Question

Describe how you would prepare each of the following aqueous solutions, starting with solid KBr: (a) \(0.75 \mathrm{~L}\) of \(1.5 \times 10^{-2} M \mathrm{KBr},\) (b) \(125 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KBr},\) (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 KBr to precipitate \(16.0 \mathrm{~g}\) of AgBr from a solution containing \(0.480 \mathrm{~mol}\) of \(\mathrm{AgNO}_{3}\)

Short answer

(a) Calculate moles of KBr needed: moles of KBr = (1.5 × 10⁻² M) * (0.75 L). Convert moles to grams: mass of KBr = (moles of KBr) * (119 g/mol). Preparation: Weigh out the mass of KBr, dissolve in a 0.75 L volumetric flask, and fill with water. (b) Calculate molality: mass of solvent = (125 g) / (0.180 m). Calculate moles of KBr: moles of KBr = (0.180 m) * (mass of solvent in kg). Convert moles to grams: mass of KBr = (moles of KBr) * (119 g/mol). Preparation: Weigh out the mass of KBr, dissolve in the calculated mass of the solvent. (c) Calculate mass of KBr: mass of KBr = 0.12 * (1.10 g/mL) * (1.85 L * 1000 mL/L). Preparation: Weigh out the mass of KBr, dissolve in a 1.85 L volumetric flask, and fill with water. (d) Determine moles of AgBr: moles of AgBr = (16.0 g) / (187.8 g/mol). Calculate moles of KBr (equal to moles of AgBr). Convert moles to grams: mass of KBr = (moles of KBr) * (119 g/mol). Find volume: volume = moles of KBr / (0.150 M). Preparation: Weigh out the mass of KBr, dissolve in a volumetric flask of the calculated volume, and fill with water.

Step-by-step solution

(a) Calculate the moles of KBr needed for the solution

As we are given the volume (0.75 L) and molarity(1.5 * 10^-2 M) of the desired solution, first, find out the number of moles of KBr required: moles of KBr = Molarity * Volume = (1.5 × 10⁻² M) * (0.75 L)

(a) Convert moles to grams

Now that we have the number of moles required, we need to convert this to grams using the molar mass of KBr (molecular weight = 39.1 g/mol for K + 79.9 g/mol for Br = 119 g/mol): mass of KBr = (moles of KBr) * (molar mass of KBr)

(a) Preparation of the solution

Finally, weigh out the calculated mass of KBr and dissolve it in a 0.75 L volumetric flask. Fill the flask with water (preferably distilled water) up to the 0.75 L mark to make the 1.5 * 10^-2 M KBr solution.

(b) Calculate molality of KBr solution

We are given the mass (125 g) and molality (0.180 m) of the desired solution. First, divide the mass by the molality to find the mass of the solvent (usually water in an aqueous solution): mass of solvent = mass of solution / molality = (125 g) / (0.180 m)

(b) Calculate the moles of KBr required

Now that we have the mass of the solvent, we can find the number of moles of KBr required: moles of KBr = (molality) * (mass of solvent in kg)

(b) Convert moles to grams

Next, convert the number of moles of KBr to grams using the molar mass of KBr (119 g/mol): mass of KBr = (moles of KBr) * (molar mass of KBr)

(b) Preparation of the solution

Weigh out the calculated mass of KBr and dissolve it in the calculated mass of the solvent (usually distilled water) to prepare the 0.180 m KBr solution.

(c) Calculate the mass of KBr required

As we are given the volume (1.85 L), mass fraction (12.0 %), and density (1.10 g/mL) of the desired solution, first, find out the mass of KBr required: mass of KBr = mass fraction * density * volume = 0.12 * (1.10 g/mL) * (1.85 L * 1000 mL/L)

(c) Preparation of the solution

Finally, weigh out the calculated mass of KBr and dissolve it in a 1.85 L volumetric flask. Fill the flask with water (preferably distilled water) up to the 1.85 L mark to make the 12.0% KBr solution.

(d) Determine moles of AgBr

We are given the mass of precipitated AgBr (16.0 g) and the number of moles of AgNO3 (0.480 mol). First, calculate the moles of AgBr using its molar mass (molecular weight = 107.9 g/mol for Ag + 79.9 g/mol for Br = 187.8 g/mol): moles of AgBr = (mass of AgBr) / (molar mass of AgBr)

(d) Calculate the moles of KBr required

Since the reaction between KBr and AgNO3 is 1:1, the number of moles of KBr required is equal to the number of moles of AgBr.

(d) Convert moles to grams

Next, convert the number of moles of KBr to grams using the molar mass of KBr (119 g/mol): mass of KBr = (moles of KBr) * (molar mass of KBr)

(d) Find the volume of the solution

The concentration of the KBr solution is given as 0.150 M. Calculate the volume of the solution using the relationship between moles, molarity, and volume: volume = moles of KBr / (0.150 M)

(d) Preparation of the solution

Finally, weigh out the calculated mass of KBr and dissolve it in a volumetric flask of the calculated volume. Fill the flask with water (preferably distilled water) up to the mark to make the desired 0.150 M KBr solution. This solution contains just enough KBr to precipitate 16.0 g of AgBr from a solution containing 0.480 mol of AgNO3.

Key concepts

Molarity

Molarity is a key concept in chemistry, measuring the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Calculating molarity involves knowing the amount of solute and the volume of the solution. The formula is:

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

Molarity enables scientists to understand how much solute is present in a given volume of solution. It is widely used in labs to prepare solutions with precise concentrations. This ensures reactions proceed as intended.
To prepare a solution of specific molarity, first calculate the required moles of solute, then measure out the corresponding mass using the molar mass. Dissolve this mass in enough solvent to reach the desired total volume.
For example, in preparing a solution of KBr with a molarity of \(1.5 \times 10^{-2}\ M\), you multiply the desired molarity by the volume in liters to find the moles of solute needed:

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

Molality

Molality is another way of expressing the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent, and is typically used when working with temperature-sensitive solutions because it doesn't change with temperature. The formula is as follows:

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

To calculate molality, you must know the mass of the solvent and the solute. Molality is not affected by changes in temperature or pressure, making it particularly useful in situations involving extreme conditions.
For instance, if you want to prepare a \(0.180\ m\) KBr solution with a total mass of \(125\ g\), simply divide the solution's mass by the molality to find the mass of the solvent. Then, find the moles of solute required by multiplying the molality by the solvent mass (in kg). This will determine the correct amount of KBr you need to add.

Mass Percentage

Mass percentage is a measure of the concentration of a component in a mixture or solution. It is expressed as the mass of a solute divided by the total mass of the solution, multiplied by 100%. The formula is:

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

Mass percentage provides a straightforward way to express concentration, useful in day-to-day applications. Unlike molarity and molality, which are typically more suited for laboratory measurements, mass percentage offers a simple way to represent component concentrations within a solution.
To find the mass percentage of a KBr solution, like in one that is \(12.0\%\) by mass, you calculate the mass of solute based on the total mass of the solution and the density provided. By multiplying the volume of the solution by its density, you find the total mass, then use that to calculate how much solute is needed to achieve the desired percentage.
This concept is particularly helpful when solutions are prepared by weight rather than volume, allowing you to ascertain exactly how much of each component is present.