Question

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50 \mathrm{~L}\) of \(0.110 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) solution, starting with solid \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\); (b) \(225 \mathrm{~g}\) of a solution that is \(0.65 \mathrm{~m}\) in \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), starting with the solid solute; (c) \(1.20 \mathrm{~L}\) of a solution that is \(15.0 \% \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass (the density of the solution is \(1.16 \mathrm{~g} / \mathrm{mL}\) ), starting with solid solute; (d) a \(0.50 \mathrm{M}\) solution of \(\mathrm{HCl}\) that would just neutralize \(5.5 \mathrm{~g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with \(6.0 \mathrm{M} \mathrm{HCl}\)

Short answer

To prepare the given solutions, follow the steps below: a) For a 1.50 L of 0.110 M (NH4)2SO4 solution, weigh 21.80 g of solid (NH4)2SO4 and add it to a 2-liter volumetric flask. Add distilled water to make up the volume to 1.50 L. b) For a 225 g of a 0.65 m Na2CO3 solution, weigh 15.52 g of solid Na2CO3 and add it to a beaker. Add 209.48 g of distilled water to the beaker and mix. c) For a 1.20 L of a 15.0% w/w Pb(NO3)2 solution with a density of 1.16 g/mL, weigh 208.8 g of solid Pb(NO3)2 and add it to a 2-liter volumetric flask. Add distilled water to make up the volume to 1.20 L. d) To prepare a 0.50 M HCl solution to neutralize 5.5 g of Ba(OH)2, calculate the required volume (0.0107 L) of 6.0 M HCl. Dilute this volume with distilled water to achieve a 0.50 M HCl concentration.

Step-by-step solution

Case (a): Preparation of 1.50 L of 0.110 M (NH4)2SO4 solution.

Step 1: Calculate the required mass of (NH4)2SO4 The molar mass of (NH4)2SO4 is (2*(14.01+4*1.01)+32.07+4*16.00) = 132.14 g/mol. We need to prepare a 0.110 M solution, thus; Required mass = (Molarity × Volume) × Molar mass Required mass = (0.110 mol/L × 1.50 L) × 132.14 g/mol ≈ 21.80 g Step 2: Prepare the solution Weigh 21.80 g of solid (NH4)2SO4 and add it to a 2-liter volumetric flask. Add distilled water to dilute the solute, and stir the mixture until the solute dissolves completely. Fill up to the 1.50 L mark with distilled water and mix gently.

Case (b): Preparation of 225 g of a 0.65 m Na2CO3 solution.

Step 1: Calculate the required mass of Na2CO3 The molar mass of Na2CO3 is (2*22.99 + 12.01 + 3*16.00) = 105.99 g/mol. We need to prepare a 0.65 m solution, thus; Required mass = (molality × mass of solvent) × Molar mass Required mass = (0.65 mol/kg × 0.225 kg) × 105.99 g/mol ≈ 15.52 g Step 2: Prepare the solution Weigh 15.52 g of solid Na2CO3 and add it to a beaker. Add 225 g - 15.52 g = 209.48 g of distilled water to the beaker and stir the mixture until the solute dissolves completely.

Case (c): Preparation of 1.20 L of a solution that is 15.0% Pb(NO3)2 by mass.

Step 1: Calculate the mass of Pb(NO3)2 15.0% by mass means that in every 100 g of solution, there are 15 g of Pb(NO3)2. To prepare 1.20 L, we have: Total mass of solution = volume × density = 1.20 L × 1.16 g/mL × 1000 mL/L ≈ 1392 g So the mass of Pb(NO3)2 in the solution = 0.15 × 1392 g ≈ 208.8 g Step 2: Prepare the solution Weigh 208.8 g of solid Pb(NO3)2 and add it to a 2-liter volumetric flask. Add distilled water to dilute the solute, and stir the mixture until the solute dissolves completely. Fill up to the 1.20 L mark with distilled water and mix gently.

Case (d): Preparation of a 0.50 M solution of HCl that would just neutralize 5.5 g of Ba(OH)2 starting with 6.0 M HCl.

Step 1: Calculate the moles of Ba(OH)2 The molar mass of Ba(OH)2 is (137.33+2*(15.99+1.01)) = 171.34 g/mol. Moles of Ba(OH)2 = mass / molar mass = 5.5 g / 171.34 g/mol ≈ 0.0321 mol Step 2: Calculate the moles of HCl needed The balanced chemical equation for the neutralization reaction is: Ba(OH)2 + 2HCl -> BaCl2 + 2H2O Moles of HCl required = moles of Ba(OH)2 × 2 = 0.0321 mol × 2 ≈ 0.0643 mol Step 3: Calculate the volume of 6.0 M HCl required volume = moles / molarity = 0.0643 mol / 6.0 M ≈ 0.0107 L Step 4: Prepare the 0.50 M HCl solution to neutralize Ba(OH)2 Add the calculated volume (0.0107 L) of 6.0 M HCl to a volumetric flask. Dilute the HCl with distilled water until the molarity of the solution reaches 0.50 M. The balanced equation shows that one mole of Ba(OH)2 reacts with two moles of HCl, which means that the 0.50 M HCl solution will just neutralize the given 5.5 g of Ba(OH)2.

Key concepts

Solution Preparation

Understanding how to prepare a particular solution in chemistry requires insight into the type of solution needed and its concentration. For instance, preparing an aqueous solution typically involves dissolving a solid solute in water to achieve a desired molarity or molality. The process starts with calculating the amount of solute required, which is derived from the desired concentration and volume of the solution. The solute is then carefully weighed, often using a balance, and added to a volumetric flask. Distilled water is gradually added to dissolve the solute completely before the solution is diluted to the final target volume with additional distilled water.

Throughout this process, several key factors must be considered. First, the purity of the solute can impact the accuracy of the solution's concentration. Second, the temperature of the water can influence the solubility of the solute. Lastly, the precision of measuring instruments, such as the volumetric flask, is crucial for ensuring the expected concentration of the solution.

Molarity and Molality

When it comes to solution composition, molarity and molality are two critical concepts. Molarity, denoted as 'M', is the number of moles of solute per liter of solution. It is temperature dependent as the volume of the solution can change with temperature. In contrast, molality, denoted as 'm', is the number of moles of solute per kilogram of solvent. It does not change with temperature, making it a valuable measure when dealing with scenarios involving temperature variations.

Understanding Conversions

It is often necessary to convert between these units based on the requirements of an experiment or chemical process. This can be done using stoichiometry calculations and understanding the density of the solution, which allows for conversion between mass and volume.

Neutralization Reactions

Neutralization reactions involve the reaction of an acid with a base to form water and a salt. They are quintessential examples of double displacement reactions. When an acid donates its protons to a base, a neutralization occurs, and the objectives generally include finding the stoichiometric ratios of reactants or the concentration of one reactant when another reactant's mass or concentration is known.

Practical Application

Preparing a solution to neutralize a given amount of base, like in our exercises, requires knowledge of these stoichiometric ratios. It's important to consider that the stoichiometry can provide the exact equivalents needed for neutralization and, therefore, the precise composition of the required acidic solution.

Stoichiometry Calculations

Stoichiometry is the field of chemistry that pertains to the quantification of reactants and products in chemical reactions. These calculations are instrumental in predicting the outcomes of reactions and in preparing solutions with the precise reactant proportions.

Stoichiometry and Solution Preparation

When preparing solutions, stoichiometry calculations help identify the amount of reactants needed. This involves using the balanced chemical equation and the concept of mole ratios. For example, in a problem where the goal is to prepare an acid solution that will neutralize a given base, the stoichiometry of the neutralization reaction will guide the proportions of acid and base required. The calculations typically involve converting grams to moles, moles to moles (using the mole ratio), and finally moles back to grams or liters depending on whether we are working with molarity or molality.