Question

How would you prepare \(60.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) from a stock solution of \(4.00 M \mathrm{HNO}_{3} ?\)

Short answer

You need to take 3.00 mL of the 4.00 M HNO3 stock solution and dilute it to 60.0 mL to create a 0.200 M HNO3 solution.

Step-by-step solution

Understand the given

Here, the initial concentration \(C_1\) is given as 4.00 M, we want to prepare 60.0 mL of a 0.200 M solution. Therefore, the final concentration \(C_2\) is 0.200 M and the final volume \(V_2\) is 60.0 mL. We need to find the volume of the concentrated solution we must use, which will be \(V_1\).

Conversion of units

Since \(M = mol/L\), for the equation \(C_1V_1 = C_2V_2\) to be correct, all variables should be in the same units. Start by converting the final volume from mL to L: \(60.0 mL = 0.060 L\).

Apply the dilution formula

Apply the formula \(C_1V_1 = C_2V_2\) and solve for \(V_1\): \(V_1 = \frac{C_2 * V_2}{C_1} = \frac{0.200 M * 0.060 L}{4.00 M} = 0.00300 L\)

Convert the result back to mL

Convert the volume we found from L back to mL since it's the more standard unit in a lab context: \(0.00300 L = 3.00 mL\). It is this volume of 4.00 M HNO3 that we need to dilute to 60.0 mL to achieve a solution of 0.200 M HNO3.

Key concepts

Concentration and Molarity

When it comes to dilution calculation in chemistry, understanding the concept of concentration and molarity is crucial. Molarity, denoted by 'M', is a common measure of concentration representing moles of solute per liter of solution. It's important to be familiar with the molarity formula, which is:
\( M = \frac{moles \ of \ solute}{liters \ of \ solution} \)
This formula helps you determine how concentrated a solution is. When preparing a diluted solution from a more concentrated one, the molarity changes, but the total amount of solute in moles remains the same. As a result, the product of the initial concentration (\(C_1\)) and the initial volume (\(V_1\)) equals the product of the final concentration (\(C_2\)) and the final volume (\(V_2\)), expressed in the equation \(C_1V_1 = C_2V_2\).

Solution Preparation

Preparing a new solution of a specific concentration from a stock solution involves a process known as dilution. The key to this process is the proper use of the dilution formula mentioned above (
\(C_1V_1 = C_2V_2\)). The process typically involves adding solvent, usually water, to a measure of stock solution to achieve the desired concentration. Doing so correctly is essential, as improper dilution can lead to inaccurate experimental results. Always add solvent to the solute when diluting, to ensure safe and accurate preparation. Additionally, mixing should be done carefully to ensure that the solute is evenly distributed throughout the solution.

Volume Conversion

In chemistry, it is essential to be comfortable with converting volumes. Different processes and instruments might require measurements in various units, so converting between milliliters (mL) and liters (L) is common practice. To convert mL to L, divide the volume in mL by 1,000. Conversely, to convert L to mL, multiply the volume in L by 1,000. In the context of our exercise, the final volume needed in liters was calculated by converting from milliliters:
\(60.0 mL = 60.0 \div 1000 = 0.060 L\).
Mastering these conversions is essential, ensuring accuracy in the preparation of solutions.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is rooted in the law of conservation of mass, where the mass of the reactants equals the mass of the products. Stoichiometry is not just limited to reactions; it applies to solution preparation as well. In the context of preparing solutions, stoichiometry involves ensuring that the number of moles before dilution remains constant after dilution. The dilution formula used for our calculations is a form of stoichiometry as it confirms that the moles of solute before and after dilution are equivalent.
In practice, stoichiometry will help determine the volume (\(V_1\)) of the concentrated stock solution required to achieve a specific molarity and volume (\(V_2\)) of the diluted solution.