The dilution factor (DF) is a number that indicates how many times a solution must be diluted to obtain a lower concentration . The solution can have either a solid, liquid or a gaseous solute dissolved. Therefore, its concentration depends on the number of particles of the solute and the total volume V.
In the field of chemistry , many expressions of concentration are used: percentage, molar (M), normal (N), among others. Each of them depends on a finite amount of solute; from grams, kilograms, or moles, to equivalents. However, when reducing such concentrations, the DF applies to all these expressions.
An example of a successive dilution of grenadine is shown in the image above. Note that from left to right the red color is becoming lighter; which is equal to a lower concentration of grenadine.
The dilution factor allows you to determine how dilute the last glass is compared to the first. Thus, instead of the simple organoleptic properties, with DF the experiment can be repeated from the same bottle of grenadine (stock solution); so that in this way it is ensured that the concentrations of the new vessels are equal.
The concentration of grenadine can be expressed in any unit; however, the volume of the vessels is constant, and to facilitate calculations, we simply resort to the volumes of grenadine dissolved in water. The sum of these will be equal to V: the total volume of liquid in the glass.
As with the grenadine in the example, it happens in the laboratory with any other reagent. Concentrated stock solutions are prepared, from which aliquots are taken, and diluted to obtain more dilute solutions. In this way, it seeks to reduce risks in the laboratory and losses of reagents.
What is the dilution factor?
Dilution is a procedure that allows the reduction of the concentration of a solution or its density. The action of reducing the intensity of the color in a solution of a colorant can also be considered as a dilution.
To successfully dilute a solution to a certain concentration, the first thing to do is to know how many times the concentration of the stock solution is greater than the concentration of the diluted solution.
Thus, it is known how many times the initial solution must be diluted to obtain a solution with the desired concentration. The number of times is what is known as the dilution factor. And in this it consists, in a dimensionless fraction, which indicates a dilution.
It is common to find a dilution expressed, for example, as follows: 1/5, 1/10, 1/100, etc. What does this mean? It simply indicates that to obtain a solution with the desired concentration, the stock solution must be diluted as many times as the denominator of the named fraction indicates.
If, for example, the 1/5 dilution is used, the initial solution must be diluted 5 times to obtain a solution with this concentration. Therefore, the number 5 is the dilution factor. This translates as follows: the 1/5 solution is five times more dilute than the mother.
How to prepare such a solution? If 1mL of the stock solution is taken, this volume must be quintupled, so that the solute concentration is diluted by a factor of 1/5. So, if it is going to be diluted with water (as in the grenadine example), 4 mL of water must be added to 1mL of this solution (1 + 4 = 5mL of final volume V F ).
Next, we will discuss how to deduct and calculate the DF.
How do you get the dilution factor?
To prepare a dilution, a volume of an initial or stock solution is taken to a volumetric flask, where water is added until the measuring capacity of the volumetric flask is completed.
In this case, when water is added to the volumetric flask, no solute mass is being added. So, the mass of solute or solution remains constant:
m i = m f (1)
m i = mass of the initial solute (in the concentrated solution).
Y m f = mass of the final solute (in the diluted solution).
But, m = V x C. Substituting in equation (1), we have:
V i x C i = V f x C f (2)
V i = volume of the stock or initial solution that was taken to make the dilution.
C i = concentration of the stock or initial solution.
V f = volume of the diluted solution that was prepared.
C f = concentration of the diluted solution.
Equation 2 can be written as follows:
C i / C f = V f / V i (3)
Two valid expressions for FD
But, C i / C f by definition is the Dilution Factor , since it indicates the times that the concentration of the stock or initial solution is greater in relation to the concentration of the diluted solution. Therefore, it indicates the dilution to be carried out to prepare the diluted solution from the stock solution.
Likewise, from the observation of equation 3 it can be concluded that the V f / V i relationship is another way to obtain the Dilution Factor. That is, any of the two expressions (C i / C f , V f / V i ) are valid to calculate FD. The use of one or the other will depend on the available data.
A 0.3 M NaCl solution was used to prepare a dilute 0.015 M NaCl solution. Calculate the value of the dilution factor.
The dilution factor is 20. This indicates that to prepare the dilute 0.015 M NaCl solution, the 0.3 M NaCl solution had to be diluted 20 times:
FD = C i / C f
0.3M / 0.015M
Knowing that the dilution factor is 15: what volume of water should have been added to 5 ml of a concentrated glucose solution to make the desired dilution?
As a first step, the volume of the diluted solution (V f ) is calculated . Once calculated, the volume of water added to make the dilution is calculated from this.
FD = V f / V i .
V f = FD x V i
15 x 5 ml
Added volume of water = 75 ml – 5 ml
Then, to prepare the diluted solution with a dilution factor of 15, 70 ml of water was added to 5 ml of the concentrated solution to make the final volume of 75 ml.
The concentration of a fructose stock solution is 10 g / L. It is desired to prepare from it, a fructose solution with a concentration of 0.5 mg / mL. Taking 20 mL of the stock solution to make the dilution: what should be the volume of the diluted solution?
The first step in solving the problem is to calculate the dilution factor (DF). Once obtained, the volume of the diluted solution (V f ) will be calculated .
But before making the proposed calculation, it is necessary to make the following observation: we must place the amounts of fructose concentrations in the same units. In this particular case, 10 g / L is equivalent to 10 mg / mL, this situation being illustrated by the following transformation:
(mg / mL) = (g / L) x (1,000 mg / g) x (L / 1,000 mL)
10 g / L = 10 mg / mL
Continuing with the calculations:
FD = C i / C f
DF = (10 mg / mL) / (0.2 mg / mL)
But since V f = FD x V i
V f = 50 x 20 mL
Then, 20 mL of the 10g / L fructose solution was diluted to 1L of 0.2g / L solution.
A method of making serial dilutions will be illustrated. There is a glucose solution with a concentration of 32 mg / 100mL, and from it, it is desired to prepare by dilution a set of glucose solutions with concentrations: 16 mg / 100mL, 8 mg / 100mL, 4 mg / 100mL, 2 mg / 100mL and 1 mg / 100mL.
5 test tubes are labeled for each of the concentrations indicated in the statement. In each of them, for example, 2 mL of water is placed.
Then to tube 1 with water, 2 mL of the stock solution is added. The content of tube 1 is shaken and 2 mL of its content is transferred to tube 2. In turn, tube 2 is shaken and 2mL of its content is transferred to tube 3; proceeding in the same way with tubes 4 and 5.
2mL of water and 2 mL of the stock solution with a glucose concentration of 32 mg / 100 mL are added to tube 1. So the final glucose concentration in this tube is 16 mg / 100mL.
To tube 2, 2 mL of water and 2 mL of the contents of tube 1 are added with a glucose concentration of 16 mg / 100 mL. Then, in tube 2 the concentration of tube 1 is diluted 2 times (DF). So the final glucose concentration in this tube is 8 mg / 100mL.
2 mL of water and 2 mL of the contents of tube 2 are added to tube 3, with a glucose concentration of 8 mg / 100 mL. And like the other two tubes, the concentration is divided into two: 4 mg / 100 mL of glucose in tube 3.
For the reason explained above, the final glucose concentration in tubes 4 and 5 is, respectively, 2mg / 100mL and 1mg / 100mL.
The DF of tubes 1, 2, 3, 4, and 5, relative to the stock solution, are: 2, 4, 8, 16 and 32, respectively.
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