Problem 53 - Chemistry: The Central Science in SI Units, Global Edition

Question

Describe how you would prepare each of the following aqueous solutions, starting with solid \(\mathrm{KBr}\) : (a) \(0.75 \mathrm{~L}\) of \(1.5 \times 10^{-2} M \mathrm{KBr},(\mathbf{b}) 125 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KBr},(\mathbf{c}) 1.85 \mathrm{~L}\) of a solution that is \(12.0 \% \mathrm{KBr}\) by mass (the density of the solution is \(1.10 \mathrm{~g} / \mathrm{mL}),\) (d) a \(0.150 \mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough \(\mathrm{KBr}\) to precipitate \(16.0 \mathrm{~g}\) of AgBr from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\).

Step-by-Step Solution

Answer

(a) Weigh out 1.34 g of solid KBr and dissolve it in approximately 0.7 L of water. Stir well and then add enough water to obtain a final volume of 0.75 L. (b) Weigh out 125 g of water and 2.68 g of solid KBr. Mix them together until the KBr is fully dissolved. (c) Weigh out 244.2 g of solid KBr and dissolve it in approximately 1.8 L of water. Stir well and then add enough water to obtain a final volume of 1.85 L. (d) Weigh out 5.74 g of solid KBr and dissolve it in approximately 0.3 L of water. Stir well, and then add enough water to obtain a final volume of 0.3217 L. Mix this solution with the AgNO3 solution to cause the precipitation of AgBr.

Step 1: (a) Calculating mass of KBr

To prepare a \(0.75 \mathrm{~L}\) of a \(1.5 \times 10^{-2} M \mathrm{KBr}\) solution, first calculate the moles of \(\mathrm{KBr}\) required using the formula: moles = molarity × volume. Moles of \(\mathrm{KBr} = 1.5 \times 10^{-2} \mathrm{M} \times 0.75 \mathrm{L} = 0.01125 \mathrm{mol}.\) Now, convert moles to mass, knowing the molar mass of \(\mathrm{KBr}\) is approximately 119 g/mol. Mass of \(\mathrm{KBr} = 0.01125 \mathrm{mol} \times 119 \mathrm{g/mol} = 1.34 \mathrm{g}.\)

Step 2: (a) Preparing the solution

Weigh out 1.34 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 0.7 L of water. Stir well and then add enough water to obtain a final volume of 0.75 L.

Step 3: (b) Calculating mass of KBr

To prepare 125 g of a 0.180 m solution of \(\mathrm{KBr}\), first, calculate the mass of KBr to add to the water using the formula: mass of solute = molality × mass of solvent × molar mass of solute. Mass of \(\mathrm{KBr} = 0.180 \mathrm{m} \times 0.125 \mathrm{kg} \times 119 \mathrm{g/mol} = 2.68 \mathrm{g}.\)

Step 4: (b) Preparing the solution

Weigh out 125 g of water and 2.68 g of solid \(\mathrm{KBr}\). Mix them together until the \(\mathrm{KBr}\) is fully dissolved.

Step 5: (c) Calculating mass of KBr

Knowing the given volume (1.85 L) and density (1.10 g/mL) of the 12.0% by mass \(\mathrm{KBr}\) solution, first, calculate the total mass of the solution. Total mass of solution = 1.85 \mathrm{L} \times 1.10 \mathrm{g/mL} \times 1000 \mathrm{mL/L} = 2035 \mathrm{g}.$ Now, find the mass of \(\mathrm{KBr}\) in the solution. Mass of \(\mathrm{KBr} = 0.120 \times 2035 \mathrm{g} = 244.2 \mathrm{g}.\)

Step 6: (c) Preparing the solution

Weigh out 244.2 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 1.8 L of water. Stir well and then add enough water to obtain a final volume of 1.85 L.

Step 7: (d) Moles of AgBr to be precipitated

Calculate the moles of AgBr that need to be precipitated: Moles of AgBr = \(16.0 \mathrm{~g}\times \frac{1 \mathrm{~mol}}{331.32 \mathrm{~g}} = 0.04826 \mathrm{~mol}\).

Step 8: (d) Moles of KBr needed

For each mole of AgBr precipitated, one mole of \(\mathrm{KBr}\) is needed. Thus, we need 0.04826 mol of \(\mathrm{KBr}\).

Step 9: (d) Calculating mass and volume of KBr

Calculate the mass of \(\mathrm{KBr}\) needed. Mass of \(\mathrm{KBr} = 0.04826 \mathrm{mol} \times 119 \mathrm{g/mol} = 5.74 \mathrm{g}.\) Calculate the volume of the \(0.150 \mathrm{M}\) solution needed. Volume of solution = \(\frac{0.04826 \mathrm{mol}}{0.150 \mathrm{M}}= 0.3217 \mathrm{L}\).

Step 10: (d) Preparing the solution

Weigh out 5.74 g of solid \(\mathrm{KBr}\) and dissolve it in approximately 0.3 L of water. Stir well, and then add enough water to obtain a final volume of 0.3217 L. Mix this solution with the \(\mathrm{AgNO}_{3}\) solution to cause the precipitation of AgBr.

Key Concepts

Molarity Calculations

Molarity refers to the concentration of a solute in a solution, represented by the number of moles of solute per liter of solution. The formula for calculating molarity is:

• \( M = \frac{n}{V} \)

Here, \( M \) is the molarity, \( n \) is the number of moles of solute, and \( V \) is the volume of the solution in liters.
To determine the mass of solute needed, you first calculate the moles of the solute using the given molarity and volume. Multiply the moles by the molar mass of the solute to find the required mass.
For example, for a 0.75 L solution of 1.5 x 10^-2 M \( \mathrm{KBr} \), the moles of \( \mathrm{KBr} \) are calculated as:

• \( \text{moles} = 1.5 \times 10^{-2} \text{ M} \times 0.75 \text{ L} = 0.01125 \text{ mol} \)

To find the mass, use the chemical's molar mass:

• \( \text{mass} = 0.01125 \text{ mol} \times 119 \text{ g/mol} = 1.34 \text{ g} \)

Thus, you have the amount of \( \mathrm{KBr} \) needed to prepare this specific molar solution. Remember, when preparing solutions, it is important to add the solute first and then adjust the solution to the desired final volume.

Mass Percent Solutions

Mass percent is used to describe the concentration of a component in a solution. It shows the percentage of the solute compared to the total mass of the solution:

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

Suppose you have a solution that is 12% \( \mathrm{KBr} \) by mass. It means 12% of the total mass of this solution is \( \mathrm{KBr} \).
To find the actual mass of \( \mathrm{KBr} \) needed, you first need the total mass of the solution, calculated from its volume and density:

• \( \text{Total mass} = \text{Volume} \times \text{Density} \)

If your solution volume is 1.85 L with a density of 1.10 g/mL, it would result in:

• \( \text{Total mass} = 1.85 \text{ L} \times 1100 \text{ g/L} = 2035 \text{ g} \)
• \( \text{Mass of } \mathrm{KBr} = 0.12 \times 2035 \text{ g} = 244.2 \text{ g} \)

In conclusion, for a solution with a particular mass percent, it's crucial to find out the total mass of the solution to determine how much of the solute you need.

Molality Calculations

Molality is another measure of solute concentration. It describes the number of moles of solute per kilogram of solvent and it is especially useful at varying temperatures because it does not change with temperature fluctuations. The formula is:

• \( m = \frac{n}{m_{\text{solvent}}} \)

Where \( m \) is the molality, \( n \) is the moles of solute, and \( m_{\text{solvent}} \) is the mass of the solvent in kilograms.
To calculate the mass of a solute for a solution of known molality, determine moles of the solute first and then convert to mass using the solute's molar mass.
For preparation, if tasked with 125 g of 0.180 m \( \mathrm{KBr} \):

• \( \text{Mass of } \mathrm{KBr} = 0.180 \text{ m} \times 0.125 \text{ kg} \times 119 \text{ g/mol} = 2.68 \text{ g} \)

This calculation shows that in a given mass of solvent, the specific amount of \( \mathrm{KBr} \) in grams can be found using the definition of molality. Remember to always ensure total uniformity by fully dissolving the solute in the solvent.