How to Dilute Solutions
Determine what you do and don't know., Plug your values into the formula C1V1 = C2V2., Account for any differences in units., Solve., Understand how to use your answer practically.
Step-by-Step Guide
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Step 1: Determine what you do and don't know.
Performing a dilution in chemistry usually means taking a small amount of a solution whose concentration you know, then adding a neutral liquid (like water) to make a new solution with a larger volume but a lower concentration.
This is done very frequently in chemistry labs, as, for efficiency's sake, reagents are often stored at relatively high concentrations which are then diluted for use in experiments.
Usually, in most real-world situations, you will know the concentration of your starting solution and both the concentration and volume you want in your second solution, but not the volume of the first solution you need to use to get there.
However, in other situations (especially in schoolwork problems), you may need to find some other part of the puzzle
- for instance, you may be given an initial volume and concentration, then asked to find the final concentration if you dilute the solution to a given volume.
In the case of any dilution, it's helpful to take stock of known and unknown variables before beginning.
Let's tackle an example problem.
Say that we're tasked with diluting a 5 M solution with water to make 1 liter (0.3 US gal) of a 1 mM solution.
In this case, we know the concentration of the solution we're starting with and the target volume and concentration we want, but not how much of the initial solution we need to add water to to get there.
Reminder:
In chemistry, M is a measure of concentration called Molarity, which indicates moles of a substance per liter. -
Step 2: Plug your values into the formula C1V1 = C2V2.
In this formula, C1 is the concentration of the starting solution, V1 is the volume of the starting solution, C2 is the concentration of the final solution, and V2 is the volume of the final solution.
Plugging your known values into this equation should allow you to find the unknown value with a minimum of difficulty.
You may find it helpful to put a question mark in front of the unit you need to determine in order to help you solve for it.
Let's continue our example.
We would plug our known values in as follows:
C1V1 = C2V2 (5 M)V1 = (1 mM)(1 L).
Our two concentrations have different units.
Let's stop here and proceed to the next step. , Because dilutions involve changes in concentration (which can sometimes be quite large), it's not uncommon for two variables in your equation to be in different units.
Though this is easily overlooked, mismatching units in your equation can cause your answer to be orders of magnitude off.
Before solving, convert all values with differing concentration and/or volume units.
In our example, we use different units for concentration M (molar) and mM (millimolar).
Let's convert our second measurement to M: 1 mM × 1 M/1,000 mM =
0.001 M , When all your units match, solve your equation.
This almost always can be done with simple algebra.
We left our example problem here: (5 M)V1 = (1 mM)(1 L).
Let's solve for V1 with our new units. (5 M)V1 = (0.001 M)(1 L) V1 = (0.001 M)(1 L)/(5 M).
V1 =
0.0002 L., or
0.2 mL. , Let's say you have found your missing value, but you're unsure of how to apply this new information to a real-world dilution you need to perform.
This is understandable
- the language of math and science sometimes doesn't lend itself to the real world.
When you know all four values in the equation C1V1 = C2V2, perform your dilution as follows:
Measure the volume V1 of the solution with concentration C1.
Then, add enough diluting liquid (water, etc.) to make a total volume V2.
This new solution will have your desired concentration (C2).
In our example, for instance, we would first measure
0.2 mL of our 5 M solution.
Next, we would add enough water to increase the volume of the solution to 1 L: 1 L
-
0.0002 L =
0.9998 L, or
999.8 mL.
In other words, we would add
999.8 mL of water to our tiny sample of solution.
Our new, diluted solution has a concentration of 1 mM, which is what we wanted in the first place. -
Step 3: Account for any differences in units.
-
Step 4: Solve.
-
Step 5: Understand how to use your answer practically.
Detailed Guide
Performing a dilution in chemistry usually means taking a small amount of a solution whose concentration you know, then adding a neutral liquid (like water) to make a new solution with a larger volume but a lower concentration.
This is done very frequently in chemistry labs, as, for efficiency's sake, reagents are often stored at relatively high concentrations which are then diluted for use in experiments.
Usually, in most real-world situations, you will know the concentration of your starting solution and both the concentration and volume you want in your second solution, but not the volume of the first solution you need to use to get there.
However, in other situations (especially in schoolwork problems), you may need to find some other part of the puzzle
- for instance, you may be given an initial volume and concentration, then asked to find the final concentration if you dilute the solution to a given volume.
In the case of any dilution, it's helpful to take stock of known and unknown variables before beginning.
Let's tackle an example problem.
Say that we're tasked with diluting a 5 M solution with water to make 1 liter (0.3 US gal) of a 1 mM solution.
In this case, we know the concentration of the solution we're starting with and the target volume and concentration we want, but not how much of the initial solution we need to add water to to get there.
Reminder:
In chemistry, M is a measure of concentration called Molarity, which indicates moles of a substance per liter.
In this formula, C1 is the concentration of the starting solution, V1 is the volume of the starting solution, C2 is the concentration of the final solution, and V2 is the volume of the final solution.
Plugging your known values into this equation should allow you to find the unknown value with a minimum of difficulty.
You may find it helpful to put a question mark in front of the unit you need to determine in order to help you solve for it.
Let's continue our example.
We would plug our known values in as follows:
C1V1 = C2V2 (5 M)V1 = (1 mM)(1 L).
Our two concentrations have different units.
Let's stop here and proceed to the next step. , Because dilutions involve changes in concentration (which can sometimes be quite large), it's not uncommon for two variables in your equation to be in different units.
Though this is easily overlooked, mismatching units in your equation can cause your answer to be orders of magnitude off.
Before solving, convert all values with differing concentration and/or volume units.
In our example, we use different units for concentration M (molar) and mM (millimolar).
Let's convert our second measurement to M: 1 mM × 1 M/1,000 mM =
0.001 M , When all your units match, solve your equation.
This almost always can be done with simple algebra.
We left our example problem here: (5 M)V1 = (1 mM)(1 L).
Let's solve for V1 with our new units. (5 M)V1 = (0.001 M)(1 L) V1 = (0.001 M)(1 L)/(5 M).
V1 =
0.0002 L., or
0.2 mL. , Let's say you have found your missing value, but you're unsure of how to apply this new information to a real-world dilution you need to perform.
This is understandable
- the language of math and science sometimes doesn't lend itself to the real world.
When you know all four values in the equation C1V1 = C2V2, perform your dilution as follows:
Measure the volume V1 of the solution with concentration C1.
Then, add enough diluting liquid (water, etc.) to make a total volume V2.
This new solution will have your desired concentration (C2).
In our example, for instance, we would first measure
0.2 mL of our 5 M solution.
Next, we would add enough water to increase the volume of the solution to 1 L: 1 L
-
0.0002 L =
0.9998 L, or
999.8 mL.
In other words, we would add
999.8 mL of water to our tiny sample of solution.
Our new, diluted solution has a concentration of 1 mM, which is what we wanted in the first place.
About the Author
Jerry Murray
Specializes in breaking down complex organization topics into simple steps.
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