Problem 60 - College Algebra with Corequisite Support

Question

For the following exercises, determine the function described and then use it to answer the question. A container holds \(100 \mathrm{~mL}\) of a solution that is 25 \(\mathrm{mL}\) acid. If \(n \mathrm{~mL}\) of a solution that is \(60 \%\) acid is added, the function \(C(n)=\frac{25+.6 n}{100+n}\) gives the concentration, \(C\), as a function of the number of \(\mathrm{mL}\) added, \(n\). Express \(n\) as a function of \(C\) and determine the number of \(\mathrm{mL}\) that need to be added to have a solution that is \(50 \%\) acid.

Step-by-Step Solution

Answer

Add 250 mL of the 60% acid solution.

Step 1: Set up the equation

We start with the given formula for the concentration: \[ C(n) = \frac{25 + 0.6n}{100 + n} \]We need to express the number of milliliters added, \( n \), as a function of the concentration \( C \). First, replace \( C \) with \( 0.5 \) since we want the concentration to be 50% acid.

Step 2: Substitute and simplify

Substitute \( C = 0.5 \) into the equation:\[ 0.5 = \frac{25 + 0.6n}{100 + n} \]Multiply both sides by \( 100 + n \) to eliminate the fraction:\[ 0.5(100 + n) = 25 + 0.6n \]This leads to:\[ 50 + 0.5n = 25 + 0.6n \]

Step 3: Rearrange the equation

To isolate \( n \), bring all terms involving \( n \) to one side of the equation. Subtract \( 0.5n \) from both sides:\[ 50 = 25 + 0.1n \]Subtract 25 from both sides:\[ 25 = 0.1n \]

Step 4: Solve for n

Now, solve for \( n \) by dividing both sides by 0.1:\[ n = \frac{25}{0.1} = 250 \]So, 250 mL of the 60% acid solution must be added.

Key Concepts

Concentration Calculation

Calculating the concentration of a solution involves determining the amount of a substance within a specified volume. This is crucial in chemistry, especially when dealing with solutions. Concentration can be expressed in various units, but often in chemistry, it is shown as a percentage or as molarity (moles per liter).

In the problem at hand, a concentration formula is provided: \[ C(n) = \frac{25 + 0.6n}{100 + n} \]This function represents the concentration of an acid solution and varies based on the amount \( n \) of another acid solution added. Understanding this formula requires recognizing each term's role:

• "25" represents the initial milliliters of acid in the container.
• "0.6n" accounts for the 60% concentration of the added solution.
• The denominator "100 + n" considers the total new volume after addition.

To find out the necessary added solution \( n \) to achieve a certain concentration, like 50%, we plug in the desired concentration and solve for \( n \). This transforms concentration calculations into an exercise of algebraic manipulation.

Function Inversion

Function inversion is a powerful algebraic technique used to solve equations where one variable needs to be expressed in terms of another. In this exercise, we wanted to invert the given concentration function to express \( n \) (the milliliters of solution added) as a function of \( C \) (the new concentration).

Starting with:\[ C(n) = \frac{25 + 0.6n}{100 + n} \]the goal is to solve for \( n \) in terms of \( C \). This involves strategic manipulation:

• Substituting the specific concentration you aim for, i.e., 50% or \( 0.5 \).
• Clearing the fraction by multiplying through by \( 100 + n \).
• Rearranging the resulting equation to isolate \( n \).

By performing these steps, we rearranged the equation to solve for \( n \), demonstrating function inversion. This technique is not just applicable here but valuable for calculations involving any variable-dependent outcomes.

Acid-Base Solutions

Acid-base solutions are central to many chemical processes and experiments. An acid solution contains a particular amount of acidic substance dissolved in water, often expressed as a percentage or molarity.

The exercise revolves around finding out how much of a 60% acidic solution needs to be added to an existing solution to reach a desired concentration. This type of problem touches upon the principles of dilution and mixing that are common in acid-base chemistry. Key points to remember include:

• The initial and added solutions contribute to the final concentration based on their volumes and concentrations.
• When combining solutions, remember that the total acid amount and total solution volume changes.
• It's important to ensure the final desired concentration, as shown here by adjusting \( n \) to achieve a 50% total acid concentration.

Understanding how to adjust mixtures to reach certain concentrations is a core skill in handling acid-base solutions effectively.