Question

To what volume must \(25.0 \mathrm{~mL}\) of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) be diluted to produce \(1.50 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short answer

The volume needed for dilution is 300.0 mL.

Step-by-step solution

Understand the Dilution Concept

Dilution is the process of decreasing the concentration of a solute in a solution by adding more solvent. The amount of solute stays the same before and after dilution. The dilution formula is given by: \(M_1V_1 = M_2V_2\), where \(M_1\) and \(M_2\) are the initial and final molarities, and \(V_1\) and \(V_2\) are the initial and final volumes respectively.

Set Up the Equation Using the Dilution Formula

We will use the dilution formula to set up an equation with the given values. \(M_1V_1 = M_2V_2\) where \(M_1 = 18.0\, M\), \(V_1 = 25.0\, mL\), and \(M_2 = 1.50\, M\). We need to find \(V_2\).

Plug in the Known Values into the Equation

The equation becomes: \((18.0\, M) \times (25.0\, mL) = (1.50\, M) \times V_2\).

Solve for the Unknown Volume (\(V_2\))

To solve for \(V_2\), rearrange the equation to isolate \(V_2\) on one side: \(V_2 = \dfrac{(18.0\, M) \times (25.0\, mL)}{1.50\, M}\).

Perform the Calculation

Calculating the above gives us: \(V_2 = \dfrac{450.0\, mL}{1.50}\), which simplifies to \(V_2 = 300.0\, mL\). Therefore, 25.0 mL of 18.0 M \(H_{2}SO_{4}\) must be diluted to 300.0 mL to produce 1.50 M \(H_{2}SO_{4}\).

Key concepts

Molarity and Dilution

Understanding the connection between molarity and dilution is a key concept in chemistry, especially when preparing solutions of desired concentrations. Molarity, denoted as 'M', is the measure of concentration of a solution expressed in moles of solute per liter of solution. When a solution is diluted, the amount of solute stays constant, but the total volume increases due to the addition of solvent.

To calculate the new concentration after dilution, we use the dilution equation, \(M_1V_1 = M_2V_2\), which expresses the relationship that the product of the initial molarity \(M_1\) and volume \(V_1\) is equal to the product of the final molarity \(M_2\) and volume \(V_2\). This formula helps in finding out how much solvent needs to be added to reach a specific diluted concentration.

For example, if we have high-concentration \(H_2SO_4\) and we want to dilute it to a lower concentration, we can use the formula to determine the final volume needed. This process is critical in various applications such as preparing laboratory reagents, medicine dosage, and industrial processes.

Concentration of Solutions

The concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution. The most common quantitative expression of concentration in chemistry is molarity. However, there are other units such as molality, weight/volume percent, and volume/volume percent, each suitable for different scenarios.

Molarity is useful because it directly relates to the number of moles in a given volume, making it particularly relevant for stoichiometric calculations and reactions occurring in solution. It's important to note that molarity changes with temperature as it is dependent on volume, which can expand or contract with temperature shifts.

In the context of dilutions, understanding the original and final concentrations allows for precise calculations of how much solvent to add without altering the amount of solute. This accuracy is essential for experiments that require a specific concentration for reactions to occur as expected or when a particular molarity is needed for a chemical application.

Chemical Calculations

Chemical calculations are integral to various processes in chemistry including determining reactant quantities for reactions, analyzing experimental data, and performing dilutions as seen in our example. These calculations often use fundamental formulas, one of which is the dilution formula discussed earlier. By mastering this form of mathematical application in chemistry, one can accurately manipulate solutions to obtain desired concentrations.

Apart from dilution calculations, other common chemical calculations include stoichiometry, which is used to calculate the amounts of reactants and products in a chemical reaction, and the use of the ideal gas law to find the relationships between pressure, volume, temperature, and moles of a gas.

For students, becoming comfortable with these calculations entails understanding the concepts behind the formulas, practicing problem-solving, and recognizing when to apply which sort of calculation. For the dilution problem provided, following the step-by-step solution methodically, while understanding each step's purpose, is critical for mastery and application in real-world scenarios.