Problem 69 - Chemistry Matter and Change

Question

You need to make a large quantity of a 5\(\%\) solution of HCl but have only 25 \(\mathrm{mL}\) HCl. What volume of 5\(\%\) solution can be made from this volume of HCl?

Step-by-Step Solution

Answer

500 mL of 5\% HCl solution can be made.

Step 1: Understand the Components of the Solution

When we say a 5\(\%\) solution of HCl, we mean that there are 5 parts of HCl for every 100 parts of the solution. This means that if you have a total solution volume of \( V_{total} \), 5\% of it is pure HCl.

Step 2: Calculate the Total Volume for 25 mL of HCl

Let's denote the total volume of the solution we want to make by \( V_{total} \). Given that the solution is 5\% HCl, we have the relationship: \[ 0.05 \times V_{total} = 25 \, \text{mL} \]Solving for \( V_{total} \), we divide both sides by 0.05:\[ V_{total} = \frac{25 \, \text{mL}}{0.05} \]

Step 3: Perform the Calculation

Now, plug in the values and compute:\[ V_{total} = \frac{25}{0.05} = 500 \, \text{mL} \]This means you can make a total of 500 mL of a 5\% HCl solution.

Key Concepts

Concentration Calculations

Concentration calculations are crucial for determining how strong a solution is. They dictatethe proportion of a solute in a given solution. For instance, when we talk about a 5% HClsolution, it means that 5% of the volume is made up of hydrochloric acid, with the remaining95% being some solvent, typically water. This percentage can also be expressed as a ratio ordetermined by mass or volume.

Performing concentration calculations involves using basic arithmetic to relate the quantity ofsolute to the total volume of the solution. For our example, the concentration expresses howmuch HCl is present in every 100 ml of solution. These calculations are fundamental inworking with solutions because they help ensure proper solution preparation for variousapplications.

When carrying out these calculations, always ensure precision in measurement for accurateresults. Even a minor error can significantly impact the solution’s effectiveness in practicalapplications.

Dilution Problems

Dilution involves reducing the concentration of a solute in a solution, typically by addingmore solvent. In the given problem, we utilize dilution to create a 5% HCl solution from aspecified quantity.

The relationship used in solving dilution problems is straightforward: the product ofconcentration and volume before dilution equals the product after dilution. This relationshipcan be represented by the equation:

• Initial Concentration × Initial Volume = Final Concentration × Final Volume

This equation simplifies finding the unknown final volume in many scenarios, such as thesolution we worked through. Given that we start with a specific amount of hydrochloric acid(25 mL), and we need to find the final volume where this amount is just 5% of the totalsolution.

Practicing these types of calculations is beneficial, as dilutions frequently occur in laboratoryand medical settings, where precise concentrations are needed to achieve the desired effect.

Chemical Solutions

Chemical solutions uniquely combine a solute and solvent to form a homogeneous mixture.They are foundational in chemistry studies and various practical applications likepharmaceuticals, cleaning agents, and more.

In our example, hydrochloric acid acts as the solute, while water is commonly the solvent. Kreyeing a 5% solution means meticulously measuring and mixing chemical substances to achieve uniformity effect. Chemical solutions serve numerous purposes, including treatment applications where specifically tailored chemical concentrations are necessitated.

Understanding the dynamics of chemical solutions also means knowing their behavior, such assolubility limits, reactivity, and temperature and pressure impact. Being adept in these areasenables more profound insights into making solutions and adjusting them as needed forparticular uses.