Mixing Sucrose Solutions: A Chemistry Problem

In the fascinating realm of chemistry, mixing solutions is a common task. Whether in a lab or a kitchen, understanding how concentrations change when substances are combined is crucial. Let's delve into a specific problem involving sucrose solutions, exploring the math behind the mixing and how to achieve a desired concentration. This exercise is not just a mathematical puzzle, but a glimpse into the practical applications of chemistry in everyday scenarios. By the end, you'll be able to work out the ideal solution.

The Problem: Diluting and Concentrating Sucrose

Imagine a chemist starting with 200 mL of a 10% sucrose solution. This means that 10% of the solution's volume is pure sucrose, while the remaining 90% is likely water or another solvent. The chemist then adds x mL of a 40% sucrose solution. The key here is to realize that we're essentially adding a more concentrated sucrose solution to a less concentrated one, aiming to find the sweet spot, the target concentration. The amount x will determine our goal.

The percent concentration, y, of the final mixture is given by a rational function:

y = (0.1(200) + 0.4x) / (200 + x) * 100

This formula is the heart of the problem. Let's break it down: The numerator, 0.1(200) + 0.4x, represents the total amount of sucrose in the final mixture. The first part, 0.1(200), calculates the amount of sucrose initially present in the 10% solution (10% of 200 mL). The second part, 0.4x, calculates the amount of sucrose added from the 40% solution (40% of x mL). The denominator, 200 + x, represents the total volume of the final mixture (the initial 200 mL plus the added x mL). Multiplying this ratio by 100 converts the fraction to a percentage.

Now, the chemist has a specific goal: to determine how much of the 40% solution (x) to add in order to achieve a particular target concentration. This is where we need to dive deeper and begin solving the problem.

Solving for the Desired Concentration

Let's assume the chemist wants to create a 25% sucrose solution. In this case, y = 25. Our task is to solve for x, the volume of the 40% solution needed. We will plug this value into our function.

25 = (0.1(200) + 0.4x) / (200 + x) * 100

First, we divide both sides by 100:

0.25 = (20 + 0.4x) / (200 + x)

Next, we multiply both sides by (200 + x) to remove the denominator:

0.25 * (200 + x) = 20 + 0.4x

Expanding the left side:

50 + 0.25x = 20 + 0.4x

Subtracting 0.25x from both sides:

50 = 20 + 0.15x

Subtracting 20 from both sides:

30 = 0.15x

Finally, dividing both sides by 0.15:

x = 200

Therefore, the chemist needs to add 200 mL of the 40% sucrose solution to obtain a 25% mixture. This method is applicable for other concentration goals, allowing the chemist to fine-tune the final solution.

Analyzing the Solution and Its Implications

In our example, x = 200 mL. This means that to achieve a 25% sucrose solution, the chemist must add an equal volume of the 40% solution to the initial 10% solution. Let's consider what happens if the chemist wants an even more concentrated solution, say, 30%. In that scenario, the chemist would need to add even more of the 40% solution, and we'd go through the same mathematical process, but with y = 30.

Understanding this equation is key to successfully mixing solutions. It highlights how the concentration of a mixture is a balance between the initial concentration, the concentration of the added solution, and their relative volumes. This kind of problem isn't unique to sucrose; it can be applied to many other types of solutions, whether the chemist works with acids, bases, or any other substances. The underlying principle remains the same: to find the right balance to get the result you want.

Let's analyze some potential scenarios that the chemist might encounter. What if the chemist wants a final concentration below 10%? To achieve this, x would likely need to be negative, which doesn't make sense physically in this context. It's impossible to