Calculating Nitric Acid Volume For A Sulfuric Acid Solution

Hey guys! Let's dive into a chemistry problem that's all about calculating the volume of concentrated nitric acid needed to create a specific sulfuric acid solution. This is a common type of calculation in chemistry, and understanding it will give you a solid foundation for tackling more complex problems. We'll break down the steps, explain the concepts, and make sure you understand the 'why' behind each calculation.

Understanding the Problem: The Core Concepts

Okay, so the question is asking us to figure out how much of a highly concentrated nitric acid solution we need to make a specific sulfuric acid solution. Think of it like this: you have a super-strong acid (nitric acid) and you want to use it to create a weaker solution of a different acid (sulfuric acid). To do this, we need to know a few key things:

• Concentration: The amount of solute (the stuff being dissolved) in a solution. In our case, we're dealing with percentages, which show how much nitric acid is present in the solution by weight (58% w/w).
• Density: This tells us how heavy the solution is for a given volume. It's measured in grams per milliliter (g/mL), and in our case, it's 1.51 g/mL for the nitric acid solution.
• Molarity: This is the concentration of a solution expressed in moles per liter (mol/L or M). We want to create a sulfuric acid (H₂SO₄) solution with a molarity of 0.4 M.
• Volume: The amount of space the solution takes up. We want to make 150 mL of the final sulfuric acid solution.

So, what's the game plan? Well, we need to connect the information about the concentrated nitric acid to the final sulfuric acid solution. This involves some stoichiometric calculations, but don't worry, it's not as scary as it sounds! It's important to understand the basics of molarity which is used to calculate the number of moles of solute present in a solution, density which is used to find the mass of a substance from its volume, and percent concentration to find the mass of a solute in the solution. We'll convert the information we have (percentage concentration and density of nitric acid) into a usable form, then use this to find the volume of the nitric acid solution needed.

The Importance of Stoichiometry

Stoichiometry is super important in this calculation. This is basically the relationship between the amounts of reactants and products in a chemical reaction. Because the question is about how much concentrated nitric acid to use, you can calculate the number of moles of sulfuric acid that you want to create (using molarity and volume). Then, you will use the balanced chemical equation to figure out how many moles of nitric acid you need to get the desired amount of sulfuric acid. This is the cornerstone of any solution-preparation calculation.

Step-by-Step Calculation: Unveiling the Answer

Alright, let's get down to the nitty-gritty and work through the problem step by step. We'll break it down into manageable chunks to make it easier to follow.

Step 1: Calculate the Moles of H₂SO₄ Needed

First, we need to figure out how many moles of sulfuric acid (H₂SO₄) are in the final solution we want to make. We know the desired molarity (0.4 M) and volume (150 mL), so we can use the following formula:

Moles = Molarity × Volume (in Liters)

First, convert the volume from mL to L: 150 mL = 0.150 L.

Moles of H₂SO₄ = 0.4 mol/L × 0.150 L = 0.06 moles

So, we need 0.06 moles of H₂SO₄.

Step 2: Determine the Moles of HNO₃ Needed

Now, we need to figure out the connection between nitric acid (HNO₃) and the sulfuric acid (H₂SO₄) we want to make. The chemical reaction between them looks like this:

2 HNO₃ + H₂SO₃ -> H₂SO₄ + 2 NO₂ + 2 H₂O

From the balanced equation, we can see that 2 moles of HNO₃ are required to produce 1 mole of H₂SO₄.

Therefore, to produce 0.06 moles of H₂SO₄, we need:

Moles of HNO₃ = 0.06 moles H₂SO₄ × (2 moles HNO₃ / 1 mole H₂SO₄) = 0.12 moles HNO₃

Step 3: Calculate the Mass of HNO₃ Needed

Next, let's find the mass of HNO₃. We need to use the molar mass of HNO₃, which is approximately 63.01 g/mol.

Mass of HNO₃ = Moles of HNO₃ × Molar mass of HNO₃

Mass of HNO₃ = 0.12 moles × 63.01 g/mol = 7.56 g

So, we need 7.56 g of pure HNO₃.

Step 4: Account for the Concentration and Density

Here's where the 58% (w/w) concentration and density of the nitric acid solution come into play. The concentrated nitric acid is not pure HNO₃, it is only 58% HNO₃ by weight. So, 100 g of the solution contains 58 g of HNO₃.

First, we will find the mass of the solution needed to obtain 7.56 g of HNO₃:

Mass of solution = (Mass of HNO₃ / % HNO₃) × 100%

Mass of solution = (7.56 g / 58%) × 100%

Mass of solution = 13.03 g

Now we can use the density of the solution (1.51 g/mL) to find the volume needed:

Volume of solution = Mass of solution / Density

Volume of solution = 13.03 g / 1.51 g/mL = 8.63 mL

Step 5: Convert the answer

The question provides the answers in Liters. We need to convert the 8.63 mL to Liters:

8. 63 mL = 0.00863 L

Conclusion: The Final Answer

Based on these calculations, the volume of 58% (w/w) concentrated nitric acid with a density of 1.51 g/mL required to make 150 mL of 0.4 M H₂SO₄ solution is approximately 8.63 mL. The closest answer choice from the original question is B. 8.63 L, but we found the correct value of the answer is 8.63 mL. This slight discrepancy could be due to rounding errors, but in any case, the answer would need to be converted to mL, or the answers in the questions corrected.

Tips and Tricks: Mastering Acid-Base Calculations

• Always balance your equations! This is the foundation for getting the mole ratios correct.
• Pay attention to units. Make sure your units are consistent throughout the calculation. Convert everything to the same units (e.g., liters for volume) before you start.
• Practice, practice, practice! The more you work through these types of problems, the easier they'll become.
• Understand the concepts. Don't just memorize formulas; understand what each term means and how it relates to the overall problem.
• Know your acids and bases. Recognizing strong and weak acids and bases will help you predict the behavior of reactions.

This calculation, while involving multiple steps, is a classic example of how chemists prepare solutions. Remember, it is necessary to convert between concentration, density, and volume. Keep practicing, and you'll be acing these calculations in no time! Keep up the great work, everyone!