Question

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) \(0.75 \mathrm{~L}\) of 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 KBr to precipitate \(16.0 \mathrm{~g}\) of \(\mathrm{AgBr}\) from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\).

Short answer

To prepare the first solution (0.75 L, 12.0% KBr by mass), weigh 99 g of solid KBr and dissolve it in water in a volumetric flask. Gradually add water to the flask until the 0.75 L mark is reached. Close the flask and shake it to ensure thorough mixing. For the second solution (0.150 M, contains enough KBr to precipitate 16 g of AgBr from a solution with 0.480 mol of AgNO₃), weigh 10.14 g of solid KBr and dissolve it in water in a volumetric flask. Gradually add water to the flask until the 0.568 L mark is reached. Close the flask and shake it to ensure thorough mixing.

Step-by-step solution

Calculate the mass of KBr for the first solution

To find the mass of KBr in this solution, we first determine the total mass of the solution by multiplying its volume and density: Total mass of solution = volume x density = 0.75 L x (1000 mL/1 L) x (1.10 g/mL) = 825 g Now, since the solution is 12.0% KBr by mass, we can calculate the mass of KBr: Mass of KBr = (12.0/100) x total mass of solution = (12.0/100) x 825 g = 99 g

Calculate the mass of KBr for the second solution

For the second solution, let's first find the moles of AgBr that will be precipitated from a solution containing 0.480 mol of AgNO₃ with the following stoichiometry: AgNO₃ + KBr -> AgBr + KNO₃ From the stoichiometry, we see that 1 mole of KBr will precipitate 1 mole of AgBr from the solution containing AgNO₃. Now, we know that 16.0 g of AgBr will be precipitated, let's convert it to moles: Moles of AgBr = (16.0 g) / (187.77 g/mol) = 0.0852 mol Since KBr and AgBr are in a 1:1 ratio, we need 0.0852 mol of KBr to precipitate 16.0 g of AgBr. Given the 0.150 M solution: Moles of KBr = 0.0852 mol Molarity of KBr = 0.150 M Volume of KBr solution = Moles of KBr / Molarity of KBr = 0.0852 mol / 0.150 M = 0.568 L Finally, let's find the mass of KBr in this solution: Mass of KBr = Moles of KBr x Molar mass of KBr = 0.0852 mol × (119.00 g/mol) = 10.14 g

Prepare the first solution

To prepare the first solution (0.75 L, 12.0% KBr by mass), weight 99 g of solid KBr and dissolve it in water in a volumetric flask. Gradually add water to the flask until the 0.75 L mark is reached. Close the flask and shake it to ensure thorough mixing. This solution is ready.

Prepare the second solution

To prepare the second solution (0.150 M, contains enough KBr to precipitate 16 g of AgBr from a solution with 0.480 mol of AgNO₃), weight 10.14 g of solid KBr and dissolve it in water in a volumetric flask. Gradually add water to the flask until the 0.568 L mark is reached. Close the flask and shake it to ensure thorough mixing. This solution is ready.

Key concepts

Stoichiometry

In chemistry, stoichiometry is the study of the quantitative relationships involved in chemical reactions. It connects the quantities of reactants and products in a balanced chemical equation, enabling you to predict how much of a substance is needed or produced under specific conditions. Consider the balanced reaction: \(\text{AgNO}_3 + \text{KBr} \rightarrow \text{AgBr} + \text{KNO}_3\). This equation tells us that one mole of potassium bromide (KBr) will react with one mole of silver nitrate (AgNO₃) to form one mole each of silver bromide (AgBr) and potassium nitrate (KNO₃). By understanding these relationships, you can calculate precisely the amount of KBr needed to react with a given amount of AgNO₃ or to produce a certain mass of AgBr. For example, the reaction needed to precipitate 16 g of AgBr can be solved by converting grams to moles, using stoichiometric coefficients, and then converting back to grams if needed. These calculations are central to any chemical process, making stoichiometry a cornerstone of solution preparation.

Remember, stoichiometry helps ensure the right proportions in any reaction, which is particularly important in both research and industrial applications where precision is key.

Molarity

Molarity is a measure of the concentration of a solution, defined as the number of moles of a solute per liter of solution. It is represented by the unit molarity \(M\), where 1 M is equal to 1 mole of solute per liter of solution. Understanding molarity is crucial when preparing solutions because it allows you to quantify how concentrated a solution is. For instance, in the exercise, we need to prepare a 0.150 M solution of KBr.

To calculate the required amount of KBr for this molarity, you can use the formula:

• Moles of KBr = Molarity \( \times \) Volume of solution.

This approach ensures accurate preparation of the solution with the desired concentration. If you know the desired molarity and the volume of your solution, simply rearrange the formula to find the number of moles, and thus the grams, of solute needed before dissolving it in a suitable amount of solvent to the desired total volume.

Molarity not only assists in predicting how much of a solution will react but also helps in practical lab settings to mix chemicals in specific proportions.

Mass Percentage

Mass percentage describes the concentration of a solute in a solution by depicting what portion of the total mass is attributed to the solute. It is calculated using the formula:

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

In our scenario, we focused on preparing a solution that is 12.0% KBr by mass. Understanding this concept implies that, in the 0.75 L solution of KBr, \(12\%\) of every 100 grams of the solution is KBr. The first step for solving such problems involves converting the volume of the solution to mass using density.

With the density provided as 1.10 g/mL, the calculation becomes straightforward:

• First, compute the total solution mass: \( \text{mass} = \text{volume} \times \text{density} \).
• From the total mass, calculate the mass of KBr using its percentage.

This knowledge allows you to properly weigh and prepare a solution with the exact concentration of a solute by mass, ensuring consistency and reliability in chemical processes.