Problem 54 - Chemistry The Central Science In Si Units

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

Step-by-Step Solution

Answer

Prepare (a) by dissolving 21.8 g of (NH4)2SO4 in water to make 1.50 L. (b) Dissolve 15.50 g Na2CO3 in water to a total of 225 g. (c) Dissolve 208.8 g Pb(NO3)2 in water to make 1.20 L. (d) Dilute 0.128 L of 0.50 M HCl from 6.0 M HCl to neutralize 5.5 g Ba(OH)2.

Step 1: Calculate Mass of (NH4)2SO4 Needed

For a 1.50 L solution with a molarity of 0.110 M of \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\), first calculate the moles required: \[\text{Moles} = 1.50 \, \text{L} \times 0.110 \, \mathrm{M} = 0.165 \, \text{mol}\].Determine the molar mass of \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\):\[\text{Molar mass} = 2(14.01 + 4 \times 1.008) + 32.07 + 4 \times 16.00 = 132.14 \, \text{g/mol}\].Calculate mass of \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\):\[\text{Mass} = 0.165 \, \text{mol} \times 132.14 \, \text{g/mol} = 21.8 \, \text{g}\].Dissolve 21.8 g of solid \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\) in enough water to make 1.50 L of solution.

Step 2: Calculate Mass of Na2CO3 for Molal Solution

For 225 g of a 0.65 molal \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution, the molality formula is:\[\text{Molality} = \frac{\text{Moles of solute}}{\text{kg of solvent}}\].First, calculate moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\):\[\text{Moles} = 0.65 \, \text{mol/kg} \times \frac{225 \, \text{g} - \text{mass of Na}_2 \text{CO}_3}{1000} \, \text{kg}\], which simplifies to \(0.14625 \, \text{mol}\).Calculate mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\):Molar mass = 2(23.00) + 12.01 + 3(16.00) = 105.99 g/mol.Thus, mass = 0.14625 mol \(\times\) 105.99 g/mol = 15.50 g.Dissolve 15.50 g of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in enough water to make a total mass of 225 g.

Step 3: Mass of Pb(NO3)2 for Percent by Mass Solution

For a 1.20 L solution with 15.0% \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) by mass:Density of solution = 1.16 g/mL means the total mass of 1.20 L = 1.20 L \(\times\) 1000 mL/L \(\times\) 1.16 g/mL = 1392 g.15.0% by mass implies:\[\text{Mass of Pb}(\mathrm{NO}_{3})_{2} = 0.15 \times 1392 \, \text{g} = 208.8 \, \text{g}\].Use 208.8 g of \(\mathrm{Pb}(\mathrm{NO}_{3})_{2}\) and dissolve in enough water to create a total volume of 1.20 L.

Step 4: Volume of HCl Solution for Neutralization

To neutralize 5.5 g of \(\mathrm{Ba}(\mathrm{OH})_{2}\), use the reaction:\[\mathrm{Ba}(\mathrm{OH})_{2} + 2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_{2} + 2 \mathrm{H}_{2}\text{O}\].Calculate moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\): Molar mass = 137.33 + 2(15.999 + 1.008) = 171.35 g/mol. Moles = 5.5 g \(\div\) 171.35 g/mol = 0.0321 mol.Since 2 moles of HCl are required for each mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\):Moles of HCl = 2 \(\times\) 0.0321 = 0.0642 mol.Using a 0.50 M solution: Volume needed = 0.0642 mol \(\div\) 0.50 M = 0.128 L.Prepare 128 mL of the 0.50 M solution from 6.0 M HCl by performing a dilution.

Key Concepts

Molarity Calculations

Molarity is a fundamental concept in chemistry that describes the concentration of a solute in a solution. It's expressed as the number of moles of solute per liter of solution (mol/L). Understanding molarity is crucial when preparing solutions since it directly affects the reactions and properties of the solution.
To calculate molarity, use the formula:

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

For instance, if you need to prepare a 1.50 L solution with a molarity of 0.110 M of \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\), start by determining the moles of solute required by multiplying the desired molarity by the volume in liters. In this case: \(1.50 \, \text{L} \times 0.110 \, \mathrm{M} = 0.165 \, \text{mol}\).

• Next, calculate the molar mass of the solute to find out how much of the solid you need. For \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\), the molar mass is 132.14 g/mol.
• This means you need 0.165 mol \(\times\) 132.14 g/mol = 21.8 g of \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\) to achieve the desired concentration.

Finally, dissolve this amount of the solute in enough water to make up the total volume of the solution, which is 1.50 L in this example.

Mass Percent Solutions

Mass percent is a way to express the concentration of a solution. It indicates how many grams of solute are present per 100 grams of solution. This can be particularly helpful in industrial and laboratory settings where precise measurements are crucial.
Calculate mass percent using the formula:

• \( \text{Mass percent} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100 \% \)

For example, if you need to prepare 1.20 L of a solution that is 15.0% \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass, you'll need the total mass of the solution. Knowing the density of the solution (1.16 g/mL), you can calculate the total mass by multiplying the volume by the density.

• Total mass = 1.20 L \(\times\) 1000 mL/L \(\times\) 1.16 g/mL = 1392 g.
• With a 15.0% mass percentage, the mass of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) needed is 0.15 \(\times\) 1392 g = 208.8 g.

Dissolve this amount of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) in water to make up the total volume of 1.20 L.

Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. These reactions are important in a variety of fields, including chemistry and biology, as they can be applied in titration and chemical synthesis.
The balanced chemical equation for a neutralization reaction that involves hydrochloric acid and barium hydroxide is:

• \(\mathrm{Ba}(\mathrm{OH})_{2} + 2 \mathrm{HCl} \rightarrow \mathrm{BaCl}_{2} + 2 \mathrm{H}_{2}\text{O}\)

To neutralize 5.5 g of \(\mathrm{Ba}(\mathrm{OH})_{2}\), calculate its moles using the molar mass (171.35 g/mol). The moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) are given by: 5.5 g \(\div\) 171.35 g/mol = 0.0321 mol.

• Since the stoichiometry of the reaction requires two moles of HCl for every mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\), you will need 2 \(\times\) 0.0321 mol = 0.0642 mol of HCl.
• Using a 0.50 M solution, determine the volume needed: 0.0642 mol \(\div\) 0.50 M = 0.128 L. This is equivalent to 128 mL of the 0.50 M solution.

Prepare this 128 mL by diluting a more concentrated 6.0 M HCl solution to achieve the desired molarity.