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Give an example of a polyprotic acid and write equations for the successive neutralizations of the acidic hydrogen atoms of the acid molecule to produce a series of salts.

Short Answer

Expert verified
Sulfuric acid (H鈧係O鈧) undergoes two stages of neutralization, first forming NaHSO鈧, then Na鈧係O鈧.

Step by step solution

01

Choose a Polyprotic Acid

A common example of a polyprotic acid is sulfuric acid (H鈧係O鈧). Sulfuric acid is a diprotic acid, meaning it has two hydrogen atoms that can be ionized.
02

First Stage of Neutralization

The first stage of neutralization involves the removal of one hydrogen ion from H鈧係O鈧 to form the bisulfate ion (HSO鈧勨伝) and water. \[ \text{H}_2\text{SO}_4 + \text{OH}^- \rightarrow \text{HSO}_4^- + \text{H}_2\text{O} \] This reaction produces the salt sodium bisulfate (if neutralized with NaOH): \[ \text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O} \]
03

Second Stage of Neutralization

The second stage of neutralization involves the removal of the second hydrogen ion from the bisulfate ion (HSO鈧勨伝) to form sulfate ion (SO鈧劼测伝) and water. \[ \text{HSO}_4^- + \text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O} \] This continues the neutralization process, producing sodium sulfate (if neutralized with NaOH): \[ \text{NaHSO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \]

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Key Concepts

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

Neutralization Reactions
Neutralization reactions are chemical processes in which an acid and a base interact to form a salt and water. This process effectively neutralizes the acidity of the acid. In many cases, this type of reaction can be expressed by a straightforward equation:
- \ ext{Acid} + \ ext{Base} \rightarrow \ ext{Salt} + \ ext{Water}When sulfuric acid \((H_2SO_4)\) is involved, it undergoes neutralization in two stages. In the first stage, a hydrogen ion \((H^+)\) is removed from the acid, and a salt with a bisulfate ion \((HSO_4^-)\) is produced. In the second stage, the remaining hydrogen ion is removed from the bisulfate ion to form a sulfate ion \((SO_4^{2-})\), completing the neutralization. Each stage results in the formation of water as one of the products.
Understanding neutralization is crucial as it plays an important role in many chemical processes, including those occurring in everyday life such as antacid use and pH balance regulation in nature.
Diprotic Acid
Diprotic acids are a type of polyprotic acid which can donate two hydrogen ions per molecule during reactions. Sulfuric acid \((H_2SO_4)\) is a well-known diprotic acid. Its ability to release two hydrogen ions means that it can participate in two separate neutralization reactions.
- **First Ionization**: When \(H_2SO_4\) releases its first hydrogen ion, it forms the bisulfate ion \((HSO_4^-)\). - **Second Ionization**: In the second stage, the bisulfate ion loses another hydrogen ion, forming sulfate \((SO_4^{2-})\).
These two stages are evident when diprotic acids undergo neutralization. The complete ionization of diprotic acids often requires strong bases to ensure total hydrogen ion removal. Diprotic acids are not only important in neutralization reactions but also play a significant role in industrial processes and environmental systems.
Sulfate Ion
The sulfate ion \((SO_4^{2-})\) is a polyatomic ion commonly known in chemistry and is the product of the complete neutralization of diprotic acids like sulfuric acid. In neutralization reactions involving diprotic acids, the sulfate ion forms after both hydrogen ions have been removed.
- The sulfate ion is formed when the bisulfate ion \((HSO_4^-)\) loses its remaining hydrogen ion.- Sulfate ions readily combine with various metal ions to form sulfate salts. These include well-known compounds such as sodium sulfate \((Na_2SO_4)\), calcium sulfate (gypsum), and magnesium sulfate (Epsom salts).
Sulfate ions are widely present in the environment, particularly in minerals and soil. Understanding the formation and properties of sulfate ions is useful for understanding both natural processes and industrial applications.

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Most popular questions from this chapter

A solution contains \(6.00 \%\) (by mass) \(\mathrm{NaBr}\) (sodium bromide). The density of the solution is \(1.046 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the molarity of \(\mathrm{NaBr}\) ?

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