Problem 34 - Algebra 2

Question

CHEMISTRY Kiara adds an 80\(\%\) acid solution to 5 milliters of solution that is 20\(\%\) acid. The function that represents the percent of acid in the resulting solution is \(f(x)=\frac{5(0.20)+x(0.80)}{5+x},\) where \(x\) is the amount of 80\(\%\) solution added. How much 80\(\%\) solution should be added to create a solution that is 50\(\%\) acid?

Step-by-Step Solution

Answer

Add 5 mL of 80% acid solution.

Step 1: Identify the Required Variable

We need to find the amount of 80% acid solution, denoted as \(x\), necessary to achieve a final solution with 50% acid concentration.

Step 2: Substitute Target Concentration

Substitute the target concentration of 50% into the function: \[\frac{5 \times 0.20 + x \times 0.80}{5 + x} = 0.50\].

Step 3: Clear the Fraction

Multiply both sides of the equation by \(5 + x\) to eliminate the fraction: \[5 \times 0.20 + x \times 0.80 = 0.50 \times (5 + x)\].

Step 4: Simplify Both Sides

Simplify the left side to get:\[1 + 0.80x\]On the right side, expand the term:\[2.5 + 0.50x\].

Step 5: Set Equation for \(x\)

Equate the simplified expressions: \[1 + 0.80x = 2.5 + 0.50x\].

Step 6: Isolate \(x\)

Rearrange the equation to isolate \(x\):Subtract 0.50x from both sides: \[1 + 0.30x = 2.5\]Subtract 1 from both sides: \[0.30x = 1.5\].

Step 7: Solve for \(x\)

Divide both sides by 0.30 to solve for \(x\):\[x = \frac{1.5}{0.30} = 5\].

Key Concepts

Acid Concentration Calculations

Understanding acid concentration calculations is essential when working with chemical mixtures. When mixing solutions of different concentrations, we aim to find the final concentration of the resulting solution. This requires juggling a few important concepts, like the amount of substance and its concentration. In this problem, Kiara is mixing an 80% acid solution with a 5 ml solution that contains only 20% acid.
The concentration of the acid in a solution is calculated as the ratio of the amount of acid to the total volume of the solution. It's often expressed in percentages. Here, we use the function: \[ f(x) = \frac{5 \times 0.20 + x \times 0.80}{5 + x} \]where \( x \) is the unknown quantity of 80% acid solution added. This function helps us understand how different amounts of the concentrated solution alter the overall acid concentration. The goal is to make the final solution have a 50% acid concentration by adjusting \( x \). Breaking it down further:

• 5 ml of the 20% acid solution starts the base.
• We add \( x \) ml of the 80% acid solution.
• The final concentration should be 50% after mixing.

These calculations help determine the amount of solution needed to achieve specific concentrations in real-world scenarios.

Rational Functions

Rational functions are expressions composed of ratios of polynomials. The equation used in this problem incorporates a rational function to model the mixing of acid solutions. Specifically, the function: \[ f(x) = \frac{5(0.20) + x(0.80)}{5 + x} \]is used to represent the concentration of acid in the resultant mixture.Rational functions are useful because they can form complex relationships between variables, like the volume of acid solutions. These relationships tell us how adding more of a concentrated solution affects the overall concentration.

• The numerator reflects the total amount of acid in both solutions combined, which is derived from multiplying volume by concentration.
• The denominator gives the total volume of the mixed solutions.

Understanding rational functions allows you to predict how systems will behave under certain conditions, like how much of a solution you need to reach a desired concentration.

Systems of Linear Equations

Systems of linear equations are mathematical tools that help solve problems involving two or more variables. They are particularly handy in mixture problems. After substituting the given values into the function for acid concentration, we can derive a linear equation that must be solved to find \( x \):Multiply to eliminate the fraction:\[5 \times 0.20 + x \times 0.80 = 0.50 \times (5 + x) \]which simplifies to:\[1 + 0.80x = 2.5 + 0.50x \]
By solving this linear equation, we isolate \( x \) and find the amount of 80% solution to add:

• Combine like terms by moving all terms involving \( x \) to one side.
• Subtract constants to solve for \( x \).

This methodological approach using linear equations lets us precisely calculate the required amounts in mixture problems, providing a clear path to achieve desired results in practical and real-world chemical solubility issues.