Problem 12 - Chemistry: Principles and Reactions

Question

Describe how you would prepare \(1.00 \mathrm{~L}\) of \(0.750 \mathrm{M}\) barium hydroxide solution starting with (a) solid barium hydroxide. (b) \(6.00 \mathrm{M}\) barium hydroxide solution.

Step-by-Step Solution

Answer

Answer: (a) To prepare the solution using solid barium hydroxide, weigh out 128.50 g of solid Ba(OH)_2 and dissolve it in roughly 800 mL of deionized water. Stir until the solid is completely dissolved. Transfer the solution to a 1.00 L volumetric flask and carefully add deionized water until the 1.00 L mark is reached, mixing thoroughly. (b) To prepare the solution using a 6.00 M Ba(OH)_2 solution, measure 125 mL of the 6.00 M Ba(OH)_2 solution and transfer it to a 1.00 L volumetric flask. Add deionized water to the flask until the 1.00 L mark is reached, mixing thoroughly.

Step 1: Calculate the moles of barium hydroxide required

To calculate the moles of barium hydroxide needed, use the formula: moles = molarity × volume, where molarity is 0.750 M and volume is 1.00 L (converted to liters). moles = 0.750 M * 1.00 L moles = 0.750 mol

Step 2: Convert moles to grams

Now, find the molar mass of Ba(OH)_2: Ba(OH)_2 = 137.33 g/mol (Ba) + 2 * (15.999 g/mol (O) + 1.008 g/mol (H)) Ba(OH)_2 ≈ 171.34 g/mol Next, convert the moles of Ba(OH)_2 to grams: 0.750 mol × 171.34 g/mol ≈ 128.50 g

Step 3: Prepare the solution

Weigh out 128.50 g of solid Ba(OH)_2 and dissolve it in roughly 800 mL of deionized water. Stir the solution until the solid is completely dissolved. Transfer the solution to a 1.00 L volumetric flask and carefully add deionized water until the 1.00 L mark is reached, mixing thoroughly. Your 0.750 M Ba(OH)_2 solution is now prepared. (b) Preparing a solution using a 6.00 M barium hydroxide solution:

Step 1: Determine the volume of 6.00 M Ba(OH)_2 solution needed

Use the dilution equation: M1 * V1 = M2 * V2 Where M1 and V1 are the molarity and volume, respectively, of the more concentrated solution (6.00 M Ba(OH)_2), and M2 and V2 are the molarity and volume of the target solution (0.750 M Ba(OH)_2 and 1.00 L). Rearrange the equation to solve for V1: V1 = (M2 * V2) / M1 V1 = (0.750 M * 1.00 L) / 6.00 M V1 ≈ 0.125 L = 125 mL

Step 2: Prepare the solution

Measure 125 mL of the 6.00 M Ba(OH)_2 solution using a graduated cylinder or pipette. Transfer the measured solution to a 1.00 L volumetric flask. Add deionized water to the flask until the 1.00 L mark is reached, mixing thoroughly. Your 0.750 M Ba(OH)_2 solution is now prepared.

Key Concepts

Understanding Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution. Molarity is often represented by the symbol "M," and its unit is moles per liter (mol/L). To calculate molarity, use the formula:

• Molarity (M) = Moles of solute / Volume of solution in liters

When preparing solutions, knowing the molarity helps ensure the right amount of substance is present in a given volume. This is crucial for chemical reactions, where precise concentrations are necessary for predictability.
By defining the specific amount of solute in a given volume of solvent, molarity provides a clear and easily understood measure of concentration, which can be quite useful in laboratory settings.

Exploring Solution Concentration

Solution concentration refers to the amount of solute present in a given quantity of solvent or solution. It provides insight into how dense or dilute a solution is, impacting chemical behavior and reactions. The higher the concentration, the more solute is present in a liter of solution.
Concentration can be expressed in various ways, but molarity is one of the most common in chemistry due to its simplicity and direct relation to volume, which is often controlled in experiments. When preparing a solution, it's essential to measure the concentration accurately. This ensures that subsequent reactions, tests, or uses of the solution are effective and reliable. Higher concentrations generally mean stronger solutions, which can have implications for how substances interact during their use.

Using the Dilution Equation

The dilution equation is a powerful tool used to calculate how to make a diluted solution from a more concentrated one. It is expressed as:

• \( M_1 \times V_1 = M_2 \times V_2 \)

Here, \( M_1 \) and \( V_1 \) are the molarity and volume of the original concentrated solution, while \( M_2 \) and \( V_2 \) are the molarity and volume of the desired diluted solution.
This equation helps preserve the quantity of solute when transitioning from different solution volumes. A common approach, once the equation is rearranged to solve for the unknown, is to measure the calculated volume of the concentrated solution, then add solvent to reach the desired final volume.
Using this method ensures that the dilution maintains the correct proportion of solute to solvent, resulting in the right concentration for your needs.

Calculating Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It bridges the gap between the microscopic world of atoms and the macroscopic world of grams and liters in the lab.To calculate molar mass, sum the atomic masses of each element in a chemical formula:

• Find the atomic mass of each element (usually in g/mol) from the periodic table.
• Multiply the atomic mass by the number of atoms of that element in the formula.
• Add the results together to obtain the total molar mass of the compound.

For example, the molar mass of barium hydroxide, \( \text{Ba(OH)}_2 \), requires finding the atomic masses of barium (Ba), oxygen (O), and hydrogen (H), and calculating as shown:

• Barium: 137.33 g/mol
• Oxygen: 15.999 g/mol
• Hydrogen: 1.008 g/mol

Then,

• \( \text{Ba(OH)}_2 = 137.33 + 2 \times (15.999 + 1.008) \approx 171.34 \, \text{g/mol} \)

Knowing the molar mass allows you to convert between moles and grams, which is crucial for preparing solutions with exact concentrations.