Question

A solution was prepared by mixing \(50.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HNO_and \(100.00 \mathrm{mL}\) of \(0.200 \mathrm{M}\) HNO \(_{3}\). Calculate the molarity of the final solution of nitric acid.

Short answer

The molarity of the final solution of nitric acid is \(0.167 M\).

Step-by-step solution

Calculate moles of HNO₃ in each solution.

To calculate moles of HNO₃ in each solution, we will use the formula moles = molarity × volume. For the first solution: Moles of HNO₃ = \(0.100 M × 50.00 mL = 5.00 mmol\) For the second solution: Moles of HNO₃ = \(0.200 M × 100.00 mL = 20.00 mmol\) Remember to convert mL to L when necessary.

Calculate total moles of HNO₃ in the final solution.

Now, we will find the total moles of HNO₃ in the final solution by adding the moles of HNO₃ in both solutions. Total moles of HNO₃ = \(5.00 mmol + 20.00 mmol = 25.00 mmol\)

Calculate the combined volume of the final solution.

In this step, we will calculate the combined volume of the final solution by adding the volumes of the two given solutions. Total Volume = \(50.00 mL + 100.00 mL = 150.00 mL\) Now, convert this volume into liters. Total Volume = \(150.00 mL × \frac{1 L}{1000 mL} = 0.150 L\)

Calculate the molarity of the final solution.

Finally, we will calculate the molarity of the final solution using the total moles of HNO₃ and the combined volume. The formula for molarity is M = moles / volume. Molarity of the final solution = \(\frac{25.00 mmol}{0.150 L} = 0.167 M\) Thus, the molarity of the final solution of nitric acid is \(0.167 M\).

Key concepts

Chemical Solution Preparation

Creating a chemical solution with a particular concentration requires meticulous technique to ensure the desired molarity. Molarity, denoted as ‘M’, is a measure of the concentration of a solute in a solution. In the context of preparing a solution, the process starts with calculating the amount in moles of the chemical of interest (in this case nitric acid, HNO₃) and then combining this with a specified volume of solvent, usually water.

To ensure accuracy, it's crucial to measure the volume of solvents properly, using a graduated cylinder or a pipette. When mixing different solutions with known molarities, like in our problem at hand, we combine the volumes and calculate the total number of moles to end up with the final molarity of the mixture. When dealing with small volumes, it's common to see measurements in milliliters (mL), but for molarity calculations, it is imperative to convert the volume to liters (L) since molarity is defined as moles per liter.

Moles and Molarity

Understanding moles and molarity is fundamental in chemistry. A mole represents a specific number of particles, usually atoms or molecules, and is equivalent to Avogadro’s number, which is approximately 6.02 x 10²³ particles. This number is akin to a dozen representing 12 items, but for a mole, it’s a much larger quantity used for tiny entities such as atoms.

The concept of molarity then ties in by quantifying the number of moles in a given volume. The formula for calculating molarity is simply:
\[M = \frac{moles}{volume}\]
where volume is always expressed in liters. This unit of concentration allows chemists to easily understand the ratio of solute to solvent in a solution, enabling consistent results in reactions or further solution preparation.

Molarity of Nitric Acid

Nitric acid (HNO₃) is a common, strong acid used in various applications, from fertilizers to explosives. When preparing a dilute solution of nitric acid or mixing it with other solutions, determining its molarity is often necessary. It provides information about the concentration of HNO₃ in a mixture, which is crucial for predicting the outcome of chemical reactions.

In our example, the molarity calculation of a final HNO₃ solution involves adding together the moles from individual solutions and then dividing by the total volume of the mixed solutions. Calculating molarity requires precision in measuring both moles and volume. Remember that practical applications might need you to account for factors such as temperature or specific gravity, which can affect density and volumes. However, for simplicity in general chemistry exercises, these factors are often excluded.