Problem 111 - Chemistry The Molecular Nature of Matter

Question

If \(25.0 \mathrm{~mL}\) of \(0.56 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted to a volume of \(125 \mathrm{~mL}\), what is the molarity of the resulting solution?

Step-by-Step Solution

Answer

The molarity of the diluted H2SO4 solution is 0.112 M.

Step 1: Understand the concept of dilution

Dilution is the process of reducing the concentration of a solute in solution by adding more solvent. The amount of solute remains the same, but the overall volume of the solution is increased, which results in a decreased molarity or concentration.

Step 2: Use the dilution equation

The dilution equation can be represented as \(M_1 \times V_1 = M_2 \times V_2\), where:\(M_1\) is the molarity of the initial concentrated solution,\(V_1\) is the volume of the initial concentrated solution,\(M_2\) is the molarity of the diluted solution,\(V_2\) is the volume of the diluted solution. Molarity after dilution can be found by rearranging the equation to solve for \(M_2\): \(M_2 = \frac{M_1 \times V_1}{V_2}\).

Step 3: Calculate the molarity of the diluted solution

First, convert the volumes from mL to L if necessary. Here, we directly plug in the given values into the dilution equation: \(M_2 = \frac{0.56 \mathrm{M} \times 25.0 \mathrm{mL}}{125 \mathrm{mL}}\). Cancel the units of volume (mL) and compute the molarity \(M_2\).

Step 4: Perform the calculation

Doing the math gives us \(M_2 = \frac{0.56 \cdot 25.0}{125} = 0.112 \mathrm{M}\). This is the molarity of the H2SO4 solution after dilution.

Key Concepts

Dilution Equation

Consider the dilution equation as a golden rule for chemists: it reliably predicts the outcome when a solution's concentration is lowered by adding more solvent. The equation is succinctly represented as \(M_1 \times V_1 = M_2 \times V_2\), serving as a mathematical symmetry between the concentration and volume before and after the dilution process.

Understanding the dilution equation is crucial because it maintains the conservation of mass: the amount of solute (the substance being dissolved) remains constant, despite the increased volume of the solution. To determine the new molarity (\(M_2\)), we rearrange the equation to \(M_2 = \frac{M_1 \times V_1}{V_2}\), where \(M_1\) and \(V_1\) are the initial molarity and volume, and \(V_2\) is the final volume after dilution.

As a tip, always ensure your volume units are the same when applying the dilution equation. In many cases, you will need to convert milliliters (mL) to liters (L) to maintain consistency, as molarity is typically expressed in moles per liter (M or mol/L).

Concentration of Solutions

The heart of many chemical processes, the concentration of a solution dictates its properties and reactivity. It indicates how much solute is present in a specific volume of solvent. Concentration isn't just a value; it's a profound expression of the solution's strength.

There are various ways to express concentration, such as molarity (M), molality (m), mass percent, and volume percent, but molarity is the most frequently used in aqueous solutions and is defined as the number of moles of solute per liter of solution. It's important to know that when we talk about diluting a solution, we're referring to the reduction of its concentration while the quantity of solute remains unchanged.

When you dilute a solution, you're essentially spreading the same amount of solute over a greater volume, leading to a less concentrated solution. This concept is central to many applications, be it in preparing laboratory reagents or adjusting the flavor intensity of a beverage.

Molarity Calculations

Performing molarity calculations is a fundamental skill in chemistry. To derive the molarity of a solution, you'll need to know the number of moles of solute and the total volume of the solution. The formula is beautifully simple: \( \text{Molarity} (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \).

Molarity calculations become more intriguing when you're dealing with dilutions. As seen in our exercise, after diluting a concentrated solution, the final molarity can be found using the dilution formula. We set the initial and final states of the solution in relation to one another and observe the shift in concentration.

For a practical example of molarity calculations, consider our textbook problem. We know the initial molarity (\(M_1\)) and volume (\(V_1\)) and the final volume (\(V_2\)). By substituting these values into the dilution equation, we can calculate the final molarity (\(M_2\)). It's all about keeping track of the quantities before and after the dilution and ensuring we align the units to avoid computational errors.