Question

You can purchase nitric acid in a concentrated form that is 70.3\(\%\) HNO \(_{3}\) by mass and has a density of 1.41 \(\mathrm{g} / \mathrm{mL} .\) How can you prepare 1.15 \(\mathrm{L}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) from the concentrated solution?

Short answer

To prepare 1.15 L of 0.100 M HNO3, measure 7.313 mL of concentrated nitric acid and dilute it with approximately 1.1427 L of water.

Step-by-step solution

Calculate the mass of the concentrated nitric acid

Since density is given, calculate the mass of 1 mL of the concentrated nitric acid by multiplying its density by 1 mL: \[ \text{mass of 1 mL} = 1.41 \, \text{g/mL} \times 1 \, \text{mL} = 1.41 \, \text{g} \]

Calculate the mass of pure HNO3 in 1 mL of the concentrated solution

Use the percentage composition to find the mass of pure HNO3 in 1 mL of the concentrated solution:\[ \text{mass of pure HNO3} = 1.41 \, \text{g} \times 0.703 = 0.9913 \, \text{g} \]

Calculate moles of HNO3 in 1 mL of the concentrated solution

Calculate moles using the molar mass of HNO3 (63.01 g/mol):\[ \text{moles of HNO3} = \frac{0.9913 \, \text{g}}{63.01 \, \text{g/mol}} = 0.01573 \, \text{mol} \]

Calculate the total moles of HNO3 needed

Using the desired molarity and volume:\[ \text{total moles of HNO3 needed} = 0.100 \, \text{M} \times 1.15 \, \text{L} = 0.115 \, \text{mol} \]

Determine the volume of the concentrated solution required

Divide the total moles needed by the moles in 1 mL of the concentrated solution:\[ \text{volume of concentrated solution} = \frac{0.115 \, \text{mol}}{0.01573 \, \text{mol/mL}} = 7.313 \, \text{mL} \]

Determine the volume of water needed

Subtract the volume of the concentrated nitric acid from the total volume required:\[ \text{volume of water} = 1.15 \, \text{L} - 7.313 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} \approx 1.1427 \, \text{L} \]

Mix the acid with water to prepare the solution

Carefully mix the 7.313 mL of concentrated HNO3 with enough water to make up the volume to 1.15 L. Always add acid to water to ensure safety and proper dilution.

Key concepts

Molarity and Dilution

Molarity, denoted by the symbol 'M', is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. To understand molarity, one must be familiar with the concept of moles, which is a basic unit in chemistry representing the amount of a substance. In the context of preparing diluted solutions, molarity becomes crucial in ensuring the correct proportion of solute in the resulting mixture.

For instance, if you start with a concentrated solution, you can create a diluted solution by adding solvent (usually water) until you achieve the desired molarity. This dilution process is represented by the formula:
\begin{align*} M_1V_1 &= M_2V_2, \ \end{align*} where \(M_1\) and \(M_2\) are the molarities of the concentrated and diluted solutions, respectively, while \(V_1\) and \(V_2\) are their respective volumes. Solving this equation provides the volume of concentrate needed to achieve the target molarity after dilution.

Concentration of Solutions

The concentration of a solution indicates the amount of solute present in a specified volume of solvent. A concentrated solution contains a large amount of solute relative to the amount of solvent. There are various ways to express concentration, such as molarity (M), which has been discussed, and percent composition by mass, which compares the mass of the solute to the total mass of the solution.

Understanding concentration is vital for preparing solutions accurately and performing stoichiometric calculations in chemistry. When making solutions, the 'stock solution'—a concentrated form—is often diluted to obtain a working concentration suitable for a particular experiment or application. To obtain the desired concentration, chemists will calculate the volume of stock solution needed and add solvent to reach the final volume.

Stoichiometry in Solution

Stoichiometry involves the quantitative relationships between reactants and products in chemical reactions. In solutions, stoichiometry applies to the reactants and products dissolved in solvents. Molarity plays an integral role in solution stoichiometry because it allows for the direct conversion between the volume of a solution and the moles of a solute.

For example, when preparing a diluted solution from a concentrated stock, stoichiometry is used to calculate the precise amount of solute required. After calculating the moles of solute needed for the desired concentration and volume, this information gets transferred into a volume of the concentrated solution using its molarity, as shown in the exercise. The accuracy of these calculations is essential in industries like pharmaceuticals, where precise chemical concentrations are crucial.

Percent Composition by Mass

Percent composition by mass provides a ratio that expresses the mass of each element within a compound relative to the total mass of the compound. This concept is useful for preparing solutions where a chemical's purity or its concentrated form is involved, like 70.3% HNO3 by mass, given in our example.

Calculating the percent composition by mass involves dividing the mass of the solute by the total mass of the solution, and then multiplying by 100 to convert it to a percentage. In the process of dilution, it's this percentage that helps determine how much of the concentrated stock solution is actually the desired chemical, allowing for accurate preparation of the diluted solution. The percent composition provides the necessary link between the actual mass of the substance used and the theoretical calculations based on pure substances.