Problem 50 - 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) \(120 \mathrm{~g}\) 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 \(6.0\) M HCl.

Step-by-Step Solution

Answer

a) To prepare 1.50 L of 0.110 M (NH4)2SO4 solution: 1. Measure and add 21.78 g of solid (NH4)2SO4 to a 1.50 L volumetric flask. 2. Add distilled water to the flask until it reaches the 1.50 L mark. 3. Mix the solution thoroughly to dissolve the solute completely. b) To prepare 120 g of a 0.65 m Na2CO3 solution: 1. Measure and add 78.23 g of solid Na2CO3. 2. Add distilled water until the total mass of the solution reaches 120 g. 3. Mix the solution thoroughly to dissolve the solute completely. c) To prepare 1.20 L of a 15.0% Pb(NO3)2 by mass solution: 1. Measure and add 208.8 g of solid Pb(NO3)2 to a 1.20 L volumetric flask 2. Add distilled water to the flask until it reaches the 1.20 L mark. 3. Mix the solution thoroughly to dissolve the solute completely. d) To prepare a 0.50 M solution of HCl that would just neutralize 5.5 g of Ba(OH)2, starting with 6.0 M HCl: 1. Measure and add 5.36 mL of 6.0 M HCl to a volumetric flask. 2. Add distilled water to the flask until the total volume reaches the volume that gives the desired 0.50 M HCl solution. 3. Mix the solution thoroughly.

Step 1: a) Calculate Moles

To prepare the aqueous solution, first calculate the moles of (NH4)2SO4 needed using the formula: Moles = Molarity × Volume Moles = 0.110 M × 1.50 L = 0.165 mol

Step 2: a) Calculate Mass

Convert moles of (NH4)2SO4 to grams using the molar mass: Molar mass of (NH4)2SO4 = 2×(14+4×1)+32+4×16 = 132 g/mol Mass = Moles × Molar mass Mass = 0.165 mol × 132 g/mol = 21.78 g

Step 3: a) Prepare the Solution

To prepare the solution: 1. Measure and add 21.78 g of solid (NH4)2SO4 to a 1.50 L volumetric flask. 2. Add distilled water to the flask until it reaches the 1.50 L mark. 3. Mix the solution thoroughly to dissolve the solute completely.

Step 4: b) Calculate Moles

In this case, we first calculate the moles of Na2CO3 using the formula: Moles = (Mass × Molality) / Molar mass Molar mass of Na2CO3 = (2×23) + 12 + (3×16) = 106 g/mol Moles = (120 g × 0.65 mol/kg) / 106 g/mol = 0.738 mol

Step 5: b) Calculate Mass

Convert moles of Na2CO3 to grams using the molar mass: Mass = Moles × Molar mass Mass = 0.738 mol × 106 g/mol = 78.23 g

Step 6: b) Prepare the Solution

To prepare the solution: 1. Measure and add 78.23 g of solid Na2CO3. 2. Add distilled water until the total mass of the solution reaches 120 g. 3. Mix the solution thoroughly to dissolve the solute completely.

Step 7: c) Calculate Mass of Pb(NO3)2

In this case, we first calculate the mass of Pb(NO3)2 using the given percentage: Mass of Pb(NO3)2 = (Solution mass × Percentage) / 100 Volume = 1.20 L = 1200 mL Solution mass = 1200 mL × 1.16 g/mL = 1392 g Mass of Pb(NO3)2 = (1392 g × 15.0%) / 100 = 208.8 g

Step 8: c) Prepare the Solution

To prepare the solution: 1. Measure and add 208.8 g of solid Pb(NO3)2 to a 1.20 L volumetric flask. 2. Add distilled water to the flask until it reaches the 1.20 L mark. 3. Mix the solution thoroughly to dissolve the solute completely.

Step 9: d) Calculate Volume of HCl

In this case, we first calculate the moles of Ba(OH)2 using the mass given: Molar mass of Ba(OH)2 = 137 + 2×(16+1) = 171 g/mol Moles of Ba(OH)2 = 5.5 g / 171 g/mol = 0.03216 mol Since 1 mol of HCl neutralizes 1 mol of Ba(OH)2, the moles of HCl needed are also 0.03216 mol. Then, calculate the volume of 6.0 M HCl needed using the equation: Volume = Moles / Molarity Volume = 0.03216 mol / 6.0 M = 0.00536 L or 5.36 mL

Step 10: d) Prepare the Solution

To prepare the solution: 1. Measure and add 5.36 mL of 6.0 M HCl to a volumetric flask. 2. Add distilled water to the flask until the total volume reaches the volume that gives the desired 0.50 M HCl solution. 3. Mix the solution thoroughly.

Key Concepts

Molarity

Molarity, often symbolized as M, is a measure of the concentration of a solute in a solution. It represents the number of moles of a substance dissolved in one liter of solution. This concept is crucial when preparing solutions because it ensures precision and consistency in the chemical makeup of the solution.

To calculate molarity, you use the formula:

• \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \)

Let's consider one of our examples: to prepare 1.50 L of a 0.110 M \((NH_4)_2SO_4\) solution, you must first calculate the moles using the given molarity and volume.

• Calculate moles = 0.110 M × 1.50 L = 0.165 mol

Then, convert these moles into grams using the compound's molar mass. This step helps measure out the precise amount of solid needed to achieve the intended molarity when mixed with water.

Mass Percent

Mass percent is a way of expressing the concentration of an element in a compound or a component in a mixture. It quantifies how much of a specific component is present on a mass basis, usually in a mass/mass percent.To calculate mass percent, use the following formula:

• \( \text{Mass Percent} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100 \)

For instance, when preparing a 1.20 L solution of 15.0% \(\text{Pb(NO}_3)_2\) by mass, you first need to compute the mass of the entire solution and then determine how much \(\text{Pb(NO}_3)_2\) corresponds to this percentage.
In our exercise:

• Total solution mass = 1.20 L \( \times 1.16 \text{ g/mL} = 1392 \text{ g} \)
• Mass of \(\text{Pb(NO}_3)_2\) = \( \frac{1392 \text{ g} \times 15.0\%}{100} = 208.8 \text{ g} \)

With this calculation, you will measure out the exact amount of lead nitrate to add to the solution.

Molality

Molality is another concentration measure, denoted as m, that is slightly different from molarity. It considers the mass of the solvent rather than the volume of the solution. This measurement is particularly useful when working with temperature-variable processes since it doesn't change with temperature fluctuations.

The formula for molality is:

• \( \text{Molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \)

In the given exercise, to create a 0.65 m solution of \(\text{Na}_2\text{CO}_3\), you initially calculate the moles of \(\text{Na}_2\text{CO}_3\) needed using its molar mass. Then, these moles are used to find out the mass of the solute required:

• Calculate moles = \( \frac{120 \text{ g} \times 0.65 \text{ mol/kg}}{106 \text{ g/mol}} = 0.738 \text{ mol} \)
• Convert moles to grams = \(0.738 \text{ mol} \times 106 \text{ g/mol} = 78.23 \text{ g}\)

From here, you will combine these 78.23 grams with enough solvent to reach the total solution mass needed, ensuring a precise concentration.

Neutralization

Neutralization is a chemical reaction between an acid and a base that results in the formation of water and a salt, effectively canceling out any excess hydrogen or hydroxide ions in the solution. This reaction is essential in titrations and preparing specific solutions.In the exercise, you need a 0.50 M solution of HCl to neutralize a given mass of \(\text{Ba(OH)}_2\). Here's how it works:Let's say you have 5.5 grams of \(\text{Ba(OH)}_2\):

• First, calculate the moles of \(\text{Ba(OH)}_2\) you have by dividing mass by molar mass: \( \frac{5.5 \text{ g}}{171 \text{ g/mol}} = 0.03216 \text{ mol} \).
• The same amount of moles of HCl will be needed to completely neutralize it, as per the stoichiometry of the neutralization reaction (1:1 ratio here).

Measuring accurately, you will then add sufficient HCl to achieve the desired concentration while neutralizing the \(\text{Ba(OH)}_2\), completing the neutralization process.

Aqueous Solutions

Aqueous solutions are solutions where water is the solvent. They are the most common type of solutions and are frequently used in various scientific and industrial processes because water is an excellent solvent. When preparing an aqueous solution, like those described in the exercise, the solid solute is dissolved in water until the target concentration is reached. Key points when working with aqueous solutions include:

• Always measure the solid and liquid components accurately to ensure the right concentration.
• Mix the solution thoroughly to ensure complete dissolution of the solute.
• Use distilled or deionized water to prevent contamination that can skew results.

Understanding the nature of aqueous solutions is fundamental because they serve as the basis for many chemical reactions and processes. This knowledge applies directly when preparing solutions of specific concentrations, ensuring reliable results in every experiment or application.