Question

(a) How would you prepare \(175.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution starting with pure solute? (b) An experiment calls for you to use \(100 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HNO}_{3}\) solution. All you have available is a bottle of \(3.6 \mathrm{M} \mathrm{HNO}_{3}\). How would you prepare the desired solution?

Short answer

(a) To prepare a 0.150 M AgNO3 solution with a total volume of 175.0 mL using pure AgNO3 solute, first calculate the moles of AgNO3 required (0.02625 mol). Then, calculate the mass of AgNO3 needed (4.4581 g) and dissolve it in distilled water, adding water until the total solution reaches 175.0 mL. (b) To prepare 100 mL of 0.50 M HNO3 solution from a stock bottle of 3.6 M HNO3 solution, use the dilution equation to find the volume of stock solution required (13.89 mL). Measure this volume, transfer it to a 100 mL volumetric flask, and add distilled water until the volume reaches 100 mL to obtain the desired 0.50 M HNO3 solution.

Step-by-step solution

Problem (a): Preparing a solution from pure solute

To prepare a 0.150 M AgNO3 solution with a total volume of 175.0 mL using pure AgNO3 solute, we need to calculate the mass of the pure solute to be added. 1. Calculate the moles of AgNO3 required: Use the formula, moles = Molarity x Volume, where Molarity = 0.150 M and Volume = 175.0 mL (0.175 L). Moles of AgNO3 = \(0.150 \times 0.175\) = 0.02625 mol 2. Calculate the mass of AgNO3 needed: To find the mass, use the formula Mass = Moles × Molar Mass, where Molar Mass of AgNO3 = 169.87 g/mol. Mass of AgNO3 = \(0.02625 \times 169.87\) = 4.4581 g 3. Prepare the solution: To prepare the solution, dissolve 4.4581 g of pure AgNO3 in distilled water, and then add water until the total volume of the solution reaches 175.0 mL.

Problem (b): Diluting a concentrated solution

To prepare 100 mL of 0.50 M HNO3 solution from a stock bottle of 3.6 M HNO3 solution, we need to determine the volume of the stock solution required and then dilute it to the desired molarity and volume. 1. Calculate the volume of stock solution required: Use the dilution equation: \(C_1V_1 = C_2V_2\), where \(C_1\) is the initial concentration (3.6 M), \(V_1\) is the volume of the initial solution to be determined, \(C_2\) is the final concentration (0.50 M), and \(V_2\) is the final volume (100 mL). Rearrange the equation to get the initial volume: \(V_1 = \frac{C_2V_2}{C_1} = \frac{(0.50)(100)}{3.6}\) Volume of stock solution (V1) = 13.89 mL 2. Prepare the solution: To prepare the solution, carefully measure 13.89 mL of the 3.6 M HNO3 stock solution using a graduated cylinder or pipette, transfer it to a 100 mL volumetric flask, and add distilled water until the volume reaches 100 mL. This will give you the desired 0.50 M HNO3 solution.

Key concepts

Molarity

Molarity is a fundamental concept in chemistry that measures the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution, and its unit is mol/L or M. Calculating molarity involves knowing both the number of moles of a solute and the volume of the solution in liters. This makes it a useful measure for preparing solutions with precise concentrations.

For instance, when preparing a solution of silver nitrate (AgNO3), knowing the required molarity allows us to determine the exact mass of solute needed. The equation used here is: \[ ext{Molarity (M)} = rac{ ext{moles of solute}}{ ext{liters of solution}} \] This relationship helps ensure that chemical reactions proceed correctly and efficiently by maintaining consistent reactant concentrations.

Remember, molarity is crucial when calculating reagents for reactions and in preparing standard solutions in laboratory settings. It provides a reliable way to express solution strength, which is essential for quantitative analysis.

Dilution

Dilution is the process of reducing the concentration of a solute in a solution, typically by adding more solvent. This concept is common in labs when working with concentrated stock solutions. To achieve a desired concentration, the dilution equation is used: \[ C_1V_1 = C_2V_2 \] where \(C_1\) and \(V_1\) are the concentration and volume of the initial solution, respectively, and \(C_2\) and \(V_2\) are the desired concentration and final volume.

This equation illustrates that the product of the initial concentration and volume equals the product of the desired concentration and volume. Thus, it aids in calculating the correct volume of stock to dilute. For example, if you have a 3.6 M HNO3 solution and need a 0.50 M solution, you can figure out how much of the stock solution is needed before adding water.

Dilution is particularly useful because it allows the transformation of small amounts of concentrated solutions into larger volumes of less concentrated solutions, which are often safer and easier to handle in experiments.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is a tool chemists use to calculate how much of each reactant is required to form a given amount of product. This concept extends to solution preparation and analysis.

In tasks like preparing solutions, stoichiometry helps determine the precise amounts of substances required. It involves balanced chemical equations, which ensure that matter is conserved in a reaction (following the Law of Conservation of Mass). For example, preparing a specific Molar solution involves stoichiometric conversions between mass, moles, and volume.

Stoichiometry also aids in scaling up laboratory experiments to industrial scales, ensuring the same proportions of reactants are used, regardless of the quantities involved. Its accuracy is indispensable for efficient chemical manufacturing, analysis, and research, allowing chemists to predict yields and optimize reactant usage.