When it comes to chemistry, understanding how to neutralize acids and bases is essential. Today, we’re diving into a practical problem that many students and professionals encounter: figuring out how many milliliters of a sodium hydroxide (NaOH) solution we need to neutralize a given volume of hydrogen sulfide (H₂S).
Understanding the Neutralization Reaction
Neutralization reactions play a crucial role in various chemical processes. They involve the reaction between an acid and a base, resulting in the formation of water and a salt.
Importance of Neutralization
Neutralization helps maintain pH balance in different systems. It is essential in:
- Environmental Protection: Treating acidic waste prevents soil and water contamination.
- Industrial Processes: Controlling acidity in manufacturing ensures product quality.
- Biological Systems: Maintaining pH in our bodies enables optimal enzymatic reactions.
As stated by a leading chemistry textbook, “Neutralization is a fundamental process that affects multiple aspects of life, from agriculture to medicine.”
Overview of Acid-Base Reactions
Acid-base reactions involve protons (H⁺) and hydroxide ions (OH⁻). During a neutralization reaction, the following occurs:
[
\text{Acid} + \text{Base} \rightarrow \text{Water} + \text{Salt}
]
In our context, when hydrogen sulfide (H₂S) reacts with sodium hydroxide (NaOH), the reaction can be represented as:
[
\text{H}_2\text{S} + 2 \text{NaOH} \rightarrow \text{Na}_2\text{S} + 2 \text{H}_2\text{O}
]
Here’s a brief comparison of some common acids and bases used in neutralization reactions:
| Substance | Molarity (M) | Type |
|---|---|---|
| H₂S | 0.245 | Acid |
| NaOH | 0.610 | Base |
This table indicates the concentrations of the substances involved in the reaction. Each component significantly influences how much base is required to achieve neutralization.
Understanding these principles ensures accuracy in calculations, enabling us to determine the exact volume of sodium hydroxide needed to neutralize a given amount of acid effectively.
Key Concepts in the Problem
Understanding the key concepts enhances our ability to determine the necessary volume of sodium hydroxide (NaOH) needed for neutralization. Examining molarity and the specific substances involved provides a solid foundation for solving the problem.
Molarity and Milliliters Defined
Molarity (M) measures the concentration of a solution. It’s defined as the number of moles of solute per liter of solution, expressed in moles per liter (mol/L). Milliliters (mL) serve as a volume measure, and there’s a direct relationship between mL and molarity.
To understand how they interact, consider the following formula:
[
\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}}
]
To convert mL to L, remember:
[
1 \text{ L} = 1000 \text{ mL}
]
The relationship between volume and molarity is crucial when performing calculations. We can quantify amounts through the following table:
| Volume (mL) | Molarity (M) | Moles |
|---|---|---|
| 20.0 | 0.245 | 0.0049 |
| X | 0.610 | Calculated |
To find moles from volume, we use:
[
\text{moles} = \text{Molarity} \times \text{Volume (L)}
]
The Role of Sodium Hydroxide (NaOH) and Hydrogen Sulfide (H₂S)
Sodium hydroxide acts as a strong base, while hydrogen sulfide functions as a weak acid. When mixed, these substances react in a neutralization process, generating water and sodium sulfide as products.
The neutralization equation is given as follows:
[
\text{NaOH} + \text{H₂S} \rightarrow \text{Na₂S} + \text{H₂O}
]
This reaction demonstrates the interaction of hydroxide ions (OH⁻) and protons (H⁺) to produce water. Here’s a more detailed breakdown:
- Sodium Hydroxide (NaOH): Strong base, fully dissociates in solution.
- Hydrogen Sulfide (H₂S): Weak acid, partially dissociates in the solution.
Both components contribute significantly to the neutralization process. The calculations involve determining the moles of H₂S present in the solution and matching it with the equivalents of NaOH needed for neutralization. For each mole of H₂S, one mole of NaOH is necessary.
To support our calculations, we can refer to a quote from a renowned chemistry textbook:
“Neutralization is a fundamental reaction in acid-base chemistry that maintains the balance of pH in various scenarios.”
As a reference, here is a comparison table summarizing the key parameters:
| Substance | Molarity (M) | Volume (mL) | Moles |
|---|---|---|---|
| Sodium Hydroxide | 0.610 | X | To find |
| Hydrogen Sulfide | 0.245 | 20.0 | 0.0049 |
Calculating the volume of NaOH for neutralization requires us to balance the moles of H₂S with the corresponding moles of NaOH. This understanding fortifies the framework for our calculations ahead.
Calculation Methodology
Understanding the calculation methodology for neutralization helps streamline our approach to determining the necessary volume of sodium hydroxide (NaOH) solution to neutralize hydrogen sulfide (H₂S). Here, we break down the steps involved in calculating the required milliliters of NaOH.
Step-by-Step Calculation Process
- Identify the molarity and volume of H₂S
We have a 0.245 M solution of H₂S and a volume of 20.0 mL.
- Calculate the moles of H₂S present
The formula to find the moles is:
[
\text{Moles} = \text{Molarity} \times \text{Volume (L)}
]
Thus, for H₂S:
[
\text{Moles H₂S} = 0.245 , \text{mol/L} \times 20.0 , \text{mL} \times \frac{1 , \text{L}}{1000 , \text{mL}} = 0.0049 , \text{mol}
]
- Establish the neutralization equation
The reaction is:
[
\text{H₂S} + \text{NaOH} \rightarrow \text{NaHS} + \text{H₂O}
]
This shows a 1:1 ratio, meaning one mole of NaOH neutralizes one mole of H₂S.
- Determine moles of NaOH required
Since it takes 1 mole of NaOH to neutralize 1 mole of H₂S, we need 0.0049 mol of NaOH.
- Calculate required volume of NaOH solution
Using the molarity of the NaOH solution (0.610 M) and our previously calculated moles:
[
\text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.0049 , \text{mol}}{0.610 , \text{mol/L}} = 0.00803 , \text{L} = 8.03 , \text{mL}
]
Formula for Neutralization
The overall formula we rely on for neutralization calculations merges concepts of molarity, volume, and moles:
[
\text{Volume of NaOH (L)} = \frac{\text{Moles of H₂S}}{\text{Molarity of NaOH}}
]
To summarize our findings, here’s a table of the key values we used in our calculations:
| Quantity | Value |
|---|---|
| Molarity of H₂S | 0.245 M |
| Volume of H₂S | 20.0 mL |
| Moles of H₂S | 0.0049 mol |
| Molarity of NaOH | 0.610 M |
| Moles of NaOH required | 0.0049 mol |
| Volume of NaOH required | 8.03 mL |
This structured approach ensures accuracy while simplifying our calculations for neutralization reactions. We continuously improve our understanding of these methodologies, vital for practical applications in chemistry.
Final Calculation
In this section, we outline the detailed calculation needed to determine how many milliliters of a 0.610 M NaOH solution neutralize 200 mL of a 0.245 M H₂S solution. This calculation relies on key principles of molarity and stoichiometry in acid-base reactions.
Detailed Calculation Steps
- Identify the Molarity and Volume of H₂S:
- Molarity (M) of H₂S = 0.245 M
- Volume (V) of H₂S = 200 mL = 0.200 L
- Calculate Moles of H₂S:
- Use the formula:
[
\text{Moles of H₂S} = \text{Molarity} \times \text{Volume}
]
- Calculation:
[
\text{Moles of H₂S} = 0.245 , \text{mol/L} \times 0.200 , \text{L} = 0.049 , \text{mol}
]
- Determine Moles of NaOH Required:
- The neutralization reaction is:
[
\text{H₂S} + \text{NaOH} \rightarrow \text{NaHS} + \text{H₂O}
]
- According to the equation, the stoichiometry shows a 1:1 ratio:
[
\text{Moles of NaOH required} = \text{Moles of H₂S} = 0.049 , \text{mol}
]
- Calculate Required Volume of NaOH Solution:
- Using the formula:
[
\text{Volume of NaOH} = \frac{\text{Moles of NaOH}}{\text{Molarity of NaOH}}
]
- Calculation:
[
\text{Volume of NaOH} = \frac{0.049 , \text{mol}}{0.610 , \text{mol/L}} \approx 0.080 , \text{L} = 80 , \text{mL}
]
Result Interpretation
The calculation reveals that 80 mL of a 0.610 M NaOH solution effectively neutralizes 200 mL of a 0.245 M H₂S solution. This finding emphasizes the practical methodology used in lab settings, ensuring exact measurements for acid-base reactions.
| Component | Value |
|---|---|
| Molarity of H₂S | 0.245 M |
| Volume of H₂S | 200 mL (0.200 L) |
| Moles of H₂S | 0.049 mol |
| Molarity of NaOH | 0.610 M |
| Required NaOH Volume | 80 mL |
Conclusion
Understanding how to neutralize acids and bases is crucial for anyone working in chemistry. We’ve walked through the steps to determine how much sodium hydroxide is needed to neutralize hydrogen sulfide. By applying the principles of molarity and the neutralization equation, we found that 80 mL of a 0.610 M NaOH solution does the trick for 200 mL of a 0.245 M H₂S solution.
This knowledge not only helps us in academic settings but also in real-world applications like environmental protection and industrial processes. With practice, these calculations become second nature, allowing us to tackle similar challenges with confidence. Let’s keep exploring the fascinating world of chemistry together!