Mixing Alcohol Solutions A Step By Step Guide

by Admin 46 views

Introduction

In the realm of chemistry and everyday problem-solving, mixing solutions of different concentrations is a common task. Whether you're a student in a chemistry lab or someone trying to prepare a specific cleaning solution at home, understanding how to calculate the required amounts of each solution is crucial. This article delves into the process of mixing alcohol solutions, specifically addressing the scenario where Bruce needs to create 50 ml of a 12% alcohol solution using 10% and 15% alcohol solutions. We will break down the equation provided, 0.10x + 0.15(50 - x) = 0.12(50), and explore the underlying principles and steps involved in solving it. By the end of this guide, you'll have a clear understanding of how to approach similar mixture problems and confidently calculate the necessary volumes for achieving your desired concentration.

When dealing with alcohol solutions, understanding concentrations and how they combine is paramount. Bruce's challenge is a classic example of a mixture problem, where different solutions with varying concentrations are combined to achieve a specific target concentration. The key to solving such problems lies in setting up the correct equation that represents the total amount of solute (in this case, alcohol) in the final mixture. The equation 0.10x + 0.15(50 - x) = 0.12(50) is the mathematical representation of this principle. Here, 'x' represents the volume (in ml) of the 10% alcohol solution needed. The term '50 - x' then represents the volume of the 15% alcohol solution, as the total volume of the mixture is 50 ml. The coefficients 0.10, 0.15, and 0.12 represent the concentrations of the respective solutions, expressed as decimal fractions. The right side of the equation, 0.12(50), calculates the total amount of alcohol in the desired 50 ml of 12% solution. By understanding each component of this equation, we can effectively solve for 'x' and determine the exact volumes of the 10% and 15% solutions required to achieve Bruce's goal.

The ability to accurately mix solutions is not just a theoretical exercise; it has practical applications across various fields. In chemistry, precise concentrations are essential for conducting experiments and reactions. In medicine, compounding medications often involves diluting or mixing solutions to achieve the correct dosage. Even in everyday life, understanding how to mix solutions can be useful for tasks such as preparing cleaning solutions, adjusting the strength of a disinfectant, or even making homemade cosmetics. The underlying principle remains the same: the total amount of solute in the final mixture must equal the sum of the amounts of solute in the individual solutions. By mastering the techniques for solving mixture problems, you gain a valuable skill that can be applied in a wide range of contexts. In the following sections, we will delve deeper into the steps involved in solving the equation and explore the practical implications of the solution. We will also discuss alternative methods for solving mixture problems and provide additional examples to solidify your understanding.

Breaking Down the Equation: 0.10x + 0.15(50 - x) = 0.12(50)

To effectively solve the equation 0.10x + 0.15(50 - x) = 0.12(50), it's essential to understand each term and its significance in the context of the problem. Let's break down the equation step by step:

  • 0. 10x: This term represents the amount of alcohol (solute) in the 10% alcohol solution. Here, 'x' is the unknown variable, representing the volume (in ml) of the 10% solution that Bruce needs to use. The coefficient 0.10 represents the concentration of the solution, meaning that 10% of the volume 'x' is pure alcohol. Multiplying the concentration (0.10) by the volume (x) gives us the total amount of alcohol contributed by this solution to the final mixture.
  • 0. 15(50 - x): This term represents the amount of alcohol in the 15% alcohol solution. The expression '(50 - x)' represents the volume (in ml) of the 15% solution. This is because Bruce wants to make a total of 50 ml of the final solution, and 'x' ml is coming from the 10% solution. Therefore, the remaining volume must come from the 15% solution. The coefficient 0.15 represents the concentration of this solution, meaning that 15% of the volume '(50 - x)' is pure alcohol. Multiplying the concentration (0.15) by the volume (50 - x) gives us the total amount of alcohol contributed by this solution.
  • 0. 12(50): This term represents the total amount of alcohol in the final 50 ml solution with a desired concentration of 12%. The coefficient 0.12 represents the desired concentration of the final solution, meaning that 12% of the total volume (50 ml) should be pure alcohol. Multiplying the concentration (0.12) by the total volume (50 ml) gives us the total amount of alcohol required in the final mixture.

By understanding the significance of each term, we can see how the equation represents the conservation of alcohol in the mixture. The left side of the equation (0.10x + 0.15(50 - x)) represents the total amount of alcohol contributed by the two solutions being mixed, while the right side of the equation (0.12(50)) represents the total amount of alcohol in the desired final solution. The equation states that these two quantities must be equal, which is the fundamental principle behind solving mixture problems. In the next section, we will delve into the algebraic steps required to solve this equation for 'x', thereby determining the required volumes of the 10% and 15% solutions.

In addition to understanding the individual terms, it's crucial to grasp the overall structure of the equation. The equation is a linear equation in one variable (x), which means it can be solved using standard algebraic techniques. The goal is to isolate 'x' on one side of the equation, thereby determining its value. This involves simplifying the equation by distributing coefficients, combining like terms, and performing operations on both sides of the equation to maintain equality. The process of solving the equation not only provides the numerical answer but also reinforces the understanding of the relationships between the different components of the mixture. By mastering this process, you can confidently tackle similar mixture problems and adapt the equation to different scenarios and concentrations. In the following sections, we will demonstrate the step-by-step solution of the equation and discuss the practical implications of the result.

Solving for 'x': The Algebraic Steps

Now that we understand the equation 0.10x + 0.15(50 - x) = 0.12(50), let's proceed with solving it to find the value of 'x', which represents the volume of the 10% alcohol solution needed. The following steps outline the algebraic process:

  1. Distribute the 0.15: The first step is to distribute the 0.15 across the terms inside the parentheses. This gives us:
    • 0.10x + 0.15 * 50 - 0.15x = 0.12(50)
      1. 10x + 7.5 - 0.15x = 0.12(50)
  2. Calculate 0.12(50): Next, we calculate the value of 0.12 multiplied by 50:
      1. 10x + 7.5 - 0.15x = 6
  3. Combine like terms: Now, we combine the terms containing 'x' on the left side of the equation:
    • (0.10x - 0.15x) + 7.5 = 6
    • -0.05x + 7.5 = 6
  4. Isolate the 'x' term: To isolate the term with 'x', we subtract 7.5 from both sides of the equation:
    • -0.05x + 7.5 - 7.5 = 6 - 7.5
    • -0.05x = -1.5
  5. Solve for 'x': Finally, we solve for 'x' by dividing both sides of the equation by -0.05:
    • -0.05x / -0.05 = -1.5 / -0.05
    • x = 30

Therefore, the value of 'x' is 30, which means Bruce needs 30 ml of the 10% alcohol solution.

By following these algebraic steps, we have successfully solved the equation and determined the required volume of the 10% solution. However, the solution is not complete until we also determine the volume of the 15% solution. Remember that the total volume of the mixture is 50 ml, and 'x' represents the volume of the 10% solution. Therefore, the volume of the 15% solution can be calculated by subtracting 'x' from 50. In the next section, we will calculate the volume of the 15% solution and verify our solution to ensure that the final mixture has the desired concentration of 12%. We will also discuss the practical implications of these results and how they can be applied in real-world scenarios.

Calculating the Volume of 15% Solution and Verification

Now that we have determined that Bruce needs 30 ml of the 10% alcohol solution (x = 30), we can calculate the volume of the 15% alcohol solution required to make the final 50 ml mixture. As we established earlier, the volume of the 15% solution is represented by (50 - x). Substituting the value of x, we get:

  • Volume of 15% solution = 50 - x
  • Volume of 15% solution = 50 - 30
  • Volume of 15% solution = 20 ml

Therefore, Bruce needs 20 ml of the 15% alcohol solution.

To ensure the accuracy of our solution, we need to verify that the mixture of 30 ml of 10% solution and 20 ml of 15% solution indeed results in a 50 ml solution with a 12% concentration. We can do this by calculating the total amount of alcohol in the mixture and comparing it to the amount of alcohol in a 50 ml solution with a 12% concentration.

  • Amount of alcohol in 30 ml of 10% solution: 0.10 * 30 ml = 3 ml
  • Amount of alcohol in 20 ml of 15% solution: 0.15 * 20 ml = 3 ml
  • Total amount of alcohol in the mixture: 3 ml + 3 ml = 6 ml

Now, let's calculate the amount of alcohol in a 50 ml solution with a 12% concentration:

  • Amount of alcohol in 50 ml of 12% solution: 0.12 * 50 ml = 6 ml

Since the total amount of alcohol in the mixture (6 ml) matches the amount of alcohol in the desired 50 ml solution with a 12% concentration (6 ml), our solution is verified. This confirms that Bruce needs to mix 30 ml of the 10% alcohol solution with 20 ml of the 15% alcohol solution to obtain 50 ml of a 12% alcohol solution.

This verification step is crucial in problem-solving, especially in situations where accuracy is paramount, such as in chemistry experiments or medical preparations. By verifying our solution, we ensure that we have not made any calculation errors and that the final mixture will have the intended properties. In the next section, we will discuss alternative methods for solving mixture problems and explore scenarios where these techniques might be particularly useful. We will also provide additional examples to further solidify your understanding of mixing solutions.

Alternative Methods for Solving Mixture Problems

While we have successfully solved Bruce's alcohol solution problem using an algebraic equation, there are alternative methods that can be employed to tackle similar mixture problems. These methods can offer different perspectives and may be more intuitive or efficient in certain situations. Let's explore some of these alternative approaches:

  1. The Alligation Method: This method is particularly useful when dealing with two solutions of different concentrations and aiming for a specific target concentration. The alligation method provides a visual and intuitive way to determine the ratio in which the two solutions should be mixed. It involves setting up a grid with the concentrations of the two solutions on the left, the desired concentration in the center, and then calculating the differences diagonally. The differences represent the relative amounts of each solution needed.

    • Example: In Bruce's case, we would set up the grid as follows:

      10%      
   12%   
15%

    • Then, we calculate the differences diagonally:

      10%      | 15 - 12 = 3
   12%   
15%      | 12 - 10 = 2

    • The differences, 3 and 2, represent the ratio in which the 10% and 15% solutions should be mixed. This means for every 3 parts of the 15% solution, we need 2 parts of the 10% solution. To obtain a total of 50 ml, we can divide 50 ml into 5 parts (3 + 2 = 5), with each part being 10 ml. Therefore, we need 2 parts * 10 ml = 20 ml of the 15% solution and 3 parts * 10 ml = 30 ml of the 10% solution, which matches our previous result.

  2. System of Equations: Another approach is to set up a system of two equations with two variables. Let 'x' be the volume of the 10% solution and 'y' be the volume of the 15% solution. We can set up the following equations:

    • Equation 1 (Total volume): x + y = 50

    • Equation 2 (Total alcohol): 0.10x + 0.15y = 0.12(50)

    • We can solve this system of equations using substitution or elimination methods. For example, we can solve Equation 1 for 'y' (y = 50 - x) and substitute it into Equation 2, which will result in the same equation we solved earlier: 0.10x + 0.15(50 - x) = 0.12(50).

By exploring these alternative methods, you gain a broader understanding of how to approach mixture problems and can choose the method that best suits your problem-solving style and the specific characteristics of the problem. The alligation method is particularly useful for quick mental calculations, while the system of equations approach is more versatile and can be applied to more complex scenarios with multiple solutions and concentrations. In the next section, we will provide additional examples of mixture problems and demonstrate how these methods can be applied in different contexts.

Additional Examples and Practice Problems

To further solidify your understanding of mixing solutions, let's explore some additional examples and practice problems. These examples will demonstrate how the principles and methods we have discussed can be applied in various scenarios:

Example 1: Mixing Acid Solutions

A chemist needs to create 100 ml of a 25% acid solution. They have a 10% acid solution and a 40% acid solution available. How much of each solution should the chemist mix?

  • Solution: We can use the algebraic equation method or the alligation method. Let's use the algebraic equation method:

    • Let 'x' be the volume of the 10% solution.
    • The volume of the 40% solution will be (100 - x).
    • The equation is: 0.10x + 0.40(100 - x) = 0.25(100)
    • Simplifying the equation: 0.10x + 40 - 0.40x = 25
    • Combining like terms: -0.30x = -15
    • Solving for 'x': x = 50 ml
    • Volume of 40% solution: 100 - 50 = 50 ml
    • Therefore, the chemist needs to mix 50 ml of the 10% solution with 50 ml of the 40% solution.

Example 2: Mixing Salt Solutions

A marine biologist needs to create 5 liters of a 3% salt solution for an experiment. They have a 1% salt solution and a 5% salt solution available. How much of each solution should the biologist mix?

  • Solution: Let's use the alligation method:

    1%      | 5 - 3 = 2
   3%   
5%      | 3 - 1 = 2
    • The ratio is 2:2, which simplifies to 1:1.
    • This means the biologist needs to mix equal parts of the 1% and 5% solutions.
    • Since the total volume is 5 liters, they need 5 liters / 2 = 2.5 liters of each solution.

By working through these examples, you can see how the different methods can be applied in various contexts. The key is to understand the underlying principles of mixture problems and to choose the method that best suits the specific problem at hand. Now, let's try some practice problems to test your understanding:

Practice Problems:

  1. How many milliliters of a 20% alcohol solution must be mixed with 40 ml of a 50% alcohol solution to produce a 30% alcohol solution?
  2. A pharmacist needs to prepare 200 ml of a 15% saline solution. They have a 10% saline solution and a 25% saline solution available. How much of each solution should they mix?
  3. A gardener wants to create 10 liters of a fertilizer solution with a concentration of 8%. They have a 5% fertilizer solution and a 10% fertilizer solution. How much of each solution should they mix?

By attempting these practice problems, you can reinforce your understanding of mixing solutions and develop your problem-solving skills. Remember to choose the method that you find most comfortable and to verify your answers to ensure accuracy. In the next section, we will summarize the key concepts and provide some final tips for tackling mixture problems.

Conclusion and Key Takeaways

In this comprehensive guide, we have explored the process of mixing solutions, focusing on the scenario where Bruce needed to create 50 ml of a 12% alcohol solution using 10% and 15% alcohol solutions. We have broken down the equation 0.10x + 0.15(50 - x) = 0.12(50), solved it algebraically, and verified our solution. We have also discussed alternative methods for solving mixture problems, such as the alligation method and the system of equations approach. By working through examples and practice problems, we have solidified our understanding of the key concepts and techniques involved in mixing solutions.

Key takeaways from this guide include:

  • Understanding Concentrations: The concentration of a solution represents the amount of solute (e.g., alcohol, acid, salt) present in a given volume of solution. It is typically expressed as a percentage.
  • Setting up the Equation: The fundamental principle behind solving mixture problems is that the total amount of solute in the final mixture must equal the sum of the amounts of solute in the individual solutions. This principle is used to set up the equation.
  • Algebraic Solution: The equation is typically a linear equation in one variable, which can be solved using standard algebraic techniques, such as distributing coefficients, combining like terms, and isolating the variable.
  • Verification: It is crucial to verify the solution by calculating the total amount of solute in the mixture and comparing it to the amount of solute in the desired final solution.
  • Alternative Methods: The alligation method and the system of equations approach provide alternative ways to solve mixture problems, which may be more intuitive or efficient in certain situations.

By mastering these key concepts and techniques, you will be well-equipped to tackle a wide range of mixture problems, whether in chemistry, medicine, or everyday life. Remember to practice regularly and to choose the method that best suits your problem-solving style and the specific characteristics of the problem.

In conclusion, mixing solutions is a fundamental skill that has practical applications across various fields. By understanding the underlying principles, setting up the correct equation, and applying appropriate problem-solving techniques, you can confidently calculate the required amounts of each solution to achieve your desired concentration. We hope this guide has provided you with a comprehensive understanding of mixing solutions and has empowered you to tackle similar problems with confidence.