Question

How much water must be added to \(500 .\) mL of 0.200 \(M \mathrm{HCl}\) to produce a \(0.150 \mathrm{M}\) solution? (Assume that the volumes are additive.)

Short answer

To produce a 0.150 M solution from 500 mL of 0.200 M HCl, 166.67 mL of water must be added.

Step-by-step solution

Write down the dilution formula

We will use the dilution formula: \[C_1V_1 = C_2V_2\]

Substitute given values into the formula

We have \(C_1 = 0.200\,{M}, V_1 = 500\,\text{mL}, \text{and } C_2 = 0.150\,{M}\). Substitute these values into the formula: \[(0.200\,{M})(500\,\text{mL}) = (0.150\,{M})(V_2)\]

Solve for the final volume V2

To find the final volume (\(V_2\)), we will divide both sides of the equation by 0.150 M: \[\frac{(0.200\,{M})(500\,\text{mL})}{(0.150\,{M})} = V_2\]

Calculate V2

Now, perform the calculation to find the value of \(V_2\): \[V_2 = \frac{(0.200\,{M})(500\,\text{mL})}{(0.150\,{M})} = 666.67\,\text{mL}\]

Determine the volume of water added

To find the volume of water added, subtract the initial volume of HCl (\(V_1\)) from the final volume (\(V_2\)): \[\text{Volume of water added} = V_2 - V_1 = 666.67\,\text{mL} - 500\,\text{mL} = 166.67\,\text{mL}\] So, 166.67 mL of water needs to be added to the 0.200 M HCl solution to produce a 0.150 M solution.

Key concepts

Dilution Formula

Understanding the dilution formula is essential when working with solutions in chemistry. Dilution involves adding a solvent, usually water, to a concentrated solution in order to decrease its concentration. The formula that represents this process is expressed mathematically as:
\[\begin{equation}C_1V_1 = C_2V_2 \br\br\br\br\br\br\br\end{equation}\] where:

• \(C_1\) is the original concentration of the solution,
• \(V_1\) is the original volume of the solution,
• \(C_2\) is the final concentration after dilution, and
• \(V_2\) is the final volume after adding the solvent.

To demonstrate, if we start with 500 mL of a 0.200 M HCl solution and wish to dilute it to 0.150 M, we keep the molarity of the original and final solution and the initial volume constant in our equation while solving for the new volume. This relationship allows us to control the concentration of a solution precisely, which is crucial in various scientific and industrial applications.

Molarity Calculation

Molarity, represented as M, measures the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The formula for calculating molarity is as follows:
\[\begin{equation}M = \frac{\text{moles of solute}}{\text{liters of solution}}\br\br\br\br\end{equation}\] For effective molarity calculation, it's important to know both the amount of substance present (in moles) and the volume of the solution (in liters).

Example:

If we have a 500 mL (or 0.500 L) solution containing 0.200 moles of HCl, the molarity would be calculated as:
\[\begin{equation}M = \frac{0.200\text{ moles}}{0.500\text{ L}} = 0.400\text{ M}\br\br\br\br\end{equation}\] By mastering the concepts of molarity, students will be able to prepare solutions of desired concentrations and understand the quantitative aspects of chemical reactions in solution.

Solution Concentration Adjustment

Solution concentration adjustment is a frequent task in laboratories that requires precision and a solid grasp of the dilution concept. To adjust the concentration of a solution, a chemist must add the correct amount of solvent or remove some solute, depending on the desired outcome.When the concentration needs to be decreased, dilution is performed. This is done by adding more solvent, which increases the volume of the solution while keeping the amount of solute the same. However, if we need to increase the concentration, we can either add more solute or evaporate some solvent to reduce the volume.

Practical Application:

Given a 0.200 M HCl solution, if we aim to achieve a 0.150 M solution, water must be added. By applying the dilution formula, we can determine the exact volume of water needed. In our case, adding 166.67 mL of water provided the desired concentration adjustment. Such ability to manipulate concentrations is vital for preparing reagents, conducting experiments, and even discussing pharmacological dosages.