Question

How many milliliters of water must be added to 120.0 \(\mathrm{mL}\) of \(1.50 \mathrm{M} \mathrm{HCl}\) to give \(1.00 \mathrm{M} \mathrm{HCl}\) ? (Assume the volumes are additive.)

Short answer

60.0 mL of water must be added.

Step-by-step solution

Identify the Dilution Formula

To dilute a solution to a desired molarity, use the dilution formula: \( C_1V_1 = C_2V_2 \), where \(C_1\) and \(C_2\) are the initial and final concentrations (molarities), and \(V_1\) and \(V_2\) are the initial and final volumes.

Substitute Known Values

Substitute the known values into the dilution formula: \(1.50 \, \mathrm{M} \cdot 120.0 \, \mathrm{mL} = 1.00 \, \mathrm{M} \cdot V_2\).

Solve for the Final Volume \(V_2\)

To find \(V_2\), divide both sides of the equation by the final concentration (\(C_2\)): \(V_2 = \frac{1.50 \, \mathrm{M} \cdot 120.0 \, \mathrm{mL}}{1.00 \, \mathrm{M}}\).

Calculate the Final Volume

Perform the calculation to determine \(V_2\): \(V_2 = \frac{180.0 \, \mathrm{mL}}{1.00}\), which simplifies to \(V_2 = 180.0 \, \mathrm{mL}\).

Determine the Volume of Water to Add

To find the volume of water needed, subtract the initial volume of the solution (\(V_1\)) from the final volume (\(V_2\)): Volume of water = \(V_2 - V_1 = 180.0 \, \mathrm{mL} - 120.0 \, \mathrm{mL} = 60.0 \, \mathrm{mL}\).

Key concepts

Solution Dilution Problems

Understanding how to approach solution dilution problems is essential in chemistry for various practical applications like preparing laboratory reagents or adjusting medication concentrations. In essence, these problems involve adding a solvent, typically water, to a solution in order to decrease its concentration to a desired level.

The key to solving dilution problems lies in the dilution formula, which is expressed as: \( C_1V_1 = C_2V_2 \). This formula indicates that the product of the initial concentration (\(C_1\)) and the initial volume (\(V_1\)) of a solution is equal to the product of the final concentration (\(C_2\)) and the final volume (\(V_2\)). The versatility of this formula allows us to solve for any one of the variables if the others are known.

For instance, in our example problem, we're asked to find how much water must be added to dilute an \(HCl\) solution. Our steps involve identifying the initial and final molarities (concentrations), and the initial volume. With these values, we manipulate the dilution formula to find the required final volume and, subsequently, determine the volume of water needed to achieve the desired dilution.

Molarity and Concentration

Molarity and concentration are fundamental concepts in chemistry that describe the amount of solute present in a given volume of solution. Molarity, denoted as \(M\), is defined as the number of moles of solute per liter of solution.

To calculate molarity, one can use the formula: \( M = \frac{moles \, of \, solute}{liters \, of \, solution} \). This formula allows for the quantification of concentration, providing a means for chemists to accurately prepare solutions with precise chemical compositions.

When adjusting concentrations through dilution, molarity provides an invaluable reference point for quantifying changes in concentration. Without a solid grasp of how molarity reflects concentration, students may struggle with accurately preparing solutions, which can have significant implications in both academic and professional laboratory settings.

Chemistry Problem Solving

Chemistry problem solving involves a systematic approach to understand and find solutions to chemical problems. This method often follows a series of steps, including identifying known and unknown variables, choosing the appropriate formulas, substituting in values, and performing calculations.

In the Context of Dilution

Specifically in dilution problems, critical thinking is used to determine the logical steps from the given information to the answer. It begins with understanding the problem statement and ends with verifying whether the calculated answer makes practical sense.

Stepwise Approach

Diligently following a stepwise approach, as demonstrated in the sample exercise, is crucial in achieving the right solution. Steps, such as identifying the correct formula, plugging in known values, solving for the unknown, and checking your work, are integral to the problem-solving process in chemistry. This technique applies not only to dilution problems but to a wide array of chemical calculations.

Fostering these problem-solving skills ensures that students gain confidence in their ability to tackle complex chemistry problems, a competence that proves to be invaluable both in the classroom and in real-world scientific endeavors.