Problem 30 - Chemistry: An Atoms-Focused Approach

Question

Chemists who analyze samples for dissolved trace elements may buy standard solutions that contain \(1.000 \mathrm{g} / \mathrm{L}\) concentrations of the elements. If a chemist needs to prepare \(0.500 \mathrm{L}\) of a working standard that has a concentration of \(5.00 \mathrm{mg} / \mathrm{L},\) what volume of the \(1.000 \mathrm{g} / \mathrm{L}\) standard is needed?

Step-by-Step Solution

Answer

Answer: 2.5 mL

Step 1: Write down the given values

We are given: - Initial concentration (C1) = 1.000 g/L - Final concentration (C2) = 5.00 mg/L (we need to convert this to g/L) - Final volume (V2) = 0.500 L - Initial volume (V1) = ?

Step 2: Convert the final concentration to g/L

Given that 1 g = 1000 mg, we need to convert the final concentration from mg/L to g/L: C2 = 5.00 mg/L × (1 g / 1000 mg) = 0.005 g/L

Step 3: Use the dilution formula to find the initial volume

The formula for dilution is: C1 × V1 = C2 × V2 Now we need to find the initial volume (V1). So, we will rewrite the formula to solve for V1: V1 = (C2 × V2) / C1

Step 4: Substitute the values and calculate the initial volume

Now, plug in the values of C1, C2, and V2 into the formula and solve for V1: V1 = (0.005 g/L × 0.500 L) / 1.000 g/L V1 = (0.0025 g) / 1.000 g/L V1 = 0.0025 L

Step 5: Convert the initial volume to mL

To convert the initial volume from L to mL, we will use the conversion factor (1 L = 1000 mL): V1 = 0.0025 L × (1000 mL / 1 L) = 2.5 mL The chemist needs to take 2.5 mL of the 1.000 g/L standard to prepare 0.500 L of the working standard with a concentration of 5.00 mg/L.

Key Concepts

Understanding Dilution in Solution Preparation

Dilution is a fundamental concept in chemistry, especially when it comes to preparing solutions.
In essence, dilution involves reducing the concentration of a solute in a solution, typically by adding more solvent.
When a chemist needs a solution at a lower concentration, they usually start with a more concentrated stock solution.

• The stock solution has a higher concentration and serves as the base.
• The desired solution is at a lower concentration and is often referred to as the working solution.

The general formula for dilution is:\[C_1 \times V_1 = C_2 \times V_2\] Where:

• \(C_1\) is the concentration of the stock solution.
• \(V_1\) is the volume of the stock solution needed.
• \(C_2\) is the concentration of the working solution.
• \(V_2\) is the volume of the working solution.

In our exercise, the chemist dilutes a solution with 1.000 g/L to make a 5.00 mg/L solution, demonstrating how the concentrated stock is adjusted to meet a specific standard.

Concentration Calculation: A Crucial Step

Calculating the concentration is an integral part of preparing solutions.
This involves determining how much solute is present in a given volume of solution, typically expressed in terms of mass per volume (e.g., g/L or mg/L).
When converting concentration units, it's important to be precise.

• Given: 5.00 mg/L needs to be converted to g/L for compatibility with the stock solution.
• Conversion factor: 1 g = 1000 mg
• So, \(5.00 \, \text{mg/L} \times \frac{1 \, \text{g}}{1000 \, \text{mg}} = 0.005 \, \text{g/L}\)

Understanding unit conversion in these calculations ensures the accuracy of the final solution's concentration.
Once the concentration is in the same units, using the dilution formula becomes straightforward, allowing for precise adjustments in experiments or applications.

Volume Conversion in Solution Preparation

Volume conversion is a common task in chemistry that often accompanies dilution processes.
When preparing solutions, you may need to switch between units of volume measurement.

• Common conversions include liters to milliliters or vice versa.
• The conversion factor most frequently used: 1 L = 1000 mL

In our example problem:

• The calculated volume for the stock solution is 0.0025 L.
• To convert to a more manageable unit for laboratory equipment, we use 0.0025 L × 1000 mL/L = 2.5 mL.

Understanding these conversions allows for seamless preparation and measurement, ensuring that volumes are accurately adjusted to provide precise concentrations.