactions. They show the starting materials, called reactants, on the left side and the products on the right side of an equation. An arrow ( \( \rightarrow \) ) separates these two sides, signifying the direction in which the reaction proceeds.
When describing a reaction, it is crucial to balance the chemical equation. Balancing ensures that the number of atoms for each element is the same on both sides of the equation, reflecting the law of conservation of mass. This concept is the foundational principle that no atoms are lost or gained in a chemical reaction, only rearranged.

• Symbols and formulas represent substances.
• Coefficients show the number of molecules or moles.
• Subscripts denote the number of atoms in a molecule.

Understanding and writing correct chemical equations is vital in predicting the outcomes of reactions and ensuring complete comprehension of chemical processes.

Ionization Steps

Ionization refers to the process by which an atom or molecule acquires a negative or positive charge. This is typically achieved by gaining or losing electrons to form ions. In the context of acids, ionization involves the donation of hydrogen ions ( \( \mathrm{H}^+ \) ) to a solution.
For a polyprotic acid, which can donate more than one proton, ionization occurs in multiple steps. Each of these steps involves the release of a single proton.

To visualize this:

• Step 1: The acid releases its first proton, resulting in the formation of a more negatively charged ion.
• Step 2: The subsequent ion sheds another proton, further increasing its negative charge.
• Step 3: The final proton is lost, leaving behind the most negatively charged ion.

Each ionization step has its unique equilibrium constant, indicating how readily each proton is donated. Understanding these steps is crucial for comprehending the behavior of polyprotic acids in solutions.

Phosphoric Acid Ionization

Phosphoric acid ( \( \mathrm{H}_3\mathrm{PO}_4 \) ) serves as a classic example of a polyprotic acid, undergoing ionization in three distinct stages. These stages demonstrate its ability to release three separate protons sequentially.

First Ionization Step

In the first ionization step, phosphoric acid loses one hydrogen ion, forming the dihydrogen phosphate ion ( \( \mathrm{H}_2\mathrm{PO}_4^- \) ).
Chemical Equation:
\[ \mathrm{H}_3\mathrm{PO}_4 (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{H}_2\mathrm{PO}_4^- (aq) \]

Second Ionization Step

The dihydrogen phosphate ion then releases another hydrogen ion, becoming the hydrogen phosphate ion ( \( \mathrm{HPO}_4^{2-} \) ).
Chemical Equation:
\[ \mathrm{H}_2\mathrm{PO}_4^- (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{HPO}_4^{2-} (aq) \]

Third Ionization Step

Finally, the hydrogen phosphate ion loses its last hydrogen ion, resulting in the formation of the phosphate ion ( \( \mathrm{PO}_4^{3-} \) ).
Chemical Equation:
\[ \mathrm{HPO}_4^{2-} (aq) \rightarrow \mathrm{H}^+ (aq) + \mathrm{PO}_4^{3-} (aq) \]
Understanding each of these steps is crucial to grasp how phosphoric acid behaves in different chemical environments. Knowing how and when these protons dissociate is key to predicting the acid's reactivity and interactions.