Problem 54 - Precalculus Mathematics for Calculus

Question

Mixture Problem What quantity of pure acid must be added to 300 mL of a \(50 \%\) acid solution to produce a \(60 \%\) acid solution?

Step-by-Step Solution

Answer

Add 75 mL of pure acid.

Step 1: Define Variables

Let \( x \) represent the volume of pure acid (100% acid) in milliliters that needs to be added to the solution.

Step 2: Calculate Current Acid Content

The 300 mL of a 50% acid solution contains \( 0.5 \times 300 = 150 \) mL of pure acid.

Step 3: Formulate the Equation

To achieve a 60% acid solution, the volume of pure acid (150 + x) divided by the total new volume (300 + x) must equal 0.6, or 60%. The equation is: \[ \frac{150 + x}{300 + x} = 0.6 \]

Step 4: Solve the Equation for x

First, clear the fraction: \[ 150 + x = 0.6(300 + x) \]Expanding on the right side gives:\[ 150 + x = 180 + 0.6x \]Next, solve for \( x \): Subtract \( 0.6x \) from both sides to get:\[ 150 + 0.4x = 180 \]Subtract 150 from both sides:\[ 0.4x = 30 \]Divide by 0.4:\[ x = 75 \]

Step 5: Verify the Solution

Adding 75 mL of pure acid, the new solution has (150 + 75) = 225 mL of acid in a total volume of (300 + 75) = 375 mL. The concentration is:\[ \frac{225}{375} = 0.6 \] which confirms that the mixture is 60% acid.

Key Concepts

Understanding Pure Acid

Pure acid refers to a substance that is entirely composed of acidic components, meaning there are no additional solvent or dilute present. When dealing with problems involving solutions, pure acid is represented as 100% acid concentration. This is crucial in mixture problems, as it allows us to precisely determine the amount needed to alter the concentration of a solution.

In our example, pure acid is what we need to add to a solution to achieve a desired concentration. By manipulating the quantity of pure acid added, we change the overall makeup of the solution. This process highlights the importance of understanding what pure acid signifies in relation to mixtures. When you think of pure acid in calculations, always remember its concentration is always 100%, as represented by the variable used in these equations.

Understanding Solution Concentration

Solution concentration describes how much acid is present in a given volume of liquid. It is expressed as a percentage, indicating the ratio of pure acid to the total volume of solution. This concept is central to mixture problems because changing the concentration is often the objective.

In our problem, we start with a 300 mL solution at 50% concentration and aim to reach 60%.

• The initial 50% concentration means there are 150 mL of pure acid in the solution (0.5 \times 300 mL = 150 mL).
• To get to a 60% concentration, the formula becomes:
  \[ \frac{\text{amount of pure acid}}{\text{total volume of solution}} = \text{desired percentage} \]

This formula is used to determine how much pure acid is needed to reach the new concentration. The concentration is key for identifying how much needs to be adjusted to reach the desired solution strength.

Equation Formulation in Mixture Problems

Creating the right equation is a foundational step in solving mixture problems. It involves setting up an equation that accurately represents the relationship between the variables in question. Let’s break down the steps necessary to formulate equations in mixture problems:

• Identify and Set Variables: Use variables to denote unknown quantities, such as the volume of pure acid to be added. In our problem, we let \( x \) be the milliliters of pure acid added.
• Determine Initial Conditions: Calculate the amount of pure acid present initially in the solution using its concentration. Here, that's 150 mL of pure acid in 300 mL of solution.
• Set the Desired Outcome: Express the desired concentration condition using an equation. For the 60% target, the equation is:
  \[ \frac{150 + x}{300 + x} = 0.6 \]
• Solve the Equation: Rearrange and solve for the unknown. Clear fractions, group like terms and solve for \( x \) to find the amount of pure acid required.
• Verify the Solution: Check that plugging the solution back into the equation results in the correct concentration.

Understanding how these elements come together guides you in crafting the right approach for a given problem and ensuring accurate results.