Question

A 35.0-mL sample of \(1.00 M \mathrm{KBr}\) and a \(60.0\) -mL sample of \(0.600 \mathrm{M} \mathrm{KBr}\) are mixed. The solution is then heated to evaporate water until the total volume is \(50.0 \mathrm{~mL}\). What is the molarity of the \(\mathrm{KBr}\) in the final solution?

Short answer

The molarity of KBr in the final solution is \(1.42 \mathrm{M}\).

Step-by-step solution

Calculate the number of moles of KBr in both samples

First, we'll calculate the number of moles of KBr in both samples using the following formula: Number of moles = Molarity * Volume For the first sample, 35.0 mL of 1.00 M KBr: Number of moles = 1.00 M * \(35.0 \times 10^{-3}\) L = 0.035 mol KBr For the second sample, 60.0 mL of 0.600 M KBr: Number of moles = 0.600 M * \(60.0 \times 10^{-3}\) L = 0.036 mol KBr

Calculate the total number of moles of KBr in the mixed solution

Next, we will add the number of moles of KBr in both samples to find the total number of moles in the mixed solution: Total moles of KBr = Moles of KBr in sample 1 + Moles of KBr in sample 2 Total moles of KBr = 0.035 mol + 0.036 mol = 0.071 mol KBr

Calculate the final molarity of KBr after evaporation of water

After the evaporation process, the final volume of the mixed solution is given as 50.0 mL or 0.050 L. Now, we will determine the final molarity of KBr using the number of moles and the final volume: Molarity of KBr = (Total number of moles of KBr) / (Final volume of solution) Molarity of KBr = (0.071 mol) / (50.0 mL * \(10^{-3}\) L/mL) Molarity of KBr = (0.071 mol) / (0.050 L) = 1.42 M So, the molarity of KBr in the final solution is 1.42 M.

Key concepts

Chemical Solution Concentration

When studying chemistry, one often encounters the concept of chemical solution concentration, which is a measure of the amount of a substance, known as a solute, that is dissolved in a given volume of solvent. This concept is pivotal for various applications, including laboratory experiments, industrial processes, and pharmacology.

There are different ways to express the concentration of a solution, with molarity being one of the most common. It is defined as the number of moles of solute per liter of solution. The formula for calculating molarity is very straightforward: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

To put it into practice, let's take the exercise provided. The concentration of potassium bromide (KBr) in two different solutions needs to be combined and then concentrated by evaporation. To find the concentration, we start by evaluating the individual molarities of the KBr solutions before they are mixed. Here, the exercise becomes a real-world application of how to manipulate and understand different concentrations in a laboratory or industrial setting. This understanding is crucial in order to predict the behavior of solutions and to control reactions.

Moles and Molarity

Moles and molarity are building blocks of chemical quantification and integral for comprehending solution concentration. A mole is a fundamental unit in chemistry that represents a specific amount of substance; it is the Avogadro number (approximately 6.022 × 10²³) of particles (atoms, molecules, ions, etc.).

Converting Between Moles and Molarity

Connecting moles to molarity is as simple as understanding the volume of the solution in which these moles are dissolved. If you know the amount of substance (in moles) and the volume of the solution (in liters), you can find the molarity through the formula:\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

In our textbook exercise, we determine the moles of solute (KBr) in two separate solutions and use these values to find the resulting molarity after the solutions are mixed and the volume is adjusted by evaporation. This calculation exemplifies the direct link between moles of a solute and molarity, and how this relationship is crucial for determining solution concentration in various scientific contexts.

Solution Dilution and Concentration

Solution dilution and concentration adjustments are common tasks in laboratories when precise chemical solution concentrations are required. Dilution involves adding more solvent to a solution, thereby reducing the solute's concentration. Conversely, concentration typically involves removing solvent, for example, by evaporation, to increase the solute concentration. Understanding this process is essential for ensuring the desired concentration in experimental and industrial practices.

Dilution Factors and Final Concentration

To illustrate, in the given exercise, we concentrate a solution by evaporating water to decrease its volume from the combined volume of two solutions to a final volume of 50.0 mL. As the total number of moles of solute (KBr) remains unchanged during evaporation, the resulting solution becomes more concentrated. We can express this change in concentration quantitatively by recalculating the molarity using the new, reduced volume.\[ \text{New Molarity} = \frac{\text{Total moles of solute}}{\text{New volume of solution in liters}} \]

The calculation from the exercise allows students to understand how to assess and adjust concentrations through different laboratory processes, highlighting the importance of mastering the concept of molarity in relation to solution volume changes.