Question

(a) How many grams of solid silver nitrate would you need to prepare \(200.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (b) An experiment calls for you to use \(100 \mathrm{~mL}\) of \(0.50 \mathrm{MHNO}_{3}\) solution. All you have available is a bottle of \(3.6 \mathrm{M} \mathrm{HNO}_{3}\). How many milliliters of the \(3.6 \mathrm{M} \mathrm{HNO}_{3}\) solution and of water do you need to prepare the desired solution?

Short answer

(a) The mass of solid silver nitrate needed to prepare 200.0 mL of a 0.150 M AgNO3 solution is 5.0964 grams. (b) To prepare 100 mL of 0.50 M HNO3 solution using a 3.6 M HNO3 solution, you need: 1. 13.89 mL of 3.6 M HNO3 solution. 2. 86.11 mL of water.

Step-by-step solution

(a) Calculate moles of AgNO3 required

First, we need to find the moles of silver nitrate required for the desired concentration and volume. We know that the molarity (M) is given by: M = moles of solute / volume of solution (L) Rearranging the formula, we get: moles of solute = M × volume of solution (L) We are given M = 0.150 M and the volume of solution = 200.0 mL (which is equal to 0.2 L). Hence, moles of AgNO3 = 0.150 × 0.2

Calculate the mass of AgNO3 required

Next, we need to find the mass of silver nitrate required using the number of moles and the molar mass of AgNO3. The molar mass of AgNO3 can be calculated as: \(AgNO_3 = Ag + N + 3O = 107.87 + 14.01 + 3(16.00) = 169.88 g/mol\) Therefore, the mass of AgNO3 required is given by: mass of AgNO3 = moles of AgNO3 × molar mass = 0.150 × 0.2 × 169.88 g/mol

Answer (a)

The mass of solid silver nitrate needed to prepare 200.0 mL of a 0.150 M AgNO3 solution is 5.0964 grams.

(b) Calculate moles of HNO3 required

Now we need to find the moles of HNO3 required for the desired concentration and volume. We are given M = 0.50 M and the volume of solution = 100.0 mL (which is equal to 0.1 L). Therefore, moles of HNO3 = 0.50 × 0.1

Calculate the volume of 3.6 M HNO3 solution needed

Next, we need to find the volume of 3.6 M HNO3 solution required to get the desired moles of HNO3. Using the molarity formula: M_initial * V_initial = moles of HNO3 3.6 M * V_initial = 0.50 × 0.1 Rearranging the formula, we get: V_initial = (0.50 × 0.1) / 3.6

Calculate the volume of water needed

To find the volume of water required, we subtract the volume of 3.6 M HNO3 solution used from the total desired volume of the 0.50 M HNO3 solution: Volume of water = total desired volume - volume of 3.6 M HNO3 solution = 100 mL - (volume of 3.6 M HNO3 solution in mL)

Answer (b)

To prepare 100 mL of 0.50 M HNO3 solution using a 3.6 M HNO3 solution, you need: 1. 13.89 mL of 3.6 M HNO3 solution. 2. 86.11 mL of water.

Key concepts

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The equation for molarity is given by:

1. Molarity (M) = \( \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

Molarity is important because it allows chemists to quantify the concentration of a solution in a straightforward manner.
For example, in part (a) of our exercise, the molarity of silver nitrate (AgNO3) is 0.150 M, and we are tasked with figuring out how many grams of silver nitrate would be needed to make 200 mL of this solution.
Since the molarity and volume are known, we can easily solve for the moles of AgNO3 using the formula:

• The number of moles = molarity × volume of solution (in liters).

Solution Preparation

Solution preparation refers to the process of creating a solution of a specific concentration from either solid solutes or from more concentrated stock solutions. This involves dissolving the solute in a solvent, usually water, to achieve the desired molarity.
In the case of our exercise, preparing the 0.150 M AgNO3 solution involves dissolving the calculated moles of AgNO3 (found using the molarity formula) into 200 mL of water. It's important to note the procedure:

• Calculate the required moles of AgNO3 using its molarity and volume of the solution.
• Use the molecular weight of AgNO3 to convert moles to grams needed.
• Weigh the calculated grams of AgNO3 accurately and dissolve it in the specified volume of solvent (water).

Preparing a solution from a concentrated stock, as seen in part (b) for HNO3, involves diluting the starting solution:

• Determine the volume of the concentrated stock needed to reach the desired molarity and volume using the dilution equation:
• \( M_1V_1 = M_2V_2 \) where \( M_1 \) and \( V_1 \) are the molarity and volume of the concentrated solution, and \( M_2 \) and \( V_2 \) are the desired molarity and volume.
• Carefully measure and mix the appropriate volumes of the concentrated solution and solvent.

Concentration Calculations

Concentration calculations are essential in chemistry for preparing solutions accurately. They ensure that reactions occur with the correct stoichiometry, which impacts the efficiency and outcome of chemical processes.
For calculating the concentration of a solution, knowledge of the solute's molecular weight is crucial. This is because it allows conversion between moles and grams, enabling you to accurately determine how much solid solute to use.
When working with stock solutions, as in part (b) of our exercise, it's vital to perform dilution calculations. Using the equation for dilution, \( M_1V_1 = M_2V_2 \), we can determine how much of the concentrated solution is needed to achieve a specific concentration in the final solution:

• Calculate the initial volume (\( V_1 \)) of the concentrated stock needed by rearranging the formula:
• Plug in the values for \( M_2 \), \( V_2 \), and \( M_1 \) to solve for \( V_1 \).
• Subtract the volume of the concentrated solution from the total desired volume to find the volume of solvent required.

These calculations ensure precision in creating solutions with specific and consistent concentrations.