Problem 54 - Chemistry The Central Science

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 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; (\boldsymbol{d} ) ~ a ~ 0.50 \(\mathrm{M}\) solution of HCl that would just neutralize 5.5 \(\mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_{2}\) starting with 6.0 \(\mathrm{M} \mathrm{HCl}\) .

Step-by-Step Solution

Answer

To prepare the four aqueous solutions: a) Measure 21.80 g of (NH₄)₂SO₄ and dissolve it in 1.50 L of distilled water to get a 0.110 M solution. b) Measure 14.29 g of Na₂CO₃ and dissolve it in enough distilled water to obtain a total mass of 225 g for a 0.65 m solution. c) Measure 208.8 g of Pb(NO₃)₂ and dissolve it in 1.20 L of distilled water to get a solution that is 15.0% Pb(NO₃)₂ by mass with a density of 1.16 g/mL. d) Dilute 10.7 mL of 6.0 M HCl with distilled water to reach a final volume of 128 mL, resulting in a 0.50 M HCl solution that can neutralize 5.5 g of Ba(OH)₂.

Step 1: To find the moles of (NH₄)₂SO₄ needed, we can use the formula: moles = Molarity (M) × Volume (L) moles = 0.110 M × 1.50 L = 0.165 mol of (NH₄)₂SO₄ Step 2: Convert moles to grams

To convert moles to grams, we use the formula: mass = moles × molar mass The molar mass of (NH₄)₂SO₄ = (2 × (1.0079 × 2 + 14.0067)) + (32.065 + 4 × 15.9994) = 132.14 g/mol mass = 0.165 mol × 132.14 g/mol = 21.80 g of (NH₄)₂SO₄ Step 3: Mix the solute and solvent

Step 2: Measure 21.80 g of solid (NH₄)₂SO₄ and dissolve it in a 1.50 L volumetric flask with distilled water up to the mark. #b) Preparing 225 g of a 0.65 m Na₂CO₃ solution from solid Na₂CO₃# Step 1: Calculate moles of solute

To find the moles of Na₂CO₃, we can use the formula: moles = molality (m) × mass of solvent (kg) Assuming the mass of the solution (mass_solution) is 225 g. Let the mass of the solute (mass_solute) be x grams, and the mass of the solvent (water) is (225−x) grams. moles = 0.65 m × (225−x) g / 1000 = (0.65 × (225−x)/1000) mol of Na₂CO₃ Step 2: Convert moles to grams

Step 3: The molar mass of Na₂CO₃ = (2 × 22.98977) + 12.0107 + (3 × 15.9994) = 105.99 g/mol mass_solute = x = moles × molar mass = (0.65 × (225−x)/1000) × 105.99 g/mol Step 3: Solve for x (mass of Na₂CO₃)

Solve the equation for x: x = (0.65 × (225−x)/1000) × 105.99 x = 14.29 g of Na₂CO₃ Step 4: Mix the solute and solvent

Step 4: Measure 14.29 g of solid Na₂CO₃ and dissolve it in a container with enough distilled water to make a total mass of 225 g of the solution. #c) Preparing 1.20 L of a solution that is 15.0% Pb(NO₃)₂ by mass; density = 1.16 g/mL# Step 1: Calculate mass of the solution

To find the mass of the solution (mass_solution), we use the formula: mass_solution = volume × density mass_solution = 1.20 L × 1000 mL/L × 1.16 g/mL = 1392 g of solution Step 2: Calculate mass of Pb(NO₃)₂

Step 5: To find the mass of Pb(NO₃)₂, we can use the formula: mass_solute = (% by mass × mass_solution) / 100 mass_solute = (15.0 × 1392) / 100 = 208.8 g of Pb(NO₃)₂ Step 3: Mix the solute and solvent

Measure 208.8 g of solid Pb(NO₃)₂ and dissolve it in a 1.20 L volumetric flask with distilled water up to the mark. #d) Preparing 0.50 M HCl solution that would just neutralize 5.5 g of Ba(OH)₂, starting with 6.0 M HCl# Step 1: Calculate moles of Ba(OH)₂

Step 6: First, we need to calculate the moles of Ba(OH)₂ using formula: moles = mass / molar mass The molar mass of Ba(OH)₂ = 137.327 + 2 × (15.9994 + 1.0079 × 2) = 171.36 g/mol moles of Ba(OH)₂ = 5.5 g / 171.36 g/mol = 0.0321 mol Step 2: Calculate moles of HCl

For the neutralization reaction: Ba(OH)₂ + 2 HCl → BaCl₂ + 2 H₂O The stoichiometry tells us that 1 mol of Ba(OH)₂ reacts with 2 mol of HCl. moles of HCl = 2 × 0.0321 mol = 0.0642 mol Step 3: Calculate volume of 6.0 M HCl

Step 7: To find the volume of 6.0 M HCl required, we can use the formula: volume = moles / molarity volume of 6.0 M HCl: 0.0642 mol / 6.0 M = 0.0107 L or 10.7 mL Step 4: Dilute the HCl solution

Dilute 10.7 mL of 6.0 M HCl with distilled water in a volumetric flask to obtain a final volume of 0.0642 mol ÷ 0.50 M ≈ 0.128 L (128 mL) (0.50 M HCl solution that would just neutralize 5.5 g of Ba(OH)₂).

Key Concepts

Molarity

Molarity is a way to express the concentration of a solution. It's one of the most common measurements used in chemistry. The definition of molarity is the number of moles of solute present in one liter of solution. This can be symbolized as M and is calculated by the formula: \( M = \frac{\text{moles of solute}}{\text{liter of solution}} \).
For instance, in the exercise, a solution of \( \text{0.110 M (NH}_4\text{)}_2\text{SO}_4 \) is prepared by dissolving 21.80 g of the solute in water to make a total solution volume of 1.50 L. The molarity indicates how "strong" or "concentrated" the solution is. This measurement helps chemists know how many particles are available in a certain volume when they conduct reactions.

• Considers volume of the solution.
• Different from molality which uses mass of solvent.

Molality

Molality is another way to measure concentration, focusing on the amount of solute per mass of solvent, not solution volume. It's represented by \( m \) and is calculated with the formula: \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). Molality does not change with temperature, unlike molarity which can vary as solution volume changes with temperature.
From the exercise, a solution prepared is \( 0.65 \text{ m Na}_2\text{CO}_3 \), meaning there are 0.65 moles of \( \text{Na}_2\text{CO}_3 \) in a kilogram of water. Here, you first calculate how much of the solid solute you need based on the mass you aim to achieve in your final solution, which is found to be 14.29 g of \( \text{Na}_2\text{CO}_3 \).

• Independent of temperature.
• Uses mass (more stable measure).

Neutralization

Neutralization is a chemical reaction where an acid and a base react to form water and a salt. This process is important in many applications including titrations, a method for determining concentrations.
In the provided exercise, to neutralize a base like \( \text{Ba(OH)}_2 \) with an acid like HCl, the molar amounts required follow the stoichiometric ratio derived from the reaction equation: \( \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \). Thus, each mole of \( \text{Ba(OH)}_2 \) needs 2 moles of HCl. For our solution, we find the amount of HCl needed to neutralize 5.5 g of \( \text{Ba(OH)}_2 \) is 10.7 mL of a 6.0 M HCl solution.

• Tells us how acids react with bases.
• Follows reaction stoichiometry.

Stoichiometry

Stoichiometry revolves around the quantitative relationships in chemical reactions, allowing us to predict the amounts of substances consumed and produced. It helps us balance reactions and ensures that the amount of reactant is adequate to produce the desired amount of product.
The exercise details how stoichiometry governs the reaction between \( \text{Ba(OH)}_2 \) and HCl. The stoichiometric coefficients from the balanced equation indicate the ratio of moles we need for complete reaction. We also calculate masses and volumes in other parts of the problem using stoichiometric principles.

• Balances equations.
• Predicts reactant and product quantities.