Problem 67 - Algebra and Trigonometry

Question

Mixture Problem A chemist has two large containers of sulfuric acid solution, with different concentrations of acid in each container. Blending 300 \(\mathrm{mL}\) of the first solution and 600 \(\mathrm{mL}\) of the second gives a mixture that is 15\(\%\) acid, 600 \(\mathrm{mL}\) of the second gives a mixture that is 15\(\%\) acid, whereas blending 100 \(\mathrm{mL}\) of the first with 500 \(\mathrm{mL}\) of the second gives a 12\(\frac{1}{2} \%\) acid mixture. What are the concentrations of sulfuric acid in the original containers?

Step-by-Step Solution

Answer

The concentrations are 25% and 10% for the first and second solutions, respectively.

Step 1: Understand the Problem

The problem states that there are two solutions with different acid concentrations. We have the volumes and resulting concentrations for two mixtures. We need to use these relationships to find the concentrations of the two original solutions.

Step 2: Define Variables

Let the concentration of the first solution be \( x \%0\). Let the concentration of the second solution be \( y \%0\). Convert the percentages into decimals: \( x = \frac{x}{100} \) and \( y = \frac{y}{100} \).

Step 3: Set up Equations from the First Mixture

For the first mixture: \(300\ \mathrm{mL}\) of the first plus \(600\ \mathrm{mL}\) of the second gives \(15\%\) acid. The equation is:\[ 300x + 600y = 900 \times 0.15 \] Simplifying gives: \[ 300x + 600y = 135. \]

Step 4: Set up Equations from the Second Mixture

For the second mixture: \(100\ \mathrm{mL}\) of the first plus \(500\ \mathrm{mL}\) of the second gives \(12.5\%\) acid. The equation is:\[ 100x + 500y = 600 \times 0.125 \] Simplifying gives: \[ 100x + 500y = 75. \]

Step 5: Solve the System of Equations

We now solve the equations:\[ 300x + 600y = 135 \] \[ 100x + 500y = 75 \] Divide the first equation by 3:\[ 100x + 200y = 45 \] Now subtract the second equation from this:\[ (100x + 200y) - (100x + 500y) = 45 - 75 \] \[ -300y = -30 \] \[ y = 0.1 \] or \( 10\% \).

Step 6: Find the Value of x

Substitute \( y = 0.1 \) back into any equation, for example, the simplified one:\[ 100x + 200(0.1) = 45 \] \[ 100x + 20 = 45 \] \[ 100x = 25 \] \[ x = 0.25 \] or \( 25\% \).

Step 7: Conclude the Solution

The original concentrations are \(25\%\) for the first solution and \(10\%\) for the second solution.

Key Concepts

Sulfuric Acid Concentration

In chemistry, the concentration of a solution refers to how much solute (in this case, sulfuric acid) is dissolved in a solvent (usually water) to form a solution. Sulfuric acid concentration in a mixture problem is often expressed as a percentage. This percentage indicates the ratio of the mass of sulfuric acid to the total mass of the solution, multiplied by 100. For instance, a 15% sulfuric acid solution means there are 15 grams of sulfuric acid in every 100 grams of solution.
Knowing the concentration is crucial because it helps chemists understand the strength of the solution, which is important when carrying out reactions or in industrial applications. To manipulate concentrations, chemists frequently blend solutions with different concentrations to achieve a desired strength, much like our problem where two solutions were blended to reach specific concentration targets.

System of Equations

A system of equations is a set of two or more equations that share common variables. In the context of mixture problems, systems of equations are incredibly useful for determining unknown values, like the concentrations (\( x \) and \( y \)) of sulfuric acid in two containers in our problem.
In this exercise, the chemist uses the relationship between volumes and their resulting concentrations from two mixtures. First, the equations are derived based on proportions from given data. These equations then represent the blends and their ultimate concentrations. To solve the problem, each equation needs to be individually simplified by applying algebra rules such as balancing variables and isolating unknowns.

• The problem initially involves two key equations formed from the given mixtures, \( 300x + 600y = 135 \) and \( 100x + 500y = 75 \).

By effectively solving this system, we can identify the concentrations of each solution.

Solution Concentration

Solution concentration is a fundamental aspect of chemistry that defines how much solute is present in a given amount of solvent. It is essential to understand this as it directly affects the properties and behavior of the solution in chemical processes.
Solution concentration is often measured in percentages but can also be expressed in molarity, molality, or other units. For the given problem, solution concentrations are provided in percentage form—the percentage by weight or volume presents an easy-to-understand measure of strength or potency for many practical and experimental situations.
To compute the concentration from experimental data, chemists will create equations based on concentration outcomes from mixtures, like in our exercise. By adjusting the initial volumes and concentrations in blends, desired solution concentrations are achieved for various applications in laboratory and industry.

Chemistry Problem-Solving

Chemistry problem-solving involves a logical and methodical approach to finding unknown variables or properties of chemical components through data and equations. Given specific parameters, chemists employ a step-by-step approach to dissect and understand the relationships between different chemical entities.
Initially, it is pivotal to understand the problem thoroughly—recognizing what is given, what is demanded, and what needs to be found. This understanding forms the basis for defining variables and setting up equations. Considering our problem, understanding the roles of the initial concentrations, the volumes used, and the resulting mixtures were key steps.
Ultimately, solving problems in chemistry often involves mastering mathematical techniques such as setting up and solving equations, logical reasoning for deductions, and making approximations where necessary. This consistent application of strategy, as exemplified in our mixture problem, underpins successful resolution of complex chemistry questions.