Balanced Equation For Sodium Hydroxide And Sulphuric Acid

The Balanced Equation for Sodium Hydroxide and Sulphuric Acid: A Deep Dive

The reaction between sodium hydroxide (NaOH), a strong base, and sulfuric acid (H₂SO₄), a strong diprotic acid, is a classic example of a neutralization reaction. Understanding this reaction, including its balanced equation and the stoichiometry involved, is fundamental in chemistry, particularly in acid-base titrations and industrial applications. This article will delve into the intricacies of this reaction, exploring its balanced equation, the stepwise process of its formation, its applications, and safety precautions.

Understanding the Reactants: Sodium Hydroxide and Sulphuric Acid

Before diving into the reaction itself, let's briefly review the properties of the individual reactants.

Sodium Hydroxide (NaOH)

Sodium hydroxide, commonly known as caustic soda or lye, is a strong, highly corrosive alkali. Its properties include:

Strong base: It completely dissociates in water to release hydroxide ions (OH⁻), significantly increasing the solution's pH.
Highly soluble: It readily dissolves in water, forming a strongly alkaline solution.
Versatile applications: It's used extensively in various industries, including chemical manufacturing, soap production, papermaking, and drain cleaning.

Sulphuric Acid (H₂SO₄)

Sulfuric acid, also known as vitriol, is a strong, highly corrosive diprotic acid. Key features include:

Strong acid: It readily donates two protons (H⁺) per molecule when dissolved in water.
Diprotic nature: This means it can undergo two separate neutralization reactions, one for each proton.
Dehydrating agent: It has a strong affinity for water and can act as a powerful dehydrating agent.
Industrial significance: It's one of the most important industrial chemicals, used extensively in fertilizer production, petroleum refining, and metal processing.

The Balanced Equation: A Step-by-Step Approach

The reaction between sodium hydroxide and sulfuric acid is a neutralization reaction, producing sodium sulfate and water. However, the balanced equation depends on the stoichiometry, meaning the relative amounts of reactants. Because sulfuric acid is diprotic, it can react in two steps with sodium hydroxide.

Step 1: Neutralization of the First Proton

In the first step, one mole of sulfuric acid reacts with one mole of sodium hydroxide to form one mole of sodium bisulfate (also known as sodium hydrogen sulfate) and one mole of water.

H₂SO₄ + NaOH → NaHSO₄ + H₂O

This equation is already balanced: one sulfur atom, two hydrogen atoms, one sodium atom, and four oxygen atoms on each side. Notice that the sulfuric acid still has one proton remaining.

Step 2: Neutralization of the Second Proton

The second step involves the neutralization of the remaining proton in sodium bisulfate. Another mole of sodium hydroxide reacts with the sodium bisulfate to produce sodium sulfate and another mole of water.

NaHSO₄ + NaOH → Na₂SO₄ + H₂O

This equation is also balanced. The final product is sodium sulfate, a neutral salt.

The Overall Balanced Equation

Combining both steps gives the overall balanced equation for the complete neutralization of sulfuric acid with sodium hydroxide:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

This shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate and two moles of water. This is crucial for stoichiometric calculations and titrations.

Stoichiometry and Calculations

The balanced equation is fundamental for performing stoichiometric calculations. For instance, if we know the amount of sulfuric acid used, we can calculate the required amount of sodium hydroxide for complete neutralization, and vice versa. Let's illustrate this with an example:

Example: If 50.0 mL of 0.100 M sulfuric acid is used, how many mL of 0.200 M sodium hydroxide are needed for complete neutralization?

Solution:

Moles of H₂SO₄: Moles = Molarity × Volume (in Liters) = 0.100 mol/L × 0.0500 L = 0.00500 mol
Moles of NaOH: From the balanced equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, 0.00500 mol H₂SO₄ requires 2 × 0.00500 mol = 0.0100 mol NaOH.
Volume of NaOH: Volume = Moles / Molarity = 0.0100 mol / 0.200 mol/L = 0.0500 L = 50.0 mL

Therefore, 50.0 mL of 0.200 M sodium hydroxide is needed to neutralize 50.0 mL of 0.100 M sulfuric acid.

Applications of the Reaction

This neutralization reaction has numerous applications across various fields:

Acid-base titrations: It's used extensively in analytical chemistry to determine the concentration of unknown acid or base solutions. The reaction's stoichiometry allows precise calculations of concentrations.
Industrial processes: Neutralization reactions are used in many industrial settings to control pH levels in wastewater treatment, chemical manufacturing, and other processes.
Chemical synthesis: The reaction can be used as a step in the synthesis of other chemicals, where a controlled pH is required.
Soap making: The saponification process involves the reaction of fats or oils with sodium hydroxide, which is a neutralization reaction.
Food processing: Neutralization may be used to adjust the pH of certain food products.

Safety Precautions

Both sodium hydroxide and sulfuric acid are highly corrosive and can cause severe burns. Appropriate safety measures must always be taken when handling these chemicals:

Wear protective gear: Always wear safety goggles, gloves, and a lab coat.
Work in a well-ventilated area: The reaction can produce heat and fumes.
Add acid to base slowly: Adding acid to base is generally safer than the reverse. This helps to control the heat generated during the exothermic reaction.
Proper disposal: Dispose of waste solutions according to local regulations.

Conclusion

The reaction between sodium hydroxide and sulfuric acid is a fundamental chemical reaction with significant theoretical and practical implications. Understanding its balanced equation, stoichiometry, applications, and safety precautions is essential for anyone working with these chemicals or studying acid-base chemistry. The detailed steps outlined in this article provide a comprehensive guide for mastering this crucial reaction. Remember always to prioritize safety when conducting experiments or working with these corrosive substances. Further research into specific applications and related chemical reactions can expand your knowledge and understanding of this important topic. This comprehensive understanding is key to success in chemistry and many related fields.