Problem 67 - Precalculus Mathematics for Calculus

Question

A chemist has two large containers of sulfuric acid solution, with different concentrations of acid in each container. Blending \(300 \mathrm{mL}\) of the first solution and \(600 \mathrm{mL}\) of the second gives a mixture that is \(15 \%\) acid, whereas blending \(100 \mathrm{mL}\) of the first with \(500 \mathrm{mL}\) of the second gives a \(12 \frac{1}{2} \%\) acid mixture. What are the concentrations of sulfuric acid in the original containers?

Step-by-Step Solution

Answer

The concentrations are 15% for the first solution and 10% for the second solution.

Step 1: Define Variables

Let's define the variables: Let \( x \) be the concentration of sulfuric acid in the first container and \( y \) be the concentration of sulfuric acid in the second container (expressed as decimals).

Step 2: Set Up the Equations

Based on the first mixture: \( 300x + 600y = 0.15 \times 900 \). Based on the second mixture: \( 100x + 500y = 0.125 \times 600 \).

Step 3: Simplify the Equations

Simplify the equations: \( 300x + 600y = 135 \) and \( 100x + 500y = 75 \).

Step 4: Solve the System of Equations

Using the first equation, we can express \( x \) in terms of \( y \): \( x = \frac{135 - 600y}{300} \). Substitute \( x \) in the second equation: \( 100\left(\frac{135 - 600y}{300}\right) + 500y = 75 \).

Step 5: Solve for \( y \)

Simplify and solve: \((135 - 600y) + 1500y = 225 \), thus \( 900y = 90 \), so \( y = 0.1 \).

Step 6: Solve for \( x \)

Substitute \( y = 0.1 \) back into the expression \( x = \frac{135 - 600 \times 0.1}{300} \). This gives \( x = 0.15 \).

Step 7: Verify the Solution

Verify by substituting \( x = 0.15 \) and \( y = 0.1 \) into the original equations to ensure both hold true.

Key Concepts

Concentration Problems

Concentration problems in chemistry often involve finding the amount of a certain substance within a solution. The goal is to determine the concentration percentage of a solute, in this case, sulfuric acid, in different solutions. The concentration is typically expressed as a percentage by volume or weight.
In problems like this one, we start by defining what each container's concentration is and then use mathematical equations to express these relationships. *It's important to remember:*

• The total volume of the solution is the sum of the individual volumes.
• The concentration is a ratio: amount of solute per total volume, expressed as a decimal or percentage.

You need to know the initial volumes and the concentration outcomes for different mixtures, as these will help set up the system of equations. With these strategies, you can effectively solve concentration problems.

Mixture Problems

Mixture problems like this one are quite common in algebra and chemistry, and they often require a systematic approach to find unknown values. When combining different solutions with known and unknown concentrations, we aim to find out the unknown concentrations in the original containers by creating equations from the given data.
Each mixture scenario described in the problem gives us an equation. For instance:

• The first mixture combines specific amounts of two solutions to achieve a certain concentration percentage.
• The second mixture does the same with different amounts, leading to another concentration outcome.

By carefully forming two equations from these two different scenarios, we can use algebraic methods to determine the unknown concentrations. This method shows how different proportions influence the final mix and helps hone problem-solving skills.

Solving Linear Equations

Solving linear equations is a fundamental part of algebra, and it becomes particularly useful in mixture and concentration problems. The process generally involves working with equations to find values that satisfy all scenarios presented in a problem.
For the given exercise:

• The first step is to write each scenario as a system of linear equations. These linear equations consist of variables representing the unknown concentrations.
• By substitution or elimination, you can solve these linear equations. This involves expressing one variable in terms of another and then substituting back, or eliminating a variable through manipulation.
• After solving, verify the values by substituting them back into the original equations to ensure they hold true.

Understanding how to systematically break down and solve these equations builds confidence and improves analytical skills. Once you master this process, you can tackle a wide range of real-world problems.