Problem 32 - Algebra for College Students

Question

Suppose that \(x\) ounces of pure acid have been added to 14 ounces of a \(15 \%\) acid solution. (a) Set up the rational expression that represents the concentration of pure acid in the final solution. (b) Graph the rational function that displays the concentration. (c) How many ounces of pure acid need to be added to the 14 ounces of a \(15 \%\) solution to raise it to a \(40.5 \%\) solution? Check your answer. (d) How many ounces of pure acid need to be added to the 14 ounces of a \(15 \%\) solution to raise it to a \(50 \%\) solution? Check your answer. (e) What concentration of acid do we obtain if we add 12 ounces of pure acid to the 14 ounces of a \(15 \%\) solution? Check your answer.

Step-by-Step Solution

Answer

(c) 6 ounces, (d) 9.8 ounces, (e) 54.23% concentration.

Step 1: Establish the Initial Setup

We start by identifying the variables and understanding the given solution. Initially, we have 14 ounces of a 15% acid solution, which means it contains \(0.15 \times 14 = 2.1\) ounces of pure acid. We add \(x\) ounces of pure acid to this solution.

Step 2: Create Rational Expression for Concentration

The total amount of pure acid in the new solution is \(x + 2.1\) ounces, and the total volume of the solution becomes \(x + 14\) ounces. Therefore, the concentration of acid in the final solution is given by the rational expression \(C(x) = \frac{x + 2.1}{x + 14}\).

Step 3: Graph the Rational Function

The graph of the function \(C(x) = \frac{x + 2.1}{x + 14}\) typically exhibits features of rational functions, such as vertical asymptotes at \(x = -14\) and a horizontal asymptote at \(y = 1\). The graph will increase as \(x\) increases.

Step 4: Solve for Part (c)

To find the amount of pure acid needed to reach a 40.5% concentration, set \(C(x) = 0.405\) and solve for \(x\):\[ \frac{x + 2.1}{x + 14} = 0.405 \]Cross-multiply to get:\[ x + 2.1 = 0.405(x + 14) \]Solve for \(x\): \[ x + 2.1 = 0.405x + 5.67 \]\[ 0.595x = 3.57 \]\[ x \approx 6 \] Thus, 6 ounces of pure acid should be added.

Step 5: Verify Part (c)

Substituting \(x = 6\) back into the concentration expression:\[ C(6) = \frac{6 + 2.1}{6 + 14} = \frac{8.1}{20} = 0.405 \]The solution is verified.

Step 6: Solve for Part (d)

To achieve a 50% concentration, set \(C(x) = 0.5\) and solve for \(x\):\[ \frac{x + 2.1}{x + 14} = 0.5 \]Cross-multiply to get:\[ x + 2.1 = 0.5(x + 14) \]Solve for \(x\):\[ x + 2.1 = 0.5x + 7 \]\[ 0.5x = 4.9 \]\[ x = 9.8 \]Thus, 9.8 ounces of pure acid should be added.

Step 7: Verify Part (d)

Substituting \(x = 9.8\) back into the concentration expression:\[ C(9.8) = \frac{9.8 + 2.1}{9.8 + 14} = \frac{11.9}{23.8} = 0.5 \]The solution is verified.

Step 8: Solve and Verify for Part (e)

Adding 12 ounces of pure acid, the concentration becomes:\[ C(12) = \frac{12 + 2.1}{12 + 14} = \frac{14.1}{26} = 0.5423076923\]Converting it to percentage, we get approximately 54.23%. The calculation confirms the concentration.

Key Concepts

Rational Functions

Rational functions are fractions where the numerator and denominator are both polynomials. In our exercise, the concentration of acid in the solution is represented by the rational function \( C(x) = \frac{x + 2.1}{x + 14} \). This function helps us calculate the acid concentration for different amounts of added pure acid. The numerator, \( x + 2.1 \), signifies the total pure acid, while the denominator, \( x + 14 \), represents the entire solution volume.
Understanding rational functions means recognizing the behavior of the function as variables change. They often have asymptotes—lines that the graph approaches but never touches. In this case, there's a vertical asymptote at \( x = -14 \) where the function is undefined and a horizontal asymptote at \( y = 1 \) showing the maximum possible concentration. As \( x \) increases, the function approaches this horizontal line.
This formulation is integral in concentration problems, as it allows transformations based on varying pure acid amounts. Keep in mind the limitations posed by the domain, which in this problem doesn't include negative \( x \) values, as adding a negative amount of acid isn't practical.

Concentration Problems

Concentration problems focus on mixtures where different substances are combined and the goal is to find the concentration of one component. When solving these, we're interested in how much of the substance—here, pure acid—is in the mixture.
To set these problems up, we study their current state: our base mixture is 14 ounces of a 15% acid solution, equivalent to \( 2.1 \) ounces of acid. When adding \( x \) ounces of pure acid, the goal is to analyze how it changes the original mixture's concentration.
Let's break down how to approach these problems:

• Identify existing quantities: We start with 14 ounces and 15% concentration.
• Define variables: How much pure acid, \( x \), will be added?
• Formulate the rational function: Combine current and added values to create \( C(x) = \frac{x + 2.1}{x + 14} \).
• Set target concentrations: Decide on desired concentration levels and solve for \( x \).

This systematic approach aids in efficiently solving for unknowns and prepares you for real-world applications.

Graphing Rational Functions

Graphing is a powerful way to visualize rational functions and understand their behavior. For the function \( C(x) = \frac{x + 2.1}{x + 14} \), plotting its values helps reveal how different amounts of pure acid change the concentration.
The graph typically has these characteristics:

• Vertical asymptote: At \( x = -14 \), denominators become zero, making the function undefined, which the graph reflects as a vertical line the function won't cross.
• Horizontal asymptote: Here, it's at \( y = 1 \), showing the highest potential concentration, which the function will approach more closely as \( x \) increases.

When plotting, focus on the graph's shape and asymptotic behavior, helping visualize the impact of added acid. Each point on the curve signifies a possible solution condition, linking theoretical computations with a tangible visual method. Graphing also illustrates gradual increases in concentration levels and their stabilization as \( x \) grows large.

Solution Verification

Verification is a critical part of solving rational function problems, ensuring your mathematical solution meets expectations. It involves checking calculations by substituting back into the original or derived equations.
Let's review this in action:

• After calculating \( x \), substitute it into \( C(x) = \frac{x + 2.1}{x + 14} \).
• For instance, when \( x = 6 \), we check: \( C(6) = \frac{6 + 2.1}{6 + 14} = 0.405 \). This satisfies the initial requirement of a 40.5% concentration.
• Continuously perform this step for different concentrations, such as 50% or any given specific value.

This process ensures accuracy and confidence in your results, validating calculations and reinforcing the understanding of the underlying principles. It supports learning, providing a feedback loop between prediction, computation, and affirmation.