Question

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 \mathrm{M}\). How much water should you add?

Short answer

Add 126.25 mL of water.

Step-by-step solution

Understand the Problem

We have a volume of HCl solution, which we want to dilute and achieve a new, lower molarity.

Use the Dilution Formula

The dilution formula is given by \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the initial concentration and volume, and \( C_2 \) and \( V_2 \) are the final concentration and volume.

Plug in Known Values

We know \( C_1 = 0.125 \mathrm{~M} \), \( V_1 = 505 \mathrm{~mL} \), and \( C_2 = 0.100 \mathrm{~M} \). Plug these values into the formula: \[ 0.125 \times 505 = 0.100 \times V_2 \]

Solve for Final Volume \( V_2 \)

Calculate \( V_2 \) by rearranging the equation: \[ V_2 = \frac{0.125 \times 505}{0.100} \]Calculate this to find \( V_2 \).

Calculate \( V_2 \)

Perform the calculation:\[ V_2 = \frac{0.125 \times 505}{0.100} = 631.25 \mathrm{~mL} \]

Find the Amount of Water to Add

The amount of water to add is the difference between the final volume \( V_2 \) and the initial volume \( V_1 \). Therefore:\[ \text{Water to add} = V_2 - V_1 = 631.25 - 505 \mathrm{~mL} \]

Calculate the Volume of Water

Perform the subtraction:\[ 631.25 - 505 = 126.25 \mathrm{~mL} \]Thus, you need to add 126.25 mL of water.

Key concepts

Dilution Formula

The dilution process helps in obtaining a lower concentration from a more concentrated solution, making it essential in various laboratory and industrial applications. The fundamental equation used in dilution is the dilution formula, expressed as \( C_1V_1 = C_2V_2 \). This equation reflects the principle of conservation of mass, where the amount of solute remains constant before and after dilution.

The terms in the formula are as follows:

• \( C_1 \): Initial concentration of the solution.
• \( V_1 \): Initial volume of the solution.
• \( C_2 \): Final concentration desired.
• \( V_2 \): Final total volume after dilution.

To find the amount of solute that will remain constant, use the left-hand side of the equation. Then, solve to find the unknown final volume or the dilution factor using the right-hand side.

Molarity

Molarity is an important concept in chemistry that measures the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, abbreviated as \( ext{M} \). For example, a 1 M solution contains one mole of solute in every liter of solution.

Represented mathematically as:
\[ ext{Molarity (M)} = \frac{ ext{moles of solute}}{ ext{liters of solution}} \]
Molarity enables chemists to mix solutions with the same stoichiometry for reactions and is critical when performing titrations or dilutions. In our original exercise, the initial molarity was 0.125 M, and the goal was to achieve a 0.100 M solution.

Solution Preparation

Preparing a solution accurately involves calculating how much initial solution and solvent are needed to reach your desired concentration and volume. Start by determining the concentration and the final volume you aim to reach. The step-by-step approach involves using the dilution formula, where the current setup dictates the original concentration and volume.

Using our exercise as an example, we started with an initial 505 mL of a 0.125 M HCl solution. After applying the dilution formula, we calculated the exact amount of water required to create a new 0.100 M solution, achieving both the demand for precision and safety while handling hazardous materials like hydrochloric acid.

Hydrochloric Acid (HCl) Solution

Hydrochloric acid (HCl) is a strong acid commonly used in laboratories and industry. It is often sold in concentrated stock solutions, which can be dangerous because of their corrosiveness. Therefore, handling and diluting HCl requires caution.

In practice, when diluting HCl from a higher concentration to a lower one, ensure proper safety measures are in place. These measures include wearing protective equipment such as gloves and goggles, and performing the dilution inside a fume hood to avoid inhaling harmful vapors.
In our discussion, we dealt with an HCl solution initially at 0.125 M, diluting it to 0.100 M by adding water. This process involves reducing the concentration safely and precisely for experimental or practical utility.

Volume Calculation

Volume calculation is a fundamental step in dilution, involving determining the initial and final volumes to understand how much solvent needs to be added. Initially, the known concentration and volume measurements can be plugged into the dilution formula.

Rearrange the formula \( V_2 = \frac{C_1V_1}{C_2} \) to calculate the desired final volume \( V_2 \). This step identifies the total volume needed to achieve the intended concentration.
In the original problem, the final volume calculation was performed as \( V_2 = \frac{0.125 imes 505}{0.100} = 631.25 \mathrm{~mL} \). Through subtraction of the original volume \( V_1 \) from \( V_2 \), you find the amount of water to be added, which in this case was 126.25 mL, finalizing the dilution process with precision.