Problem 159 - Introductory Chemistry Atoms First

Question

How would you prepare \(250.0 \mathrm{~mL}\) of a \(0.350 \mathrm{M}\) \(\mathrm{NaOH}\) solution from a \(6.00 \mathrm{M} \mathrm{NaOH}\) stock solution?

Step-by-Step Solution

Answer

To prepare \(250.0 \mathrm{~mL}\) of a \(0.350 \mathrm{M}\) NaOH solution from a \(6.00 \mathrm{M}\) NaOH stock solution, we need to mix \(14.6 \mathrm{~mL}\) of the stock solution with an appropriate volume of water (250.0 mL - 14.6 mL).

Step 1: List the Information and Formulate the Equation

The concentration and volume of the final NaOH solution are given as \(0.350 \mathrm{M}\) and \(250.0 \mathrm{~mL}\) respectively. The concentration of the stock NaOH solution is given as \(6.00 \mathrm{M}\). We denote the stock solution concentration as \(C_1\), its volume as \(V_1\), the final solution concentration as \(C_2\), and its volume as \(V_2\). We will use the dilution formula to determine the required volume of the stock solution: \(C_1V_1 = C_2V_2\)

Step 2: Substitute the Known Values into the Equation

Now, substitute the values of \(C_1\), \(C_2\), and \(V_2\) into the equation: \((6.00 \mathrm{M})(V_1) = (0.350 \mathrm{M})(250.0 \mathrm{~mL})\)

Step 3: Solve for the Volume of the Stock Solution, \(V_1\)

Now, we need to adjust the above equation to solve for \(V_1\): \(V_1 = \frac{C_2V_2}{C_1}\) Next, we can plug the known values for \(C_1\), \(C_2\), and \(V_2\) back into the equation: \(V_1 = \frac{(0.350 \mathrm{M})(250.0 \mathrm{~mL})}{6.00 \mathrm{M}}\)

Step 4: Calculate the Volume of the Stock Solution, \(V_1\)

Now, perform the calculations to find out the required volume of the stock solution: \(V_1 = \frac{(0.350)(250.0)}{6.00}\) \(V_1 = 14.5833 \mathrm{~mL}\)

Step 5: Round and Present the Final Answer

Round the final volume to an acceptable number of significant figures based on the information given: \(V_1 \approx 14.6 \mathrm{~mL}\) Thus, to prepare \(250.0 \mathrm{~mL}\) of a \(0.350 \mathrm{M}\) NaOH solution, we need to mix \(14.6 \mathrm{~mL}\) of the \(6.00 \mathrm{M}\) NaOH stock solution with an appropriate volume of water (250.0 mL - 14.6 mL).

Key Concepts

Molarity

Molarity is a measure of concentration, indicating the number of moles of a solute per liter of solution. It's a key concept in chemistry that helps describe how much of a substance is present in a given volume of liquid. For solutions, molarity is expressed as mol/L or M.
To calculate molarity, you need two values:

• The number of moles of solute, which is the substance being dissolved.
• The volume of the solution in liters, which includes both the solute and the solvent, typically water.

Molarity is crucial for solution preparation because it ensures the chemical reactions are based on a precise quantity of reactants. In this exercise, understanding the molarity of both the stock and final NaOH solutions is central to calculating how much stock solution is needed. For instance, a stock solution with a molarity of 6.00 M means it has 6 moles of NaOH per liter.

Solution Preparation

Preparing a diluted solution from a concentrated stock is a common practice in labs. It involves using a specific quantity of a stock solution, then adding enough solvent to achieve the desired concentration and volume. Here's how:

• Start by determining the final volume and molarity needed.
• Apply the dilution formula: \[ C_1V_1 = C_2V_2 \]Where: - \(C_1\): Initial concentration (stock solution) - \(V_1\): Volume of stock solution needed - \(C_2\): Final concentration (desired) - \(V_2\): Final total volume
• Calculate \(V_1\), the volume of the stock solution required.
• Mix \(V_1\) with the solvent (usually water) to achieve \(V_2\).

In the exercise, the preparation of 250.0 mL of 0.350 M NaOH involves selecting the correct volume of a 6.00 M NaOH solution, calculated as \(14.6 \mathrm{mL}\), then diluting it.

Concentration Calculation

Concentration calculations allow us to understand how much solute is in a solution. They are vital for achieving desired chemical reactivity. In dilution exercises, the concentration calculation helps us find how to transform a concentrated solution into a more diluted form.
In our exercise, we use the dilution equation: \[ C_1V_1 = C_2V_2 \]

1. Identify known values: initial concentration \(C_1 = 6.00 \mathrm{M}\), final concentration \(C_2 = 0.350 \mathrm{M}\), and final volume \(V_2 = 250.0 \mathrm{~mL}\).
2. Plug these values into the equation to solve for \(V_1\).
3. Calculate \(V_1\) to find it equals approximately \(14.6 \mathrm{~mL}\).

These calculations are crucial for properly scaling reactions or adjusting concentrations in experimental settings.

Significant Figures

Significant figures are crucial in scientific calculations. They represent the precision of the measured values and help convey how accurately the result is known. In chemistry, retaining the correct number of significant figures is important to maintain consistency and reliability in results.
Key points regarding significant figures:

• Determine based on the data provided. For the exercise, both concentration and volume values given have three significant figures.
• Ensure all calculations maintain this level of precision until the final result is determined.
• When rounding the final answer, the number of significant figures should match the least precise measurement used in calculations.

In the exercise, the calculated stock solution volume, \(14.5833 \mathrm{~mL}\), is rounded to three significant figures, resulting in \(14.6 \mathrm{~mL}\), aligning with the precision of the given data.