The sulfuric acid solution that is purchased for a stockroom has a molarity of \(17.8\) M; all sulfuric acid solutions for experiments are prepared by dilution of this stock solution. (a) Determine the volume of \(17.8 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) that must be diluted to \(250 \mathrm{~mL}\) to prepare a \(2.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. (b) An experiment requires a \(0.50 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) solution. The stockroom manager estimates that \(6.0\) L of the acid is needed. What volume of \(17.8 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) must be used for the preparation?

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, and is expressed in units of moles per liter (M). In chemistry, molarity is one of the most commonly used units to describe the concentration because it directly relates the volume of the solution to the amount of substance dissolved in it. This concentration unit provides a clear understanding of how many particles of the solute are available in a given volume to react or to exhibit their characteristics.

For example, a solution with a molarity of 2.0 M has 2 moles of solute dissolved in every liter of the solution. To calculate molarity, you can use the formula: \[ M = \frac{moles\ of\ solute}{liters\ of\ solution} \] Understanding molarity is crucial when preparing solutions for experiments or when performing titrations and other chemical analyses.

Dilution refers to the process of reducing the concentration of a solute in a solution, typically by adding more solvent. When working with sulfuric acid, which is a strong and corrosive acid, safety is a primary concern, and diluting the acid carefully is crucial. The sulfuric acid dilution process involves adding a concentrated acid (stock solution) to water to achieve a desired lower concentration. This step must be done carefully by adding the acid to the water, not the other way around, to prevent violent reactions and splashing.

The stock sulfuric acid solution may have a high molarity, such as 17.8 M, meaning it is highly concentrated. Laboratories often require more diluted solutions for experiments, so they dilute the stock to achieve lower concentrations. The problem sometimes lies in calculating the precise volume needed to reach a particular molarity for experimental needs. Dilution does not change the number of moles of solute; it only increases the volume of the solution, which lowers the molarity.

The concentration-volume relationship is a core concept in chemistry that demonstrates the relationship between the concentration of a solution and its volume. This relationship is crucial when performing dilutions, as it dictates that the amount of solute in the solution remains constant during the dilution process. Instead, it is the volume that increases, leading to a decrease in the concentration of the solution.

The product of concentration and volume before dilution (\( C_1V_1 \) ) is equal to the product of these after dilution (\( C_2V_2 \) ). This relationship is central to the dilution process because it allows chemists to calculate how much of a concentrated stock solution they need to achieve a desired concentration in a certain volume of diluted solution. Remembering that the 'amount' of solute does not change during dilution simplifies many calculations and can help prevent the wasting of reagents and the need for additional adjustments.

The dilution formula, \( C_1V_1 = C_2V_2 \), is a key tool in solving dilution problems in chemistry. It relates the concentrations and volumes of the stock solution and the diluted solution. The variables \( C_1 \) and \( V_1 \) represent the initial concentration and volume of the stock solution, while \( C_2 \) and \( V_2 \) represent the final concentration and volume of the diluted solution.

To solve dilution problems:

• Identify the concentrations and volumes of both the stock and desired diluted solutions.
• Rearrange the dilution formula to solve for the unknown quantity—usually the volume of the stock solution needed (\( V_1 \)).
• Perform the calculations, maintaining consistency in the units used for volume and ensuring appropriate conversion between milliliters and liters if necessary.

Using the problem provided as an example, if you need to prepare 250 mL of a 2.0 M \( H_2SO_4 \) solution from a 17.8 M stock solution, you insert the known values into the formula and solve for \( V_1 \). Such calculations are fundamental for the accurate preparation of solutions used in experimental procedures.