Problem 74 - Chemistry The Central Science

Question

(a) How many milliliters of a stock solution of \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) would you have to use to prepare \(110 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?\) (b) If you dilute \(10.0 \mathrm{~mL}\) of the stock solution to a final volume of \(0.250 \mathrm{~L}\), what will be the concentration of the diluted solution?

Step-by-Step Solution

Answer

(a) To prepare \(110 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3}\), you need to use approximately \(9.17 \mathrm{mL}\) of the \(6.0 \mathrm{M} \mathrm{HNO}_{3}\) stock solution. (b) After diluting \(10.0 \mathrm{mL}\) of the stock solution to a final volume of \(0.250 \mathrm{L}\), the concentration of the diluted solution is \(0.24 \mathrm{M}\).

Step 1: Identify the given values

The initial concentration (\(M_1\)) is given as 6.0 M, and the desired final concentration (\(M_2\)) is 0.500 M. The final volume (\(V_2\)) is given as 110 mL. We need to find the initial volume (\(V_1\)) required.

Step 2: Apply the dilution formula

Using the formula \(M_1 V_1 = M_2 V_2\), we have: \(6.0\mathrm{M} * V_1 = 0.500\mathrm{M} * 110\mathrm{~mL}\)

Step 3: Solve for the initial volume

We can now solve for \(V_1\): \(V_1 = \frac{0.500\mathrm{M} * 110\mathrm{~mL}}{6.0\mathrm{M}}\)

Step 4: Calculate V1

Now, carry out the calculation: \(V_1 = \frac{55\mathrm{~mL}}{6} = 9.1667\mathrm{~mL}\) So, approximately 9.17 mL of the 6.0 M HNO3 stock solution is required to prepare 110 mL of 0.5 M HNO3. For part (b):

Step 1: Identify the given values

In this case, the initial concentration (\(M_1\)) is given as 6.0 M and the initial volume (\(V_1\)) as 10.0 mL. The final volume (\(V_2\)) is given as 0.250 L. We need to find the final concentration (\(M_2\)).

Step 2: Convert final volume to the same unit

Convert the final volume (\(V_2\)) from liters to milliliters, as both volumes should be in the same unit: \(V_2 = 0.250\mathrm{~L} * 1000\mathrm{\frac{mL}{L}} = 250\mathrm{~mL}\)

Step 3: Apply the dilution formula

Using the formula \(M_1 V_1 = M_2 V_2\), we have: \(6.0\mathrm{M} * 10.0\mathrm{~mL} = M_2 * 250\mathrm{~mL}\)

Step 4: Solve for the final concentration

We can now solve for \(M_2\): \(M_2 = \frac{6.0\mathrm{M} * 10.0\mathrm{~mL}}{250\mathrm{~mL}}\)

Step 5: Calculate M2

Now, carry out the calculation: \(M_2 = \frac{60\mathrm{~mL\cdot M}}{250\mathrm{~mL}} = 0.24\mathrm{M}\) So, the concentration of the diluted HNO3 solution is 0.24 M after diluting 10.0 mL of the stock solution to a final volume of 0.250 L.

Key Concepts

Molarity

Understanding molarity is essential when dealing with chemical solutions. It represents the concentration of a solution, measured in moles of solute per liter of solution. Expressed using the symbol 'M', it quantitatively describes how many moles of a substance are dissolved in a given volume of liquid. For instance, a 1 M solution contains 1 mole of solute per liter of solution.

To calculate molarity, we use the formula:
\[ Molarity (M) = \frac{moles \, of \, solute}{volume \, of \, solution \, in \, liters} \]
In applying this to our exercise, if we know the volume of the solution and the amount of substance in moles, we can calculate the molarity. For example, when preparing a dilute solution from a concentrated stock, such as diluting a 6.0 M HNO3 to 0.500 M HNO3, molarity helps us determine the volume of stock solution needed for the dilution process.

Concentration

The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent or solution. It gives us an idea of how 'strong' or 'weak' the solution is. While molarity is one way to express concentration, there are various other units like molality, normality, and parts per million (ppm).

In our example solution, concentration is expressed as molarity (moles/liter), which simplifies the process of scaling the amount of solute up or down when preparing solutions of different volumes. Precise concentration measurements are vital in fields like medicine and chemistry where specific reactions require accurate ratios of reactants.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the quantity of each element does not change in the course of a chemical reaction. Thus, the same number of atoms of an element must be present before and after a chemical change.

In the context of our dilution problem, stoichiometry isn't directly involved in calculating volumes and concentrations, but understanding it can help with more complex tasks such as calculating how much product will form from given reactants, or determining the limiting reactant in a chemical reaction. Moreover, stoichiometric principles form the foundation for understanding and applying the dilution formula correctly, helping ensure the molar amounts before and after dilution remain constant.

Solution Preparation

Preparing a solution with a desired concentration from a stock solution involves the process of dilution. Dilution means adding more solvent to a solution to decrease the concentration of solute. The dilution formula, \( M_1V_1 = M_2V_2 \), relates the molarity and volume of the initial concentrated solution (stock solution) to the molarity and volume of the final diluted solution.

Dilution calculations are important for accurately preparing solutions, especially in research and healthcare settings. In our exercise, using the dilution formula allows us to calculate the exact volume of the concentrated stock solution we need to use to achieve a final solution of a lower concentration. Accurate measurements and conversions (e.g., milliliters to liters) are critical when preparing solutions to ensure the desired molarity is achieved, and calculations are correctly performed.