Question

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) 0.75 L of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (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 \(\mathrm{g} / \mathrm{mL}\) , ( \(\mathrm{d}\) ) a 0.150 \(\mathrm{M}\) solution of \(\mathrm{KBr}\) that contains just enough KBr to precipitate 16.0 \(\mathrm{g}\) of AgBr from a solution containing 0.480 mol of \(\mathrm{AgNO}_{3} .\)

Short answer

(a) Dissolve 1.33625 g of solid KBr in enough water to make a 0.75 L solution of \(1.5 \times 10^{-2} M\) KBr. (b) Dissolve 2.68125 g of solid KBr in 122.31875 g of water to make a 0.180 m KBr aqueous solution of 125 g. (c) Dissolve 244.2 g of solid KBr in 1790.8 g of water to make 1.85 L of a 12.0% by mass KBr solution. (d) Dissolve 0.10161 mol of KBr in enough water to make a 0.6774 L of a 0.150 M KBr solution. This solution contains enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

Step-by-step solution

(a) 1.5x10-2 M KBr Solution

First, find the moles of KBr needed using the concentration and volume of the solution: Moles of KBr = Molarity * Volume = \(1.5 \times 10^{-2} M \times 0.75 L \) = 0.01125 mol Next, convert the moles to mass using the molar mass of KBr (39.10 g/mol for K and 79.90 g/mol for Br): Mass of KBr = Moles of KBr * Molar Mass = 0.01125 mol * (39.10 g/mol + 79.90 g/mol) = 1.33625 g Dissolve 1.33625 g of solid KBr in enough water to make a 0.75 L solution.

(b) 0.180 m KBr Solution

First, find the moles of KBr needed using the molality and mass of the solution: Moles of KBr = molality * mass of the solution = \(0.180 \frac{mol}{kg} \times 0.125 kg\) = 0.0225 mol Next, convert the moles to mass using the molar mass of KBr: Mass of KBr = Moles of KBr * Molar Mass = 0.0225 mol * (39.10 g/mol + 79.90 g/mol) = 2.68125 g Now, calculate the mass of water required: Mass of water = Mass of solution - Mass of KBr = 125 g - 2.68125 g = 122.31875 g Dissolve 2.68125 g of solid KBr in 122.31875 g of water to make a 0.180 m KBr aqueous solution of 125 g.

(c) 12.0% KBr Solution

First, find the total mass of the solution using the volume and density: Mass of solution = Volume * Density = 1.85 L * 1.10 g/mL * \( \frac{1000 mL}{1 L} \) = 2035 g Next, calculate the mass of KBr and water required using mass percentages: Mass of KBr = mass percent * mass of the solution = 0.12 * 2035 g = 244.2 g Mass of water = Mass of solution - Mass of KBr = 2035 g - 244.2 g = 1790.8 g Dissolve 244.2 g of solid KBr in 1790.8 g of water to make 1.85 L of a 12.0% by mass KBr solution.

(d) 0.150 M KBr Solution

First, use stoichiometry to calculate the moles of KBr required to precipitate 16.0 g of AgBr: 1 mol AgBr = 1 mol KBr (1:1 stoichiometry) Moles of KBr = moles of AgBr = \( \frac{16.0 g}{107.87 g/mol + 79.90 g/mol} \) = 0.10161 mol Next, find the volume of the solution needed using the molarity: Volume of solution = moles of KBr / M = \( \frac{0.10161 mol}{0.150 M} \) = 0.6774 L Dissolve the calculated moles (0.10161 mol) of KBr in enough water to make a 0.6774 L of a 0.150 M KBr solution. This solution contains enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

Key concepts

molarity

Molarity is one of the most common ways to express the concentration of a solution. It represents the number of moles of solute (the substance being dissolved) per liter of solution. This unit is often denoted by the symbol \( M \) and is used widely in chemistry to calculate how much solute is present in a given volume of solvent.
To calculate molarity, you can use the formula:

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

For example, if you have a solution with 0.01125 moles of KBr in 0.75 liters of water, the molarity of this KBr solution would be \( 1.5 \times 10^{-2} M \).
This calculation is useful in determining the precise concentration of a solution, which is vital in various chemical reactions and processes.

molality

Molality is another way to express the concentration of a solution, focusing on the number of moles of solute per kilogram of solvent rather than per liter of solution. It is denoted by the symbol \( m \), and it provides a concentration measure that is temperature-independent because it depends on the mass rather than the volume.
To find molality, we use the formula:

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

For instance, if you dissolve 0.0225 moles of KBr in 0.1223 kg (or 122.3 grams) of water, the molality would be \( 0.180 \, \text{m} \). In this calculation, we ensure that our solution is exactly made using the specified mass of the solvent, providing a consistent concentration despite environmental changes.

stoichiometry

Stoichiometry is a critical concept in chemistry that involves using relationships from balanced chemical equations to determine the relative amounts of reactants and products in a chemical reaction. Essentially, it allows chemists to predict the result of reactions by knowing the quantities of involved substances.
For example, in precipitation reactions where an exact 1:1 molar ratio exists between KBr and AgBr, we can use stoichiometry to calculate the precise amount of KBr needed to precipitate a specific amount of AgBr from a solution.

• In our case, we find the moles of AgBr by dividing its mass by its molar mass, then use the 1:1 mole ratio to find the moles of KBr required.

This calculation helps determine how much of a reactant is necessary to fully react with another, ensuring no excess unreacted substances remain.

mass percent

Mass percent (or weight percent) expresses the concentration of a component in a mixture or solution, calculated as the ratio of the mass of the solute to the total mass of the solution, multiplied by 100.
The formula to determine the mass percent is:

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

For example, if we have a solution where 244.2 grams of KBr is dissolved in a total mass of 2035 grams of solution, the mass percent of KBr is 12%.
This representation is often used to describe solutions in industrial, laboratory, and commercial contexts, providing an easy understanding of solution composition based on weight distribution.

precipitation reactions

Precipitation reactions involve the formation of a solid, known as a precipitate, when two solutions react together. It occurs when an insoluble compound forms as a product of a chemical reaction.
In our provided case, when a solution of KBr is mixed with AgNO₃, silver bromide (AgBr) precipitates out of the solution.

• Using stoichiometry, if we know the amount of AgNO₃ present, we can predict the amount of KBr needed to precipitate a given mass of AgBr.

These reactions are essential in analytical chemistry and can be used to isolate or remove certain ions from solutions. The process highlights the importance of solubility rules in predicting the occurrence of solid formation from ionic solutions.