Question

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

Short answer

The molarity of the final nitric acid solution after mixing the two given solutions is approximately 0.1667 M.

Step-by-step solution

Calculate the moles of acid in each solution

First, calculate the moles of \(\mathrm{HNO}_{3}\) in the \(50.00 \mathrm{~mL}\) \(0.100 \mathrm{M}\) solution and the moles of \(\mathrm{HNO}_{3}\) in the \(100.00 \mathrm{~mL}\) \(0.200 \mathrm{M}\) solution using the formula: moles = molarity × volume For the \(50.00 \mathrm{~mL}\) \(0.100 \mathrm{M}\) solution: \\ moles = \(0.100 \mathrm{M} \times 50.00 \mathrm{~mL}\) = 5.00 mmol For the \(100.00 \mathrm{~mL}\) \(0.200 \mathrm{M}\) solution: \\ moles = \(0.200 \mathrm{M} \times 100.00 \mathrm{~mL}\) = 20.00 mmol

Add the moles of nitric acid in both solutions

Now, add the moles of \(\mathrm{HNO}_{3}\) from the first solution and the moles of \(\mathrm{HNO}_{3}\) from the second solution to find the total moles of nitric acid in the mixed solution. total moles = 5.00 mmol + 20.00 mmol = 25.00 mmol

Calculate the total volume of the mixed solution

Add up the volumes of both solutions to obtain the final volume of the mixed solution: final volume = \(50.00 \mathrm{~mL} + 100.00 \mathrm{~mL}\) = 150.00 mL

Calculate the molarity of the final solution

Now, divide the total moles of \(\mathrm{HNO}_{3}\) by the final volume of the mixed solution to find the final molarity: final molarity = \(\frac{25.00 \mathrm{~mmol}}{150.00 \mathrm{~mL}}\) = 0.1667 M So, the molarity of the final nitric acid solution is approximately 0.1667 M.

Key concepts

Mole Concept

Understanding the mole concept is essential when dealing with chemical reactions and solution preparation. A mole is a unit in chemistry that signifies a specific quantity of a substance. One mole is equivalent to Avogadro's number, which is approximately 6.022 \times 10^{23} particles — be it atoms, molecules, or ions.

For solutions, moles can be computed using the formula: \( \text{moles} = \text{molarity} \times \text{volume in liters} \).

In the given exercise, moles of \(HNO_3\) are calculated by multiplying the molarity by the volume (converted to liters). This conversion is based on the fact that 1 liter equals 1000 milliliters. Therefore, to correctly connect molarity and volume, it's important to express the volume in liters, not milliliters, so that the units align with the molarity. For example:

\( \text{moles} = 0.100M \times (50.00mL \times \frac{1L}{1000mL}) \) = 0.00500 moles

Remember that the concentration given is already in molarity (\text{M}), which is moles per liter, so we aim to find the moles of the substance first to later find the molarity of the mixed solution.

Solution Concentration

Solution concentration often refers to how much of a solute is contained within a solution. Molarity (M) is one of the most common units for expressing concentration, defined as moles of solute per liter of solution. It is vital for calculating concentrations when mixing different solutions, as in the example we are examining.

To find the molarity, one must divide the number of moles of solute by the total volume of the solution in liters (\(M = \frac{moles}{volume}\)).

Applying this knowledge to the exercise, after finding the moles of \(HNO_3\) in each solution individually, you would then add these moles together to get the total moles in the combined solution. You then need to convert the volume, originally in milliliters, into liters by dividing by 1000. With both total moles and volume in liters, you can calculate the new molarity of the combined solutions.

Stoichiometry

Stoichiometry is a subdivision of chemistry that involves the quantitative relationships between reactants and products in a chemical reaction. It's the roadmap for understanding the ratios and proportions needed to combine different chemical substances. In context with solution preparation and molarity, stoichiometry plays a crucial role in ensuring that the correct proportions of substances are mixed to achieve a desired concentration.

Application in Molarity Calculations:

When two solutions are mixed, stoichiometry helps us understand that the number of moles of a solute in the resultant solution is the sum of the moles of that solute from each individual solution. In addition, the total volume is the sum of the individual volumes. Following the stoichiometric principles assures that the final concentration will be accurate as it considers the additive properties of moles and volumes.

For instance, in the given exercise, the moles of \(HNO_3\) from both solutions are added, and the volume is simply the sum of both volumes. This straightforward method is underpinned by stoichiometric concepts that dictate that substances combine in definite proportionate amounts.