Question

To what volume must \(50.0 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HCl}\) be diluted to produce \(0.200 \mathrm{M}\mathrm{HCl}\) ?

Short answer

The final volume must be 375 mL.

Step-by-step solution

Understand the Dilution Concept

Dilution involves adding solvent to a solution to decrease the concentration of solutes. The amount of substance present does not change, only the volume it is contained in increases, which in turn lowers the concentration.

Use Dilution Equation

The dilution equation relates the concentrations and volumes before and after the dilution follows: \( C_1V_1 = C_2V_2 \). Here, \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume.

Insert Known Values

Insert the known values into the dilution equation. Since we want to find the final volume \(V_2\), we rearrange the equation to solve for \(V_2\): \( V_2 = \frac{C_1V_1}{C_2} \), and then insert the known values \( V_2 = \frac{(1.50 \, \mathrm{M})(50.0 \, \mathrm{mL})}{0.200 \, \mathrm{M}} \).

Calculate the Final Volume

Perform the calculation: \( V_2 = \frac{(1.50)(50.0 \, \mathrm{mL})}{0.200} = 375 \, \mathrm{mL} \). The final volume is 375 mL.

Key concepts

Solution Dilution

When working with chemical solutions, it's often necessary to adjust concentrations by the process of dilution. Dilution is simply the adding of solvent, which is usually water, to a solution to decrease the concentration of solutes - the dissolved substances. Dilution doesn't involve altering the amount of solutes, just the total volume of the solution which, as a result, dilutes the concentration. This practice is common in preparatory chemistry labs and various applications like achieving proper concentration for reactions or reducing the strength of pharmaceuticals. It’s important to understand that dilution maintains the integrity of the solution; it just lessens the intensity by dispersing the solute particles throughout a greater volume.

Understanding how to execute a dilution is not only a practical lab skill but also an essential part of experimenting safely and effectively. When diluting an acid, as in our exercise, it is crucial to always add the acid to water, never the reverse, to prevent exothermic reactions that can cause splashing or even explosions. Knowing the final concentration you wish to achieve enables you to calculate the necessary volume of solvent to add, ensuring precise experimental conditions.

Concentration of Solutes

The concentration of solutes within a solution is a measure of how much of a substance is dissolved within a given volume of solvent. It’s a key concept in chemistry because it affects reactivity, product formation, and energy changes in reactions. Concentration can be expressed in various ways, including molarity, molality, normality, and percentage compositions. Among these, molarity is the most commonly used in a laboratory setting. It is defined as the number of moles of solute per liter of solution.

Having a strong grasp of concentration concepts ensures you can predict how reactants will behave, whether in producing a desired product in synthesis or titrating to determine unknown concentrations. It also allows for the safe handling of reagents, as some reactions depend critically on the concentration of the reactants and can be dangerous if not properly monitored. In practical situations, such as environmental testing or the pharmaceutical industry, understanding solute concentrations can be critical to ensuring compliance with safety and potency standards.

Molarity Calculations

Molarity calculations are fundamental in chemistry, representing the moles of solute per liter of solution, and denoted as 'M'. When performing molarity calculations, you’re usually seeking to find out one of three things: the number of moles, the volume of solution, or the molarity itself. Knowing any two of these values allows you to calculate the third. The formula for molarity is represented by \( M = \frac{n}{V} \) where \( M \) is the molarity, \( n \) is the number of moles, and \( V \) is the volume of the solution in liters.

Equipped with the formula, you can tackle problems involving the preparation of solutions with a specific molarity or the dilution of a solution to a desired concentration, as demonstrated in the exercise provided. For example, if a laboratory protocol requires a 0.200 M HCl solution, but you only have a concentrated stock solution of 1.50 M HCl, molarity calculations guide you in adding the correct amount of solvent to achieve the desired concentration without trial and error. Understanding molarity is not just about solving mathematical problems; it's about being able to accurately predict and control the chemical and physical behavior of solutions in real-world scenarios.