Question

Describe how you would prepare each of the following aqueous solutions: (a) \(1.50\) 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 \(\mathrm{HCl}\).

Short answer

To prepare the four aqueous solutions: (a) Measure \(21.76\ \mathrm{g}\) of solid \((\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}\), add to a \(1.50\ \mathrm{L}\) volumetric flask, fill with water, and mix until dissolved. (b) Measure \(17.97\ \mathrm{g}\) of solid \(\mathrm{Na}_2\mathrm{CO}_3\) and \(102.03\ \mathrm{g}\) of water, combine and stir until dissolved. (c) Measure \(207\ \mathrm{g}\) of solid \(\mathrm{Pb}(\mathrm{NO}_3)_2\), add to a \(1.20\ \mathrm{L}\) volumetric flask, fill with water, and mix until dissolved. (d) Measure \(8.33\ \mathrm{mL}\) of \(6.0\ \mathrm{M}\) \(\mathrm{HCl}\) solution, add to a beaker containing \(5.5\ \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_2\), and stir until the reaction is complete and the solid is dissolved.

Step-by-step solution

Case (a)

To prepare \(1.50\) L of \(0.110\ \mathrm{M}\) \((\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}\) solution, we first need to find out how many grams of the solid is required. The formula to find mass is: Mass = Molarity \(\times\) Volume (in liters) \(\times\) Molar mass of solute The molar mass of \((\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}\) is \(2\times(14.0 + 4\times1.01) + 32.07 + 4\times16.00 = 132.14\ \mathrm{g/mol}\). Now let's calculate the mass required: Mass = \((0.110\ \mathrm{M})\times (1.50\ \mathrm{L})\times(132.14\ \mathrm{g/mol}) = 21.76\ \mathrm{g}\) So, to prepare the solution: 1. Measure \(21.76\ \mathrm{g}\) of solid \((\mathrm{NH}_{4})_{2}\mathrm{SO}_{4}\) using a balance. 2. Add the solid in a volumetric flask. 3. Add water to the flask up to the \(1.50\ \mathrm{L}\) mark. 4. Stopper the flask and mix the solution until the solid is dissolved.

Case (b)

For the second case, we need to prepare \(120\ \mathrm{g}\) of a solution that is \(0.65\ \mathrm{m}\) in \(\mathrm{Na}_2\mathrm{CO}_3\). Since molality (\(m\)) is given, we can use the formula: Molality = moles of solute / mass of solvent (in kg) We need to find out the moles of \(\mathrm{Na}_2\mathrm{CO}_3\), and the mass of solvent required. Molar mass of \(\mathrm{Na}_2\mathrm{CO}_3 = 2\times22.99 + 12.01 + 3\times16.00 = 105.99\ \mathrm{g/mol}\). Let the mass of \(\mathrm{Na}_2\mathrm{CO}_3\) required be 'x'. Then mass of solvent (water) would be \(120 - x\ \mathrm{g}\). From the given molality, we have: \(0.65\ \mathrm{m} = \frac{x\ \mathrm{moles}}{(120-x)\ \mathrm{g}}\) We need to convert the mass of solute (x) to moles: \(\frac{x}{105.99\ \mathrm{g/mol}}=\frac{x\ \mathrm{moles}}{(120-x)\ \mathrm{g}} \times 0.65\ \mathrm{m}\) By solving the equation, we get \(x \approx 17.97\ \mathrm{g}\), which gives the mass of water as \(120 - 17.97 = 102.03\ \mathrm{g}\). Preparing the solution: 1. Measure \(17.97\ \mathrm{g}\) of solid \(\mathrm{Na}_2\mathrm{CO}_3\) using a balance. 2. Measure \(102.03\ \mathrm{g}\) of water. 3. Add the solid to the water. 4. Stir the mixture until the solid is dissolved.

Case (c)

For the third case, we are given the mass percentage of solute and density of the solution, and we need to prepare \(1.20\ \mathrm{L}\) of the solution. Let the mass of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) be 'x', then the mass of the solution is \(\frac{x}{0.150}\). The volume of the solution can be calculated using the given density: \(1.16\ \mathrm{g/mL}=\frac{x}{0.150\ \mathrm{g}} / (1.20\ \mathrm{L} \times 1000\ \mathrm{mL})\) By solving the equation, we get \(x \approx 207\ \mathrm{g}\). Preparing the solution: 1. Measure \(207\ \mathrm{g}\) of solid \(\mathrm{Pb}(\mathrm{NO}_3)_2\) using a balance. 2. Add the solid in a volumetric flask. 3. Add water to the flask up to the \(1.20\ \mathrm{L}\) mark. 4. Stopper the flask and mix the solution until the solid is dissolved.

Case (d)

For the last case, we need to find the volume of \(6.0\ \mathrm{M}\) \(\mathrm{HCl}\) that would neutralize \(5.5\ \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_2\). From the balanced equation: \(\mathrm{Ba}(\mathrm{OH})_2 + 2\mathrm{HCl} \longrightarrow \mathrm{BaCl}_2 + 2\mathrm{H_2O}\) Number of moles of \(\mathrm{Ba}(\mathrm{OH})_2\) = \(\frac{5.5\ \mathrm{g}}{137.33 + 2\times(15.999 + 1.01)\ \mathrm{g/mol}} = 0.025\ \mathrm{moles}\) From the stoichiometry, \(2\ \mathrm{moles}\) of \(\mathrm{HCl}\) are needed to neutralize \(1\ \mathrm{mole}\) of \(\mathrm{Ba}(\mathrm{OH})_2\). Therefore, number of moles of \(\mathrm{HCl}\) needed = \(2\times0.025 = 0.050\ \mathrm{moles}\). Volume of \(6.0\ \mathrm{M}\) \(\mathrm{HCl}\) = \(\frac{0.050\ \mathrm{moles}}{6.0\ \mathrm{M}} = 0.00833\ \mathrm{L}\) or \(8.33\ \mathrm{mL}\) To prepare the solution: 1. Measure \(8.33\ \mathrm{mL}\) of \(6.0\ \mathrm{M}\) \(\mathrm{HCl}\) solution using a pipette. 2. Add the \(\mathrm{HCl}\) solution to a beaker containing \(5.5\ \mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_2\). 3. Stir the mixture until the reaction is complete and the solid is dissolved.

Key concepts

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. This unit is often used in chemistry for reactions in solutions and is represented by the symbol \( M \). For example, if you have a \(1.50\) L of a \(0.110 \mathrm{M}\) \((\mathrm{NH}_4)_2 \mathrm{SO}_4\) solution, this means there are \(0.110\) moles of \((\mathrm{NH}_4)_2 \mathrm{SO}_4\) per liter of solution. To calculate the amount of solid needed to prepare a solution of a certain molarity, you can use the formula:

• Mass = Molarity \( \times \) Volume (L) \( \times \) Molar mass \( (\mathrm{g/mol}) \)

This calculation tells you how much of the solid solute you need to dissolve in a specific volume of solvent to achieve the desired molarity. For example, in our case, you calculate the mass to find that \(21.76 \ \,\mathrm{g}\) of \((\mathrm{NH}_4)_2 \mathrm{SO}_4\) is needed.It is important to measure the solute accurately and dissolve it completely to ensure the solution is homogeneously mixed, providing uniform concentration throughout.

Molality

Molality is another way to describe the concentration of a solution, especially useful when temperature changes are involved because it does not change with temperature. It is defined as the number of moles of solute per kilogram of solvent, and it is represented by the symbol \( m \).In the preparation of solutions where molality is the given concentration, such as a \(0.65 \mathrm{m}\) \(\mathrm{Na}_2\mathrm{CO}_3\) solution, you need to determine the moles of solute and the mass of solvent required. The formula is:

• Molality = Moles of solute / Mass of solvent (kg)

Knowing that we need to prepare a \(120\ \, \mathrm{g}\) solution, and using the molality, you first determine the mass of the solute (in this case, \( \approx 17.97 \ \,\mathrm{g}\) of \(\mathrm{Na}_2\mathrm{CO}_3\)) and then subtract this from the total mass to get the solvent mass. Ensure the solute is well-dissolved, and the total solution mass aligns with the given requirements.

Mass Percentage

Mass percentage is a way to express the concentration of a component in a mixture as a percentage of the total mixture mass. It is calculated by dividing the mass of the solute by the total mass of the solution and then multiplying by 100. This measurement is essential when dealing with solutions in which the density is not routinely defined in molar concentration units.For instance, preparing a \(15.0\%\) \(\mathrm{Pb}(\mathrm{NO}_3)_2\) solution involves calculating the mass of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) and determining how much of this mass is relative to the \(1.20 \mathrm{L}\) solution volume. Using the formula:

• Mass percentage = (Mass of solute / Total mass of solution) \( \times 100\% \)

Given the solution's density of \(1.16 \ \, \mathrm{g/mL}\), you can find out how much of the solid solute to add to achieve the desired concentration. This concept helps in industries like pharmaceuticals, where precise solution concentrations are critical.

Stoichiometry

Stoichiometry involves the calculation and measurement of the quantities of reactants and products in chemical reactions. It ensures that reactions are balanced according to the laws of conservation of mass.For example, when neutralizing \(5.5\ \,\mathrm{g}\) of \(\mathrm{Ba}(\mathrm{OH})_2\) using \(\mathrm{HCl}\), stoichiometry enables you to determine how much \(\mathrm{HCl}\) is needed. The balanced chemical reaction is:\[ \mathrm{Ba}(\mathrm{OH})_2 + 2\mathrm{HCl} \rightarrow \mathrm{BaCl}_2 + 2\mathrm{H_2O} \]From the equation, \(2\) moles of \(\mathrm{HCl}\) are necessary to neutralize \(1\) mole of \(\mathrm{Ba}(\mathrm{OH})_2\). By using the stoichiometric ratios and knowing the molar mass of each reactant, you calculate that you need \(0.050\ \,\mathrm{mol}\) of \(\mathrm{HCl}\). To find the volume of \(6.0\ \, \mathrm{M}\) \(\mathrm{HCl}\) required, use the formula:

• Volume = Moles / Molarity

Once calculated, this information ensures enough \(\mathrm{HCl}\) is present to complete the neutralization, vital for processes such as titrations and industrial reactions.