Presentation on theme: "SKILLS Project Dilution Calculations. Why should I care? Solution preparation is one of the most common activities is chemistry lab. After all, most chemical."— Presentation transcript:
Why should I care? Solution preparation is one of the most common activities is chemistry lab. After all, most chemical reactions occur in water. This unit will teach you: –How to dilute more concentrated solutions to produce a certain molarity.
How do we do it? Dilutions are achieved by adding additional solvent, usually water, to more concentrated solutions. As a result, a greater volume of solution is produced with a lower molarity.
The Equation Where C 1 and C 2 are the starting and ending concentrations (usually molarities) and…. V 1 and V 2 are the starting and ending volumes. C 1 V 1 = C 2 V 2
Example 1: Dilution of HCl C 1 C 2 V 2 V 1 = First thing’s first: set up your equation! Remember, you can use this equation whenever you’re changing between two different molarities and volumes. How much 16.2 M HCl will be required to produce 2.44L of 0.200 M HCl solution? (16.2 M) (V 1 ) (2.44 L) (0.200 M) = According to the problem, we’re starting wih a C 1 of 16.2 M. Then, we’re diluting it down to a final concentration of 0.200 M and volume of 2.44L. We’re watering down some amount (volume) of the concentrated HCl. This volume, V 1, is what we are solving for. V 1 = 0.0301L
The concentration of our starting solution is 1.00 M. Remember to pair up your volumes and concentrations correctly. Example 2: Dilution of NaF C 1 C 2 V 2 V 1 = In general, its often easier to always place the concentration and volume of the more concentrated solution at the front. How much 0.240 M NaF solution could be made from 0.250L of a concentrated 1.00 M solution of NaF? (1.00 M) (0.250 L) (V 2 ) (0.240 M) = V 2 = 1.04L Once again, we’ll use the dilution equation. The volume of this same, starting solution, is 0.250L. We’re making a solution with a concentration of 0.240 M NaF. Finally, we’re looking for the volume of the resulting solution. This is the “how much” we’ll be solving for.
Our starting solution has a concentration (C 1 ) of 5.60 M NH 3. Our starting solution has a volume (V 1 ) of 0.300 L. We’ll be adding to this volume to produce a lower molarity. The “ending” solution should be diluted down to a molarity of 0.100 M NH 3. Lastly, we’re looking for the “ending” volume. Note: this is the total final volume. Start out with the dilution equation. As always, this will form the framework for solving this problem. Example 3: Ammonia C 1 C 2 V 2 V 1 = How much water has to be added to 0.300L of 5.60 M NH 3 to make the concentration 0.100 M? (5.60 M) (0.300 L) (V 2 ) (0.100 M) = V 2 = 16.8L. We’ll need to add 16.5 L of water.
Once again, we’ll start with the dilution equation as our general “framework.” Example 4: Milliliters C 1 C 2 V 2 V 1 = What is the initial concentration of a solution if 350 ml can be diluted to a final volume of 1.40L and concentration of 0.150 M? (C 1 ) (350 ml) (1400 ml) (0.150 M) = C 1 = 0.600 M The problem states that we’re searching for the initial concentration, C 1. This time around, our volume is in milliliters instead of liters. We can use “ml” as long as we are consistent as use ml in V 2 as well. The final concentration, C 2, is stated to be 0.150 M. For our V 2, we need to be sure to match units with our V 1. So, we’ll use 1400 ml instead of 1.40 L. Remember: 1.0L = 1000 ml.
We’re trying to produce a final concentration, C 2, of 0.0200 M HI. And lastly, we’re trying to find the final volume so we can determine how much water will need to be added. As usual, we start with our dilution equation. Example 5: Preparations C 1 C 2 V 2 V 1 = If 50.0g HI is dissolved in 500. ml of water, how much additional water must be added to produce a final concentration of 0.0200 M HI? (0.781 M) (500 ml) (V 2 ) (0.0200 M) = V 2 = 19525 ml. An additional 19025 ml (or 19.025L) will be added. To find our C 1, we’ll have to solve for molarity. Using M = mol/L, we find that the starting solution has a concentration of 0.781 M HI. We can also assume that the total volume of our starting solution is 500 ml.
Practice Problems: What is the final concentration of a solution made by adding 250 ml of water to 10 ml of 12.0M H 2 SO 4 ? 0.46 M H 2 SO 4 7.2 M HCl What is the initial concentration of an HCl solution if 0.50L can be diluted down to 3.00L with a concentration of 1.20 M?