Question

What volume of \(0.25 \mathrm{M}\) HCl solution must be diluted to prepare 1.00 L of \(0.040 M\) HCl?

Short answer

To prepare 1.00 L of 0.040 M HCl solution from a 0.25 M HCl solution, you need to dilute 0.16 L (or 160 mL) of the 0.25 M HCl solution.

Step-by-step solution

List the given information

We are given the following information: - Initial molarity of HCl solution, \(M_1 = 0.25 M\) - Final molarity of HCl solution, \(M_2 = 0.040 M\) - Final volume of the diluted HCl solution, \(V_2 = 1.00 L\) Our goal is to find the initial volume of HCl solution, \(V_1\).

Use the dilution formula to solve for \(V_1\)

Using the dilution formula, we can solve for the initial volume of the HCl solution, \(V_1\): \[M_1V_1 = M_2V_2\] \[V_1 = \frac{M_2V_2}{M_1} \] Now, plug in the given values: \[V_1 = \frac{(0.040 \mathrm{M})(1.00 \mathrm{L})}{0.25 \mathrm{M}}\]

Calculate the initial volume \(V_1\)

Now, perform the calculation: \[V_1 = \frac{0.040 \mathrm{L}}{0.25}\] \[V_1 = 0.16 \mathrm{L}\]

Present the final result

So, the volume, \(V_1\), of the \(0.25 \mathrm{M}\) HCl solution that must be diluted to prepare \(1.00 \mathrm{L}\) of a \(0.040 \mathrm{M}\) HCl solution is \(0.16 \mathrm{L}\) (which is equivalent to \(160 \mathrm{mL}\)).

Key concepts

Molarity

Molarity is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute divided by the volume of the solution in liters. The molarity formula is expressed as:
\[ M = \frac{n}{V} \]
where \(M\) is the molarity, \(n\) is the number of moles of solute, and \(V\) is the volume of the solution in liters. When preparing a chemical solution, understanding molarity is vital as it helps determine the exact amount of substance needed to achieve a specific concentration. In our exercise, we have a starting molarity (initial concentration) of HCl, and we want to reach a different, lower molarity (final concentration) by adding solvent, typically water, which is an example of dilution.

Dilution Formula

The dilution formula is a straightforward equation that shows the relationship between the concentrations and volumes before and after the dilution process:
\[ M_1V_1 = M_2V_2 \]
In this equation, \(M_1\) and \(M_2\) represent the initial and final molarities, while \(V_1\) and \(V_2\) represent the initial and final volumes, respectively. This formula is based on the principle of conservation of moles because diluting a solution with solvent doesn't change the amount of solute, just its concentration. Using the dilution formula, we can manipulate the variables to find the unknown, as shown in the exercise to determine the needed volume of the more concentrated solution to achieve a specific dilution.

Concentration

Concentration refers to the amount of a substance in a given volume of solution. Besides molarity, there are other ways to express concentration, such as mass percent, mole fraction, and parts per million (ppm). However, molarity is especially useful in stoichiometry and dilutions due to its direct relationship with volume, which most laboratory measurements involve. When we dilute a solution, we are reducing its concentration, effectively spreading out the solute particles within a larger volume of solvent, typically water. As shown in the provided exercise, understanding how concentration changes during dilution is critical for various scientific and industrial applications.

Solution Preparation

Preparing a solution with a specific molarity involves dissolving the desired moles of solute into a solvent, normally water, up to a precise volume. Standard laboratory practice includes using a volumetric flask to ensure that the volume of the resulting solution is exact. The steps typically involve adding the solute to the flask, partially filling with solvent, mixing until the solite is dissolved completely, and then bringing the volume up to the desired level. With the principles of molarity and concentration in mind, these solutions can then be diluted to a desired lower concentration if required, as demonstrated by our exercise, without altering the total moles of solute present in the solution.