Problem 65 - Chemistry The Molecular Science

Question

Commercial concentrated nitric acid contains 69.5 mass percent \(\mathrm{HNO}_{3}\) and has a density of \(1.42 \mathrm{~g} / \mathrm{mL}\). (a) Calculate the molarity of this solution. (b) Calculate what volume of the concentrated acid must be used to prepare \(10.0 \mathrm{~L}\) of \(6.00-\mathrm{M} \mathrm{HNO}_{3}\).

Step-by-Step Solution

Answer

(a) 15.68 M; (b) 3.83 L

Step 1: Calculate Molarity of Concentrated HNO3 Solution

Start by determining the mass of \( ext{HNO}_3 \) in \( 1 \text{ L} \) of the solution. Given that the density is \( 1.42 \text{ g/mL} \), the mass of \( 1 \text{ L} \) of solution is:\[ 1.42\, \text{g/mL} \times 1000\, \text{mL} = 1420\, \text{g} \]The mass of \( \text{HNO}_3 \) is \( 69.5\% \) of this, so:\[ 0.695 \times 1420\, \text{g} = 987.9\, \text{g} \]Next, you need to convert this mass to moles using the molar mass of \( \text{HNO}_3 \) (which is approximately \( 63.01 \text{ g/mol} \)):\[ \frac{987.9\, \text{g}}{63.01 \text{g/mol}} = 15.68 \text{ mol} \]Thus, the molarity (\( M \)) of the solution is:\[ 15.68 \text{ mol/L} \]

Step 2: Determine the Volume of Concentrated Acid for Dilution

To prepare \( 10.0 \text{ L} \) of a \( 6.00 \text{ M} \) solution, use the dilution formula: \[ M_1V_1 = M_2V_2 \]Where \( M_1 = 15.68 \text{ M} \), \( V_1 \) is the volume of concentrated acid needed, \( M_2 = 6.00 \text{ M} \), and \( V_2 = 10.0 \text{ L} \).Plug the values into the formula:\[ 15.68 \cdot V_1 = 6.00 \cdot 10.0 \]Solving for \( V_1 \):\[ V_1 = \frac{6.00 \times 10.0}{15.68} \approx 3.83 \text{ L} \]

Step 3: Conclusion

The molarity of the commercial concentrated \( \text{HNO}_3 \) solution is \( 15.68 \text{ M} \). To prepare \( 10.0 \text{ L} \) of \( 6.00 \text{ M} \) solution, use \( 3.83 \text{ L} \) of the concentrated acid.

Key Concepts

Dilution Formula

The dilution formula is a handy tool for chemists when they need to obtain a desired concentration of a solution. It's based on the principle of conservation of mass, where the amount of solute remains the same before and after dilution. This formula is expressed as:\[ M_1V_1 = M_2V_2 \]

• \( M_1 \) is the molarity of the concentrated solution.
• \( V_1 \) is the volume of the concentrated solution needed.
• \( M_2 \) is the molarity of the diluted solution.
• \( V_2 \) is the volume of the diluted solution.

The main goal of using the dilution formula is to find out how much of the concentrated solution is required to achieve a particular molarity in a specific volume of liquid. For instance, if a chemist wants to prepare 10 liters of a 6 M nitric acid solution using a concentrated stock solution with a molarity of 15.68 M, they would rearrange the formula to solve for \( V_1 \), the unknown volume of the concentrated solution:\[ V_1 = \frac{M_2V_2}{M_1} \]This approach helps to ensure accuracy and efficiency in lab preparations.

Chemical Density

Chemical density plays a crucial role in determining how much mass a specific volume of a substance contains. It is measured in grams per milliliter (g/mL) and indicates how tightly the substance's molecules are packed together. The formula for density is:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]When dealing with liquid solutions, knowing the density allows chemists to convert between the mass and volume of the solution accurately. This conversion is vital when calculating the total mass of a solution from its volume. In our example with nitric acid:- The density is given as 1.42 g/mL.- For 1 liter (1000 mL) of the solution, the total mass is calculated as:\[ 1.42 \, \text{g/mL} \times 1000 \, \text{mL} = 1420 \, \text{g} \]This mass is used to determine the amount of solute, such as nitric acid, in the solution, enabling further calculations like molarity.

Molar Mass

The concept of molar mass is fundamental in chemistry, serving as a bridge between the mass of a substance and the amount of substance in moles. Molar mass is expressed in grams per mole (g/mol) and represents the mass of one mole of a given substance.To calculate the molar mass, one adds up the atomic masses of all the atoms in the formula of the compound. For an example with nitric acid (HNO\(_3\)):

• Hydrogen (H) has an atomic mass of approximately 1.01 g/mol.
• Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
• Oxygen (O) has an atomic mass of approximately 16.00 g/mol.

Adding these together for HNO\(_3\):\[ 1.01 + 14.01 + (3 \times 16.00) = 63.01 \, \text{g/mol} \]Knowing the molar mass allows chemists to convert between mass and moles. For instance, when 987.9 g of nitric acid is measured, the moles of \( \text{HNO}_3 \) can be calculated:\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{987.9 \, \text{g}}{63.01 \, \text{g/mol}} = 15.68 \, \text{mol} \]This information is essential for determining the solution's molarity or working with chemical reactions.