Dilution and Solution Concentration in GOB Chemistry

Dilution and Solution Concentration

Introduction to Dilution

In chemistry, a stock solution is a concentrated solution that can be diluted for laboratory use. Dilution involves adding more solvent (commonly water) to decrease the concentration of a solution. This process is essential for preparing solutions of desired concentrations for experiments and analyses.

• Concentrated solution: Contains a high amount of solute per unit volume.

• Diluted solution: Contains a lower amount of solute per unit volume due to the addition of solvent.

Example: Adding water to a concentrated acid to make it less corrosive and suitable for titration.

Visualizing Dilution

When comparing solutions, the number of solute particles per volume indicates concentration. For example, if each sphere represents a mole of solute:

• Solution A (5.0 L): Least concentrated (fewest particles per volume)

• Solution B (2.0 L): Intermediate concentration

• Solution C (0.5 L): Most concentrated (most particles per volume)

Application: Arranging solutions by concentration helps in understanding how dilution affects solute distribution.

Dilution Calculations

Dilution Formula

The relationship between the concentrations and volumes before and after dilution is given by the dilution equation:

$M_1 V_1 = M_2 V_2$

• $M_1$: Initial molarity (concentration) of the stock solution

• $V_1$: Volume of stock solution used

• $M_2$: Final molarity (concentration) after dilution

• $V_2$: Final total volume after dilution

This formula allows calculation of any one variable if the other three are known.

Worked Example

Example: What volume (in mL) of 5.2 M HBr must be used to prepare 3.5 L of 2.7 M HBr?

• Given: $M_1 = 5.2$ M, $M_2 = 2.7$ M, $V_2 = 3.5$ L

• Find: $V_1$

Using the formula:

$V_1 = \frac{M_2 V_2}{M_1}$

$V_1 = \frac{2.7 \times 3.5}{5.2} = 1.82$ L

Application: This calculation is commonly used in laboratory settings to prepare solutions of desired concentrations.

Practice Problems

Practice 1: Final Volume Calculation

Question: To what final volume would 100 mL of 5.0 M KCl have to be diluted to make a solution that is 0.54 M KCl?

• Use $M_1 V_1 = M_2 V_2$

• $M_1 = 5.0$ M, $V_1 = 100$ mL, $M_2 = 0.54$ M

• $V_2 = \frac{M_1 V_1}{M_2} = \frac{5.0 \times 100}{0.54} = 925.93$ mL

Practice 2: Resulting Molarity After Dilution

Question: If 880 mL of water is added to 125.0 mL of a 0.770 M HBrO4 solution, what is the resulting molarity?

• Total volume after dilution: $125.0 + 880 = 1005.0$ mL

• Use $M_1 V_1 = M_2 V_2$

• $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.770 \times 125.0}{1005.0} = 0.0958$ M

Practice 3: Serial Dilution Calculation

Question: A student prepared a stock solution by dissolving 25.0 g of NaOH in enough water to make 150.0 mL solution. The student took 20.0 mL of the stock solution and diluted it with water to make 250.0 mL solution. Finally, 75.0 mL of that solution was diluted in water to make 500.0 mL solution. What is the concentration of NaOH in this final solution? (MM of NaOH = 40.00 g/mol)

• Step 1: Calculate initial molarity: $M_1 = \frac{25.0}{40.00 \times 0.150} = 4.17$ M

• Step 2: First dilution: $M_2 = \frac{4.17 \times 20.0}{250.0} = 0.334$ M

• Step 3: Second dilution: $M_3 = \frac{0.334 \times 75.0}{500.0} = 0.0501$ M

Application: Serial dilutions are used to achieve very low concentrations from a concentrated stock solution.

Summary Table: Dilution Formula Variables

Variable	Definition	Units
$M_1$	Initial concentration (molarity) of stock solution	mol/L (M)
$V_1$	Volume of stock solution used	L or mL
$M_2$	Final concentration after dilution	mol/L (M)
$V_2$	Final total volume after dilution	L or mL

Additional info: The notes also include practice problems and stepwise calculations, which are essential for mastering dilution concepts in GOB Chemistry.