Problem 53 - Precalculus: Mathematical 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 \(100 \mathrm{mL}\) of the first mixed 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 of sulfuric acid are 250% and 100% in the respective containers.

Step 1: Set Up Variables

Let the concentration of sulfuric acid in the first container be \( x \) and in the second container be \( y \). These concentrations will be in percentage form.

Step 2: Form the First Equation

Using the first scenario where 300 mL of the first solution and 600 mL of the second solution give a 15% mixture:\[0.3x + 0.6y = 0.15(300 + 600) = 135\]

Step 3: Form the Second Equation

Using the second scenario where 100 mL of the first solution and 500 mL of the second solution give a 12.5% mixture:\[0.1x + 0.5y = 0.125(100 + 500) = 75\]

Step 4: Solve the System of Equations

Now, solve the system of linear equations:1. \(0.3x + 0.6y = 135\)2. \(0.1x + 0.5y = 75\)Multiply the second equation by 3 to align the coefficients of \(x\):\[0.3x + 1.5y = 225\]Subtract the first equation from this modified equation:\[ (0.3x + 1.5y) - (0.3x + 0.6y) = 225 - 135 \]\[ 0.9y = 90 \]\[ y = 100 \]

Step 5: Find the First Variable

Substitute \( y = 100 \) back into the first equation:\[0.3x + 0.6(100) = 135\]\[0.3x + 60 = 135\]\[0.3x = 75 \]\[x = 250\]

Step 6: Conclusion

The concentration of sulfuric acid in the first container is 250%, and in the second container, it is 100%.

Key Concepts

Concentration Calculation

Concentration calculation in chemistry is a fundamental concept, especially when dealing with mixtures of solutions. When we discuss concentration, we're referring to the amount of a substance (solute) present in a given volume of solution. In simpler terms, it tells us how much of something is dissolved in a certain amount of liquid.

In the context of the problem at hand, concentration is measured in percentage form. This means that the concentration value represents the parts of sulfuric acid per 100 parts of solution. For example, a 15% acid solution means 15 parts of acid and 85 parts of water per 100 parts of solution. Concentration calculations often involve the use of equations to determine how different mixtures affect overall concentrations.

To solve concentration problems, you need to keep track of the volume of each solution and their respective concentrations. Use these details to formulate mathematical equations that relate the different concentrations to the final desired concentration.

Mixture Problems

Mixture problems in chemistry involve combining two or more solutions to achieve a desired concentration. This exercise is a classic example, where we have two solutions with unknown concentrations, and by mixing them in different ratios, we obtain mixtures with known concentrations. The challenge is to find out what the original concentrations were.

Key elements in solving mixture problems are:

• Understanding the proportions of the mixtures: Knowing it involves adding specific volumes of each solution.
• The concentration of the resulting mixture: This helps in setting equations based on the final concentration.

Setting up your information clearly using variables makes it easier. Often, these problems are solved by translating the word-based problem into mathematical equations. The given scenario tells us the final concentration after mixing different volumes of solutions. Our job is to backtrack and figure out what initial concentrations led to this result.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving these equations simultaneously allows us to find the values of those variables. In our example, we use two equations to determine the concentrations of sulfuric acid in each container.

Here's how systems of equations work in this context:

• Assign variables to unknowns: Here, we have \( x \) and \( y \) for the concentrations of the solutions.
• Use the information provided: Formulate two equations using different scenarios given by the mixture problems.
• Solve the equations: This can be done using substitution, elimination, or other algebraic methods to find the values of \( x \) and \( y \).

In our solution, we used elimination where one equation was manipulated to allow subtraction, leading to the isolation and solution of one variable, and then substitution was used to find the other variable. Solving systems of equations is crucial for these kinds of mixture problems because it provides the means to untangle the values needed for precise concentration calculations.