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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

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

Step by step solution

01

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.
02

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 = \(5.00 mmol + 20.00 mmol = 25.00 mmol\)
03

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 = \(150.00 mL × \frac{1 L}{1000 mL} = 0.150 L\)
04

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 \(0.167 M\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the reaction of \(19.0 \mathrm{g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{g}\) of solid metal is present. Calculate the mass of each metal in the 29.0 -g mixture.

A 500.0 -mL sample of \(0.200 \mathrm{M}\) sodium phosphate is mixed with \(400.0 \mathrm{mL}\) of \(0.289 M\) barium chloride. What is the mass of the solid produced?

You are given a \(1.50-g\) mixture of sodium nitrate and sodium chloride. You dissolve this mixture into \(100 \mathrm{mL}\) of water and then add an excess of 0.500 \(M\) silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.64 \(1 \mathrm{g}\). a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the percent sodium chloride in the original unknown mixture.

For the following chemical reactions, determine the precipitate produced when the two reactants listed below are mixed together. Indicate "none" if no precipitate will form.

Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{Cl}_{2}(g)+\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathbf{O}_{2}(g)+\mathbf{H}_{2} \mathbf{O}(l)+\mathbf{P b}(s) \rightarrow \mathbf{P b}(\mathbf{O H})_{2}(s)\) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(i)\)

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