Problem 55 - Chemistry The Molecular Science

Question

Calculate the mass \((\mathrm{g})\) of \(\mathrm{KI}\) required to prepare \(100 . \mathrm{mL}\) of \(0.0200-\mathrm{M}\) KI. How many milliliters of this solution are required to produce \(250 . \mathrm{mL}\) of \(0.00100-\mathrm{M} \mathrm{KI} ?\)

Step-by-Step Solution

Answer

0.332 g of KI is needed, and 12.5 mL of this solution is required.

Step 1: Understand the Problem

We want to calculate the mass of potassium iodide (KI) required to make a specific concentration in a given volume, and then determine the volume needed to dilute it to another concentration.

Step 2: Calculate Moles of KI Needed

For a 0.0200 M solution of KI in 100 mL (0.1 L), we can use the molarity formula: \( \text{moles} = \text{molarity} \times \text{volume} = 0.0200 \, \text{M} \times 0.100 \, \text{L} = 0.002 \text{ moles of KI} \).

Step 3: Calculate Mass of KI Required

Use the moles of KI and its molar mass (166.0028 g/mol) to find the mass: \( \text{mass} = \text{moles} \times \text{molar mass} = 0.002 \times 166.0028 \, \text{g/mol} = 0.332 \text{ g} \).

Step 4: Determine Initial Solution Volume for New Concentration

To find how many mL of the initial solution is required to make 250 mL of 0.00100 M KI, use the dilution equation: \( C_1V_1 = C_2V_2 \). Here, \( C_1 = 0.0200 \, \text{M} \), \( V_2 = 0.250 \, \text{L} \), and \( C_2 = 0.00100 \, \text{M} \).

Step 5: Solve for \( V_1 \) in the Dilution Equation

Rearrange the dilution equation to solve for \( V_1 \): \( V_1 = \frac{C_2V_2}{C_1} = \frac{0.00100 \, \text{M} \times 0.250 \, \text{L}}{0.0200 \, \text{M}} = 0.0125 \, \text{L} = 12.5 \, \text{mL} \).

Key Concepts

Mole Calculations

When working in chemistry, we often need to calculate the number of moles of a substance, as reactions occur in terms of moles. To find moles, we use the relationship between molarity, moles, and volume. Molarity ( M ) is defined as moles of solute per liter of solution. Here's a simple formula you can use:

• Moles = Molarity × Volume

Let's consider our problem. We need to find out how many moles of potassium iodide (KI) are necessary for a 0.0200 M solution in 100 mL of water. First, convert milliliters to liters, as molarity uses liters:

• 100 mL = 0.100 L

Now plug the values into the formula:

• Moles = 0.0200 M × 0.100 L = 0.002 moles

This tells us that 0.002 moles of KI are needed for the desired solution.

Mass Calculation

Once we have the moles of a compound, the next step often involves converting moles to grams, since mass is a more tangible measurement used in labs. To do this conversion, use the molar mass of the substance. The molar mass is the mass of one mole of a compound and is measured in g/mol.
For potassium iodide (KI), the molar mass is 166.0028 g/mol. With 0.002 moles of KI calculated earlier, we can find its mass by multiplying the moles by the molar mass:

• Mass = Moles × Molar Mass
• 0.002 moles × 166.0028 g/mol = 0.332 g

Thus, you need 0.332 grams of KI to prepare the solution.
This step highlights the ease of converting between moles and grams, which is crucial when preparing solutions in chemistry.

Dilution Equation

The dilution equation is a handy tool for when you want to decrease the concentration of a solution while maintaining the amount of solute. It is described by the following formula:

• \[ C_1V_1 = C_2V_2 \]

Here, \( C_1 \) and \( V_1 \) are the concentration and volume of the initial solution, while \( C_2 \) and \( V_2 \) are for the diluted solution.
In our example, the solution starts at 0.0200 M (\( C_1 \)) and needs to end up at 0.00100 M (\( C_2 \)) in a 250 mL total volume (\( V_2 \)). We need to find out the initial volume (\( V_1 \)) required:

• \[ V_1 = \frac{C_2 \times V_2}{C_1} \]
• \[ V_1 = \frac{0.00100 \, \mathrm{M} \times 0.250 \, \mathrm{L}}{0.0200 \, \mathrm{M}} = 0.0125 \, \mathrm{L} = 12.5 \mathrm{\, mL} \]

You need to use 12.5 mL of the initial solution to prepare the diluted solution. This shows how the dilution equation helps in determining precise volumes needed for preparing specific concentrations.