Chapter 4: Problem 148
How many milliliters of \(0.10 \mathrm{M} \mathrm{HCl}\) must be added to \(50.0 \mathrm{~mL}\) of \(0.40 \mathrm{M} \mathrm{HCl}\) to give a final solution that has a molarity of \(0.25 M ?\)
Short Answer
Expert verified
83.33 mL of 0.10 M HCl must be added.
Step by step solution
01
Understanding the Concept of Dilution
Dilution involves adding solvent to a solution to lower the concentration of solutes. In this case, we're adding a dilute HCl solution to a more concentrated one, aiming to achieve a final specific molarity. The equation for dilution is: \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( C_2 \) are the molarities of the concentrated and final solutions, respectively, and \( V_1 \) and \( V_2 \) are the corresponding volumes.
02
Set up the Dilution Equation
The equation relates the concentration and volume before and after the dilution. Here, \( C_1 = 0.10 \text{M} \), \( V_1 \) is the volume of the diluted HCl we are solving for, \( C_2 = 0.25 \text{M} \) is the final concentration, and \( V_2 = 50.0 \text{mL} + V_1 \) is the final total volume.
03
Substitute Known Values Into the Equation
Substitute the known quantities into the dilution equation to find \( V_1 \): \[ 0.10 \text{M} \times V_1 = 0.25 \text{M} \times (50.0 \text{mL} + V_1) \].
04
Solve for \( V_1 \)
Simplify and solve the equation for \( V_1 \). Multiply each side by 100 for easier calculation: \( 10V_1 = 25 \times 50 + 25V_1 \), then collect V1 terms: \( 15V_1 = 25 \times 50 \), and solve: \( V_1 = \frac{25 \times 50}{15} \text{mL} = \frac{1250}{15} \text{mL} \).
05
Calculate the Volume of Diluted HCl
Perform the division to find the volume of the diluted HCl solution needed: \( V_1 = 83.33 \text{mL} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemistry Dilution Concept
The concept of dilution in chemistry refers to the process of reducing the concentration of a solute in a solution by adding more solvent. This procedure is crucial in various laboratory applications, including preparing solutions for experimentation and achieving desired concentrations for reactions.
Understanding dilution is fundamental for working with chemical solutions because it allows the laboratory technician or chemist to manipulate the concentration without changing the solute's amount. The formula typically used for dilution is expressed as \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( C_2 \) are the molar concentrations of the original and diluted solutions, while \( V_1 \) and \( V_2 \) represent their respective volumes.
To improve the ease of understanding for students, the dilution concept can be likened to adding water to a concentrated fruit juice to lessen its intensity, without altering the actual amount of juice being served. The principle remains the same; more solvent leads to a lower concentration of the solute, thereby diluting the solution.
Understanding dilution is fundamental for working with chemical solutions because it allows the laboratory technician or chemist to manipulate the concentration without changing the solute's amount. The formula typically used for dilution is expressed as \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( C_2 \) are the molar concentrations of the original and diluted solutions, while \( V_1 \) and \( V_2 \) represent their respective volumes.
To improve the ease of understanding for students, the dilution concept can be likened to adding water to a concentrated fruit juice to lessen its intensity, without altering the actual amount of juice being served. The principle remains the same; more solvent leads to a lower concentration of the solute, thereby diluting the solution.
Molarity
Molarity, denoted as \( M \), is a commonly used unit for expressing the concentration of a solution in chemistry. It is defined as the amount of solute (usually measured in moles) present in one liter of solution.
The formula for molarity is given by \( M = \frac{n}{V} \), with 'n' representing the number of moles of the solute and 'V' the volume of the solution in liters. This measurement allows chemists to conduct precise calculations for reactions, as molarity directly correlates to the number of reactive particles in a given volume.
The formula for molarity is given by \( M = \frac{n}{V} \), with 'n' representing the number of moles of the solute and 'V' the volume of the solution in liters. This measurement allows chemists to conduct precise calculations for reactions, as molarity directly correlates to the number of reactive particles in a given volume.
Importance of Molarity in Dilution
When performing a dilution, molarity acts as a bridge between the solute's concentration and the total volume of the solution. Precise molarity calculations ensure that reactions proceed correctly, especially when exact concentrations are needed, as in the case of the provided exercise.Acid-Base Titration
Acid-base titration is a quantitative analytical technique used to determine the concentration of an unknown acid or base by neutralizing it with a base or acid of known concentration. The point at which neutralization occurs is known as the equivalence point and is often indicated by a color change due to an indicator added to the solution.
The relationship between the volume and concentration of the titrant (the solution of known concentration) and the analyte (the solution of unknown concentration) is described by the titration formula: \( C_aV_a = C_bV_b \), where \( C_a \) and \( C_b \) are the concentrations of the acid and base, and \( V_a \) and \( V_b \) are the respective volumes. By knowing three of these values, you can calculate the fourth, which is a vital aspect of this analytical method.
The relationship between the volume and concentration of the titrant (the solution of known concentration) and the analyte (the solution of unknown concentration) is described by the titration formula: \( C_aV_a = C_bV_b \), where \( C_a \) and \( C_b \) are the concentrations of the acid and base, and \( V_a \) and \( V_b \) are the respective volumes. By knowing three of these values, you can calculate the fourth, which is a vital aspect of this analytical method.