Reaction between 0.1 M Sodium Carbonate and 3 M Sulfuric Acid
Introduction
In this article, we will explore the reaction between 0.1 M sodium carbonate (Na2CO3) and 3 M sulfuric acid (H2SO4). This reaction is an example of a neutralization reaction, where an acid reacts with a base to form salt and water.
The Reaction
The reaction between 0.1 M sodium carbonate and 3 M sulfuric acid can be represented by the following equation:
Na2CO3 (aq) + H2SO4 (aq) → Na2SO4 (aq) + H2CO3 (aq)
Ionization of the Reactants
Before we dive into the reaction, let's take a look at the ionization of the reactants:
- Sodium carbonate (Na2CO3) is a strong electrolyte, which means it completely dissociates in water to form sodium ions (Na+) and carbonate ions (CO32-):
Na2CO3 → 2Na+ + CO32-
- Sulfuric acid (H2SO4) is also a strong electrolyte, which means it completely dissociates in water to form hydrogen ions (H+) and sulfate ions (SO42-):
H2SO4 → 2H+ + SO42-
The Reaction Mechanism
Now, let's examine the reaction mechanism:
- The sodium ions (Na+) from the sodium carbonate react with the sulfate ions (SO42-) from the sulfuric acid to form sodium sulfate (Na2SO4):
2Na+ + SO42- → Na2SO4
- The hydrogen ions (H+) from the sulfuric acid react with the carbonate ions (CO32-) from the sodium carbonate to form carbonic acid (H2CO3):
H+ + CO32- → H2CO3
Products and Stoichiometry
The products of the reaction are sodium sulfate (Na2SO4) and carbonic acid (H2CO3). The stoichiometry of the reaction is 1:1, meaning one mole of sodium carbonate reacts with one mole of sulfuric acid to form one mole of sodium sulfate and one mole of carbonic acid.
Conclusion
In conclusion, the reaction between 0.1 M sodium carbonate and 3 M sulfuric acid is a neutralization reaction that forms salt and water. The products of the reaction are sodium sulfate and carbonic acid, with a stoichiometry of 1:1. This reaction is an important example of acid-base chemistry and has many applications in various industries.
