Chapter 6: Problem 41
An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?
Short Answer
Expert verified
42.02 mL of 11.9 M HCl are required to make 250 mL of 2.00 M HCl.
Step by step solution
01
Understanding the Concept of Molarity and Dilution
First, understand that molarity (M) is a way of expressing the concentration of a solution, defined as the number of moles of solute per liter of solution. The problem is asking for the volume of a more concentrated solution (11.9 M HCl) needed to create a certain volume (250 mL) of a less concentrated solution (2.00 M HCl). This is a dilution problem, where the formula M1V1 = M2V2 can be used, where M1 and V1 are the molarity and volume of the concentrated solution and M2 and V2 are the molarity and volume of the diluted solution.
02
Setting Up the Equation
Use the dilution formula, M1V1 = M2V2, where M1 = 11.9 M (initial concentration), M2 = 2.00 M (final concentration), V2 = 250 mL (final volume), and V1 is the unknown volume of the concentrated solution that we want to find.
03
Solving for the Unknown Volume (V1)
Rearrange the dilution formula to solve for V1: V1 = (M2V2) / M1. Substitute the known values into the equation: V1 = (2.00 M * 250 mL) / 11.9 M.
04
Calculating the Volume Required
Perform the calculations: V1 = (500 M*mL) / 11.9 M. V1 = 42.02 mL. Therefore, 42.02 mL of 11.9 M HCl is required.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
Molarity is a fundamental concept in chemistry that defines the concentration of a solution. It's expressed as the number of moles of a solute contained in one liter of solution, and it's depicted as moles per liter (M or mol/L). A mole is a unit that represents a specific number of particles, usually atoms or molecules, and that number is Avogadro's number, which is approximately \(6.022 \times 10^{23}\).
When you come across a problem that mentions a '2.00-M solution', it indicates there are 2 moles of solute in every liter of that solution. Understanding molarity can help you perceive how much of a substance is present in a certain volume of liquid, which is crucial for reactions and experiments that need precise concentrations.
When you come across a problem that mentions a '2.00-M solution', it indicates there are 2 moles of solute in every liter of that solution. Understanding molarity can help you perceive how much of a substance is present in a certain volume of liquid, which is crucial for reactions and experiments that need precise concentrations.
Concentration of Solution
The concentration of a solution conveys the amount of solute dissolved in a given quantity of solvent. It's essential for predicting the outcome of reactions and for preparing solutions for lab experiments. Besides molarity, there are other ways to express concentration, like mass percent, volume percent, or mole fraction. Still, molarity is widely used because it directly relates the volume of the solution to the amount of substance in moles.
To determine the concentration, you must know the ratio of the amount of solute to the volume of the solution. This comes handy in various chemical calculations, where understanding the relationship between solute, solvent, and their combined volume will allow for accurate formulations and manipulations of chemical substances.
To determine the concentration, you must know the ratio of the amount of solute to the volume of the solution. This comes handy in various chemical calculations, where understanding the relationship between solute, solvent, and their combined volume will allow for accurate formulations and manipulations of chemical substances.
Dilution Calculation
Dilution calculation is a process used to reduce the concentration of a solution. It's a vital procedure in chemistry for preparing solutions of desired molarity. The concept is straightforward: You are essentially adding more solvent to your solution, which decreases the solute's concentration without changing the amount of solute present in the mix.
Dilution problems are commonly resolved using the formula \(M1V1 = M2V2\), which is known as the dilution equation. Here, \(M1\) and \(V1\) represent the initial molarity and volume, and \(M2\) and \(V2\) are the final molarity and volume, respectively. This is an invaluable equation because it allows you to calculate how much of a concentrated stock solution is needed to create a new diluted solution. It's crucial to recognize that in a dilution, the total amount of solute remains unchanged; it's the volume that increases, hence decreasing the molarity. By using this method, chemists can prepare a variety of solutions with precision to facilitate their experiments and research.
Dilution problems are commonly resolved using the formula \(M1V1 = M2V2\), which is known as the dilution equation. Here, \(M1\) and \(V1\) represent the initial molarity and volume, and \(M2\) and \(V2\) are the final molarity and volume, respectively. This is an invaluable equation because it allows you to calculate how much of a concentrated stock solution is needed to create a new diluted solution. It's crucial to recognize that in a dilution, the total amount of solute remains unchanged; it's the volume that increases, hence decreasing the molarity. By using this method, chemists can prepare a variety of solutions with precision to facilitate their experiments and research.