Problem 91 - Elementary Algebra

Question

Concentrated sulfuric acid is \(98 \%\) sulfuric acid. The liquid mixture in lead batteries is \(33.5 \%\) sulfuric acid. Use a system of linear equations to find the amount of concentrated sulfuric acid and the amount of water needed to make \(5 \mathrm{~L}\) of the liquid mixture for a lead battery. Round to the nearest tenth.

Step-by-Step Solution

Answer

1.7 L of concentrated sulfuric acid and 3.3 L of water.

Step 1 - Define Variables

Let x be the amount of concentrated sulfuric acid (in liters) and y be the amount of water (in liters).

Step 2 - Set Up Equations

The first equation represents the total volume, and the second equation represents the concentration: \[ x + y = 5 \] \[ 0.98x = 0.335 \times 5 \]

Step 3 - Substitute and Solve

Solve the second equation for x: \[ 0.98x = 1.675 \]\[ x = \frac{1.675}{0.98} \]\[ x \thickapprox 1.7 \] Then substitute x into the first equation: \[ 1.7 + y = 5 \]\[ y = 5 - 1.7 \]\[ y \thickapprox 3.3 \]

Step 4 - Verify Solution

Verify the solution by checking the concentrations: \[ 0.98 \times 1.7 \thickapprox 1.666 \] \[ 0.335 \times 5 = 1.675 \] The values are close enough, rounding to the nearest tenth gives x = 1.7 and y = 3.3

Key Concepts

Sulfuric Acid Concentration

Sulfuric acid concentration refers to the percentage of sulfuric acid present in a solution. For example, if we say a solution is 98% sulfuric acid, it means 98% of the solution's volume is sulfuric acid while the remaining 2% is other substances, usually water.

Concentration is essential in chemistry, especially in mixture problems where you need to combine substances of different concentrations to achieve a specific desired concentration.

For instance, lead-acid batteries use a mixture that is 33.5% sulfuric acid. Achieving this exact concentration is crucial for the proper functioning of the battery.

Therefore, understanding concentration helps to ensure the products we use perform as expected, and knowing how to adjust concentrations through mixing is a vital skill in both academic and real-world chemistry.

Mixture Problems

Mixture problems involve combining two or more substances to achieve a desired concentration or quantity. These problems are common in chemistry, cooking, and various industrial processes.

An example of a mixture problem is the exercise asking for 5 liters of a 33.5% sulfuric acid solution using concentrated sulfuric acid (98%) and water (0% sulfuric acid).

To approach these problems, we:

• Define variables to represent the amounts of each component.
• Set up equations based on total volume and concentration requirements.
• Solve these equations systematically.

Mixture problems help in understanding the principles of dilution and concentration, both crucial concepts in various scientific and practical fields. Successfully solving these problems requires a strong grasp of algebra and an understanding of how different substances interact when mixed.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra. It involves finding the value of variables that make an equation true.

In the context of mixture problems, we usually deal with a system of linear equations. This system typically includes two equations based on the total quantity and concentration of the mixture.

The steps to solve these equations are:

• Define the variables representing the unknown quantities.
• Write down the equations using the given conditions.
• Simplify and solve one of the equations for one variable.
• Substitute this value into the other equation to find the second variable.
• Verify the solution for accuracy.

For example, with the sulfuric acid mixture problem, we:

• Let x be the amount of concentrated sulfuric acid and y the amount of water.
• Used the equations: \( x + y = 5 \) and \( 0.98x = 0.335 \times 5 \).
• Solved for x and substituted back to find y.

Mastering the solution of linear equations enables handling a broad range of mathematical and real-world problems effortlessly.