Dilution Dilemma: Finding The Final Concentration

Hey there, chemistry enthusiasts! Today, we're diving into a common yet crucial concept: dilution. Let's tackle a classic problem: A chemistry teacher adds 50.0 mL of a 1.50 M $H_2SO_4$ solution to 200 mL of water. Our mission? To calculate the final concentration of the sulfuric acid ($H_2SO_4$) solution. Sounds interesting, right? Don't worry, we'll break it down step by step, making it super clear and easy to follow. Get ready to flex those chemistry muscles!

Understanding the Basics: Dilution and Molarity

Before we jump into the calculations, let's make sure we're all on the same page. What exactly does dilution mean in chemistry? Essentially, it's the process of reducing the concentration of a solute (in our case, sulfuric acid) in a solution by adding more solvent (water). Think of it like making a glass of juice. If you add more water to a concentrated juice, the juice becomes less strong, right? Dilution works the same way. The molarity (M) is the concentration of a solution, defined as the number of moles of solute per liter of solution. For instance, a 1.50 M $H_2SO_4$ solution means that there are 1.50 moles of sulfuric acid in every liter of solution. Molarity is a crucial concept in chemistry, allowing us to perform quantitative analyses and predict the outcomes of chemical reactions. We need to remember this because it's at the heart of our question. Now, let's explore this formula and its practical significance.

Now, how do we approach dilution problems? The key is to remember that the number of moles of solute remains constant during dilution. The solute is just spreading out into a larger volume of solvent. This crucial principle is what allows us to calculate the new concentration. The initial concentration multiplied by the initial volume equals the final concentration multiplied by the final volume. Hence, the formula used here is $M_1V_1 = M_2V_2$. So, if we know the initial concentration ($M_1$) and volume ($V_1$), along with the final volume ($V_2$), we can easily calculate the final concentration ($M_2$). Pretty straightforward, isn't it? Let’s put this knowledge to work. The formula is our trusty companion. We'll use it to unveil the mysteries of our concentration calculations. It's really the backbone of solving any dilution question. Now that we understand the core concepts, let's get into the specifics of solving the problem.

Why is Dilution Important?

You might be wondering, why is this important? Dilution is a fundamental technique in chemistry with a wide range of applications. In the lab, it's used to prepare solutions of desired concentrations from stock solutions. For instance, you might have a concentrated acid solution, but you need a much less concentrated solution for an experiment. Dilution allows you to achieve this easily and accurately. Beyond the lab, dilution plays a role in various industries, from pharmaceuticals to environmental science. In pharmaceuticals, it's essential for preparing dosages of medications. In environmental science, dilution is used to analyze the concentration of pollutants in water samples. Without this process, making precise measurements and understanding chemical reactions would be incredibly difficult. Remember this as we move forward. Now that we know why this topic is important, let's solve our problem!

Step-by-Step Solution to the Dilution Problem

Alright, let's roll up our sleeves and solve the problem. Here's a breakdown of the steps involved:

1. Identify the Given Information:

   • Initial molarity ($M_1$) of $H_2SO_4$ = 1.50 M
   • Initial volume ($V_1$) of $H_2SO_4$ = 50.0 mL
   • Final volume ($V_2$) of the solution = 200 mL (water) + 50 mL (solution) = 250 mL

   Make sure you know all of the starting parameters. You need to know what you are looking for and what you already have. This step is a must before you can start to solve any problem.

2. Choose the Right Formula:

   • As we discussed, the key equation for dilution is: $M_1V_1 = M_2V_2$

   This formula expresses the conservation of moles during dilution. It's the central idea behind dilution calculations.

3. Rearrange the Formula and Plug in the Values:

   • We want to find the final molarity ($M_2$), so rearrange the formula to solve for $M_2$: $M_2 = (M_1V_1) / V_2$
   • Now, plug in the known values: $M_2 = (1.50 M * 50.0 mL) / 250 mL$

   Remember, when performing the calculations, make sure your units are consistent. For example, your volumes should be in the same units, like mL or L. This ensures that the units cancel out properly, and your final answer is correct. Let's move on to the next step, where we'll calculate the final answer.

4. Calculate the Final Molarity:

   • Perform the calculation: $M_2 = (1.50 M * 50.0 mL) / 250 mL = 0.300 M$

   So, the final concentration of the $H_2SO_4$ solution is 0.300 M.

   This is the final step, and we have our answer. This value represents the new concentration of the sulfuric acid after it has been diluted. Does this value seem reasonable? Yes! The final concentration is lower than the initial concentration because we diluted the solution by adding water. This is why dilution is so important, because it allows us to adjust concentrations.

Choosing the Correct Answer and Why

Looking at the multiple-choice options, we can see that:

A. 0.300 M is the correct answer.

B. 0.375 M is incorrect. This value might be obtained if the student incorrectly calculates the final volume or makes a mistake in the formula.

C. 6.00 M is incorrect. This answer is far too high and indicates a significant error in calculations, such as misusing the formula or misinterpreting the initial conditions.

D. 7.50 M is incorrect. Like option C, this is an extremely high concentration that suggests major errors in the application of the dilution formula.

Therefore, the correct answer is indeed A. 0.300 M. Congratulations, you've successfully solved the dilution problem!

Quick Recap

To recap, in dilution problems, the key is to understand that the number of moles of solute remains constant. Use the formula $M_1V_1 = M_2V_2$ to find the final concentration. Make sure your units are consistent. Dilution is a super important concept in chemistry, allowing us to prepare solutions of desired concentrations. Understanding dilution is essential for a solid foundation in chemistry. The ability to work with dilutions opens the door to more advanced chemistry concepts. Keep practicing, and you'll master these types of problems in no time. Chemistry can be fun!

Tips for Success

Here are a few extra tips to help you ace dilution problems:

• Units: Always pay attention to the units. Make sure they are consistent throughout your calculations. Convert them if necessary (e.g., mL to L).
• Final Volume: Remember that the final volume is the total volume of the solution after the dilution, which includes the initial solution volume plus the volume of the added solvent.
• Practice: The more you practice, the better you'll become at solving dilution problems. Try different examples with varying initial concentrations and volumes.
• Review: Always review your work to make sure you haven't made any calculation mistakes.

Conclusion: You've Got This!

Congratulations on working through this dilution problem! Now you have the tools to tackle similar challenges. Keep exploring the fascinating world of chemistry, and remember that with practice and understanding, you can conquer any chemical concept. Chemistry is a lot like a puzzle, you just have to figure out the steps to solve it. Now that you have learned how to solve this problem, you can tackle other questions that may come your way. Best of luck with your chemistry endeavors, and keep up the great work. Happy experimenting!

For further exploration, you might find this link helpful:

• Khan Academy - Dilution