Resource Materials for the Biology Core Courses ------- - -- - -------------------Bates College
How to Make Simple Solutions and Dilutions RESOURCE MATERIALS INDEX | simple dilution | serial dilution | VC=VC method | molar solutions | percent concentrations | molar - % conversions
A simple dilution is one in which a unit volume of a liquid material of interest is combined with an appropriate volume of a solvent liquid to achieve the desired concentration. The dilution factor is the total number of unit volumes in which your material will be dissolved. The diluted material must then be thoroughly mixed to achieve the true dilution. For example, a 1:5 dilution (verbalize as "1 to 5" dilution) entails combining 1 unit volume of diluent (the material to be diluted) + 4 unit volumes of the solvent medium (hence, 1 + 4 = 5 = dilution factor).
Example: Frozen orange juice concentrate is usually diluted in 4 additional cans of cold water (the dilution solvent) giving a dilution factor of 5, i.e., the orange concentrate represents one unit volume to which you have added 4 more cans (same unit volumes) of water. So the orange concentrate is now distributed through 5 unit volumes. This would be called a 1:5 dilution, and the OJ is now 1/5 as concentrated as it was originally. So, in a simple dilution, add one less unit volume of solvent than the desired dilution factor value.
A serial dilution is simply a series of simple dilutions which amplifies the dilution factor quickly beginning with a small initial quantity of material (i.e., bacterial culture, a chemical, orange juice, etc.). The source of dilution material for each step comes from the diluted material of the previous. In a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step up to it.
Final dilution factor (DF) = DF1 * DF2 * DF3 etc.
Example: In the microbiology lab in Bio 201 the students perform a three step 1:100 serial dilution of a bacterial culture (see figure below). The initial step combines 1 unit volume culture (10 ul) with 99 unit volumes of broth (990 ul) = 1:100 dilution. In the next step, one unit volume of the 1:100 dilution is combined with 99 unit volumes of broth now yielding a total dilution of 1:100x100 = 1:10,000 dilution. Repeated again (the third step) the total dilution would be 1:100x10,000 = 1:1,000,000 total dilution. The concentration of bacteria is now one million times less than in the original sample.
Very often youll need to make a specific volume of known concentration due to limited availability of liquid materials (some chemicals are very expensive and are only sold and used in small quantities, e.g., micrograms) or to limit the amount of waste. The formula below is a quick approach to calculating such dilutions where:
V = volume, C = concentration; in whatever units you are working. (source solution attributes) V1C1=V2C2 (new solution attributes)
Example 1: Suppose you have 3 ml of a stock solution of 100 mg/ml ampicillin (= C1) and you want to make 200 ul (= C2)of solution having 25 mg/ ml (= V2). You need to know what volume (V1) of the stock to use as part of the 200 ul total volume needed.
V1 = the volume of stock youll start with. This is your unknown.
C1 = 100 mg/ ml in the stock solution
V2 = total volume needed at the new concentration = 200 ul = 0.2 ml
C2 = the new concentration = 25 mg/ ml
By algebraic rearrangement:
V1 = (V2 x C2) / C1
V1 = (0.2 ml x 25 mg/ml) / 100 mg/ml
and cancel the units,
= 0.05 ml
= 50 ul
So, you would take 0.05 ml = 50 ul of stock solution and dilute it with 150 ul of solvent to get the 200 ul of 25 mg/ ml solution needed (remember that the amount of solvent used is based upon the final volume needed, so you have to subtract the starting volume form the final to calculate it.)
Sometimes it may be more efficient to use molarity when calculating concentrations. A 1.0 Molar (1.0 M) solution is equivalent to 1 formula weight (FW) (g/mole) of chemical dissolved in 1 liter (1.0 L) of solvent (usually water). Formula weight is always given on the label of a chemical bottle (use molecular weight if it is not given).
Example 1: To prepare a liter of a simple molar solution from a dry reagent:
Multiply the formula weight (or MW) by the desired molarity to determine how many grams of reagent to use:
Chemical FW = 194.3 g/mole; to make 0.15 M solution use
194.3 g/mole * 0.15 moles/L = 29.145 g/L
Example 2: To prepare a specific volume of a specific molar solution from a dry reagent:
A chemical has a FW of 180 g/mole and you need 25 ml (0.025 L) of 0.15 M (M = moles/L) solution. How many grams of the chemical must be dissolved in 25 ml water to make this solution?
#grams/desired volume (L) = desired molarity (mole/L) * FW (g/mole)
by algrebraic rearrangement,
#grams = desired volume (L) * desired molarity (mole/L) * FW (g/mole)
#grams = 0.025 L * 0.15 mole/L * 180 g/mole
after cancelling the units,
#grams = 0.675 g
So, you need 0.675 g/25 ml
Many reagents are mixed as percent concentrations. When working with a dry chemical it is mixed as dry mass (g) per volume where #g/100 ml = percent concentration. A 10% solution is equal to 10 g dissolved in 100 ml of solvent.
Example 1: If you want to make 3 % NaCl you would dissolve 3.0 g NaCl in 100 ml water (or the equivalent for whatever volume you needed).
When using liquid reagents the percent concentration is based upon volume per volume, i.e., # ml/100 ml.
Example 2: If you want to make 70% ethanol you would mix 70 ml of 100% ethanol with 30 ml water (or the equivalent for whatever volume you needed).
To convert from % solution to molarity, multiply the percent solution value by 10 to get grams/L, then divide by the formula weight.
Molarity = (% solution) * 10
Example 3: Convert a 6.5 % solution of a chemical with FW = 325.6 to molarity,
[(6.5 g/100 ml) * 10] / 325.6 g/L = 0.1996 M
To convert from molarity to percent solution, multiply the molarity by the FW and divide by 10:
% solution = molarity * FW
Example 4: Convert a 0.0045 M solution of a chemical having FW 178.7 to percent solution:
[0.0045 moles/L * 178.7 g/mole] / 10 = 0.08 % solution
Modified 9-20-01 gja