A concentrated solution has a higher amount of solute dissolved in a given amount of solvent compared to a diluted solution. This means a concentrated solution is stronger than a diluted solution. Understanding the difference between concentrated and diluted solutions, and how to prepare dilutions, is fundamental in chemistry and various scientific applications.
Understanding Solution Concentration
Solution concentration refers to the amount of solute present in a given amount of solvent or solution. A common way to express concentration is molarity (M), defined as the number of moles of solute per liter of solution.
When diluting a solution, more solvent (often water) is added, increasing the total volume while the amount of solute remains constant. This process decreases the concentration of the solution. Conversely, a concentrated solution has a higher ratio of solute to solvent.
The Dilution Equation
A fundamental principle governing dilutions is the dilution equation:
M1V1 = M2V2Where:
- M1 = initial concentration (molarity) of the stock solution
- V1 = initial volume of the stock solution
- M2 = final concentration (molarity) of the diluted solution
- V2 = final volume of the diluted solution
This equation highlights the inverse relationship between concentration and volume. As the volume increases (during dilution), the concentration decreases proportionally, and vice versa.
Calculating Dilutions
The dilution equation allows for the calculation of any one of the four variables if the other three are known. This is crucial for preparing solutions with specific concentrations.
Example:
Let’s say you have a stock solution of hydrochloric acid (HCl) with a concentration of 2.0 M and a volume of 100 mL. You need to prepare a diluted solution of HCl with a concentration of 0.40 M. What volume of the diluted solution will you obtain?
Using the dilution equation:
(2.0 M)(100 mL) = (0.40 M)(V2)Solving for V2:
V2 = (2.0 M * 100 mL) / 0.40 M = 500 mLTherefore, you will obtain 500 mL of the 0.40 M diluted HCl solution.
Preparing Dilutions from Stock Solutions
Another common application of the dilution equation is determining the volume of stock solution required to prepare a specific volume and concentration of a diluted solution.
Example:
You need to prepare 8.00 L of a 0.50 M nitric acid (HNO3) solution from a stock solution of 16 M HNO3. How much of the stock solution do you need?
| Steps | Solution |
|---|---|
| 1. Identify knowns and unknowns | M1 = 16 M V2 = 8.00 L M2 = 0.50 M V1 = ? |
| 2. Rearrange the dilution equation | V1 = (M2 * V2) / M1 |
| 3. Calculate V1 | V1 = (0.50 M * 8.00 L) / 16 M = 0.25 L |
You would need 0.25 L (or 250 mL) of the 16 M HNO3 stock solution to prepare the desired diluted solution. Remember to always add the concentrated acid to the water slowly and carefully, with constant stirring.
Conclusion
A concentrated solution has a higher amount of solute compared to the solvent than a diluted solution. The dilution equation, M1V1 = M2V2, provides a quantitative relationship between the concentration and volume of solutions, enabling accurate calculations for preparing dilutions. Understanding these principles is essential for various scientific and practical applications.
