Problem 37 - Chemistry An Atoms First Approach

Question

Describe how you would prepare 2.00 L of each of the following solutions. a. \(0.250 \mathrm{M}\) NaOH from solid \(\mathrm{NaOH}\) b. \(0.250 M\) NaOH from \(1.00 M\) NaOH stock solution c. \(0.100 M K_{2} C r O_{4}\) from solid \(K_{2} C r O_{4}\) d. \(0.100 M K_{2} C r O_{4}\) from \(1.75 M K_{2} C r O_{4}\) stock solution

Step-by-Step Solution

Answer

a. Weigh out 20 g of solid NaOH and dissolve it in approximately 1.90 L of water, then adjust the volume to 2.00 L. b. Measure out 0.50 L of 1.00 M NaOH stock solution, pour it into a container, and add water until the total volume is 2.00 L. c. Weigh out 58.8 g of solid \(K_{2}CrO_{4}\) and dissolve it in approximately 1.90 L of water, then adjust the volume to 2.00 L. d. Measure out 0.114 L of 1.75 M \(K_{2}CrO_{4}\) stock solution, pour it into a container, and add water until the total volume is 2.00 L.

Step 1: a. Preparing \(0.250M\) NaOH Solution from Solid NaOH

: 1. Calculate the moles of NaOH needed: Since the volume is 2.00 L and the concentration is \(0.250 M\), first find the moles of NaOH using the formula \(moles = Molarity × Volume\). \(moles = 0.250 M × 2.00 L = 0.50 moles\) 2. Convert moles to grams: Now we need to convert 0.50 moles of NaOH to grams. The molar mass of NaOH is approximately 40 g/mol, so the mass of NaOH required is: \(mass = moles × molar\ mass\) \(mass = 0.50\ moles × 40\ g/mol = 20\ g\) 3. Dissolve NaOH in water: To prepare the solution, weigh out 20 grams of solid NaOH and dissolve it in approximately 1.90 L of water. 4. Adjust the final volume: After dissolving the NaOH, add the remaining water to the solution until the total volume reaches 2.00 L.

Step 2: b. Preparing \(0.250 M\) NaOH Solution from \(1.00 M\) NaOH Stock Solution

: 1. Calculate the volume of stock solution needed: We'll use the formula \(M_1V_1 = M_2V_2\) \(V_1 = \dfrac{M_2V_2}{M_1} = \dfrac{0.250 M × 2.00 L}{1.00 M} = 0.50 L\) 2. Dilute the stock solution: To prepare the \(0.250 M\) NaOH solution, measure out 0.50 L of the \(1.00 M\) NaOH stock solution and pour it into a container. 3. Add water: Fill the container with distilled water until the total volume reaches 2.00 L.

Step 3: c. Preparing \(0.100 M K_{2}C r O_{4}\) Solution from Solid \(K_{2}C r O_{4}\)

: 1. Calculate the moles of \(K_{2}C r O_{4}\) needed: Since the volume is 2.00 L and the concentration is \(0.100 M\), we have: \(moles = molarity × volume = 0.100 M × 2.00 L = 0.20 moles\) 2. Convert moles to grams: Now we need to convert 0.20 moles of \(K_{2}C r O_{4}\) to grams. The molar mass of \(K_{2}C r O_{4}\) is approximately 294 g/mol, so: \(mass = moles × molar\ mass\ = 0.20\ moles × 294\ g/mol\ = 58.8\ g\) 3. Dissolve \(K_{2}C r O_{4}\) in water: To prepare the solution, weigh out 58.8 grams of solid \(K_{2}C r O_{4}\) and dissolve it in approximately 1.90 L of water. 4. Adjust the final volume: After dissolving the \(K_{2}C r O_{4}\), add the remaining water to the solution until the total volume reaches 2.00 L.

Step 4: d. Preparing \(0.100 M K_{2} C r O_{4}\) Solution from \(1.75 M K_{2} C r O_{4}\) Stock Solution

: 1. Calculate the volume of stock solution needed: We'll use the formula \(M_1V_1 = M_2V_2\) \(V_1 = \dfrac{M_2V_2}{M_1} = \dfrac{0.100 M × 2.00 L}{1.75 M} = 0.114 L\) 2. Dilute the stock solution: To prepare the \(0.100 M K_{2}C r O_{4}\) solution, measure out 0.114 L of the \(1.75 M K_{2}C r O_{4}\) stock solution and pour it into a container. 3. Add water: Fill the container with distilled water until the total volume reaches 2.00 L.

Key Concepts

Understanding Molarity

Molarity is an essential concept in chemistry, particularly when preparing solutions. It refers to the concentration of a solution and is expressed in terms of moles of solute per liter of solution. The formula for molarity is given as:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]To calculate molarity, you need to know two main components: the amount of solute (in moles) and the total volume of the solution (in liters). This measurement is crucial for accurate solution preparation.

• In the given exercise, the molarity values are used to calculate the quantity of solute required to achieve a specific concentration in a set volume.
• For example, preparing a 0.250 M solution of NaOH involves calculating the moles of NaOH required for a 2.00 L solution, which came out to be 0.50 moles.

Understanding molarity allows chemists to create solutions with precise concentrations, ensuring experiments and reactions proceed as expected.

The Process of Dilution

Dilution is a fundamental technique in chemistry used when a less concentrated solution is needed from a more concentrated one. This process involves adding more solvent to decrease the solute concentration to a certain level without altering the solute's total amount. The key formula used in dilution calculations is:\[ M_1V_1 = M_2V_2 \]where:

• \( M_1 \) and \( V_1 \) are the molarity and volume of the concentrated solution, respectively.
• \( M_2 \) and \( V_2 \) are the molarity and volume of the diluted solution.

In the exercise, dilution is applied to prepare solutions from stock solutions:

• For instance, a 0.250 M NaOH solution is prepared by diluting a 1.00 M stock solution. The calculation shows that 0.50 L of the stock solution is needed.
• Similarly, a 0.100 M \( K_2CrO_4 \) solution is obtained by diluting a 1.75 M stock solution, which requires 0.114 L of the original concentrated solution.

By carefully measuring and adding the diluent (often water), the desired solution concentration is achieved, making dilution a practical method in laboratory settings.

Applying Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions and is crucial when preparing chemical solutions. It involves calculations based on the balanced equation of the reaction or the makeup of compounds within a solution. This field allows chemists to predict how much of a substance is required or produced in a given reaction. Even when not directly performing a chemical reaction, stoichiometry plays a vital role in solution preparation as evident in this exercise:

• To prepare a solution from a solid, you'll first convert the desired concentration from moles to grams using the compound's molar mass.
• For example, to make a 0.250 M NaOH solution, stoichiometry is used to determine that 20 grams of NaOH is needed, based on its molar mass of 40 g/mol.

These stoichiometric calculations ensure that the correct amount of reactants and products are involved, supporting experiments with precision and accuracy.