Problem 44 - General Chemistry: Principles and Modern Applications

Question

What volume of \(0.750 \mathrm{M} \mathrm{AgNO}_{3}\) must be diluted with water to prepare \(250.0 \mathrm{mL}\) of \(0.425 \mathrm{M} \mathrm{AgNO}_{3} ?\)

Step-by-Step Solution

Answer

To prepare \(250.0 \, mL\) of \(0.425 \, M \, \mathrm{AgNO}_{3}\), one must dilute 142 mL of \(0.750 \, M \, \mathrm{AgNO}_{3}\) solution.

Step 1: Identify Known Variables

The initial molarity, \(M_1\) of \(0.750 \, M\), the final volume, \(V_2\) of \(250.0 \, mL\) or \(0.250 \, L\), and the final molarity, \(M_2\) of \(0.425 \, M\), were provided.

Step 2: Use the Dilution Formula to Solve for \(V_1\)

Substitute the known values into the equation, \(M_1 \cdot V_1 = M_2 \cdot V_2\), and solve for \(V_1 = \frac{M_2 \cdot V_2}{M_1} = \frac{0.425 \, M \cdot 0.250 \, L}{0.750 \, M} = 0.142 \, L\).

Step 3: Convert \(V_1\) to Milliliters (mL)

Convert \(V_1\) from liters to milliliters, using the conversion fact that \(1 L = 1000 mL\), thus \(0.142 \, L = 142 \, mL\).

Key Concepts

Molarity

Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, and is denoted by the symbol M. This unit of measurement allows chemists to understand how much solute is present in a given volume of solution. For example, a solution with a molarity of 0.750 M contains 0.750 moles of solute (such as AgNO₃) in every liter of that solution.
This concept is crucial for quantitative chemistry, especially in reactions where the amount of reactant determines the progression and outcome of the reaction. The high or low molarity of a solution can affect reaction rates and the equilibrium state. Calculating the molarity involves determining the moles of solute and dividing by the solution volume in liters.

Volume Conversion

In chemistry, it's often necessary to convert between different units of volume, such as milliliters (mL) and liters (L). This helps in inter-relating the quantities in practical laboratory operations. The most common conversion factor used is:

• 1 Liter (L) = 1000 Milliliters (mL)

Understanding how to convert between these units is essential, especially when applying formulas that require volume in a specific unit. For instance, in the given problem, after calculating the required volume in liters, it had to be converted to milliliters for practical use.
Such conversions ensure precision and consistency in measurements, enabling accurate calculations of concentrations and reactions.

Chemical Concentration

Chemical concentration refers to the amount of solute present in a given amount of solvent or solution. It's a critical concept as it affects how substances interact in a solution. There are various ways to express concentration, but molarity is one of the most common.
The concentration impacts key properties of a solution such as boiling point, freezing point, and vapor pressure, which, in turn, are important for both theoretical calculations and practical lab work. Understanding concentration allows chemists to predict how solutions will behave under different conditions and to ensure that chemical reactions proceed effectively.
Accurately diluting solutions, as seen in the problem, ensures that the correct concentration of solutes is achieved, which is vital for experiments where precise conditions must be met.

Dilution Formula

The dilution formula is a handy tool for chemists and students alike. It is used to calculate how much of a solution you need to dilute to achieve a desired concentration. The formula is:

• \( M_1 \cdot V_1 = M_2 \cdot V_2 \)

where \(M_1\) and \(M_2\) are the initial and final molarities, and \(V_1\) and \(V_2\) are the initial and final volumes, respectively.
In the exercise, we used this formula to determine the initial volume of the more concentrated solution needed to make a desired less concentrated solution. By rearranging the formula to solve for \(V_1\), we inserted the known values and calculated the required volume.
This formula helps to quickly adjust the concentration of solutions without lengthy trial-and-error, making it indispensable in both industrial applications and academic settings.