Dilution concept, how to do it, examples, exercises

The dilution It is a procedure by which the concentration of a solution is decreased, generally with the addition of a diluent. However, dilution can also occur by a process that removes solute from solution..

This last procedure, strange as it sounds, is a routine practice in the kitchen when adding a potato to a very salty food to remove excess salt. The food will taste less salty because the potatoes absorb its salt content.

The dilution carried out, or to be carried out, is expressed in terms such as: 1/5. This means that to carry out the dilution, a volume of the concentrated solution is taken and four volumes of the diluent are added; usually water. In this case, the number 5 represents the dilution factor.

The dilution factor is the quotient between the initial concentration of the solution and the final concentration of the diluted solution. Likewise, the dilution factor is the quotient between the volume of the diluted solution and the volume of the concentrated solution that was taken to make the dilution..

Article index

• 1 How do you do a dilution?
  • 1.1 Theoretical reasoning
  • 1.2 Application
  • 1.3 Procedure
• 2 Serial dilution
• 3 Examples of dilutions
• 4 Exercises
  • 4.1 Exercise 1
  • 4.2 Exercise 2
  • 4.3 Exercise 3
• 5 References

How do you do a dilution?

Theoretical reasoning

To prepare a dilution, a certain volume of the concentrated solution is taken, and it is taken to a container, adding diluent until the volume calculated for the diluted solution is reached..

The mass of solute that was taken from the concentrated solution to make the dilution is exactly equal to the mass of solute that is placed in the container used to make the dilution..

m_i = m_F

Being m_i the mass of solute in the concentrated solution used to make the dilution, and m_F the solute mass of the diluted solution. We also know that:

m_i = v_i C_i

m_F  = v_F C_F

Then substituting:

v_i C_i = v_F C_F

Rewriting the equation:

c_i  / c_F = v_F / v_i

c_i / c_F is the dilution factor (how many times it is necessary to dilute the concentrated solution). However, v_F / v_i also counts as a dilution factor.

App

If you want to prepare a dilution, you must know the times to dilute the concentrated solution to obtain the desired concentration of the diluted solution (dilution factor). To do this, divide the concentration of the concentrated solution by the concentration of the diluted solution.

But: What volume of the concentrated solution should be taken to make the dilution? If the final volume of the diluted solution (v_F) and the dilution factor, it is easy to know the volume of the concentrated solution (v_i), necessary to carry out the desired dilution:

v_i = v_F / FD

Process

Measure the volume of the calculated stock solution (v_i) through a pipette or graduated cylinder, and pour into a volumetric flask. Then, diluent is added until the flask has a capacity that indicates the volume of the diluted solution (v_F).

Serial dilution

This type of dilution is frequently used in volumetric analysis. To do this, test tubes are arranged in series and in each one the same volume of deionized water is added; for example 2 mL.

A 1/5 dilution of the serum can be prepared separately. Then, 2 mL of the serum dilution is added to the first tube containing 2 mL of water. The tube is shaken properly and 2 mL of this mixture is transferred to tube 2.

Next, tube 2 is mixed well and 2 mL of its contents are transferred to tube 3, and so on until the series of tubes is completed. As a consequence of this procedure, there are test tubes with serum dilutions 1/10, 1/20, 1/40 ...

Examples of dilutions

Some examples of dilution are:

-Dilute a 5 M NaCl solution 1/10 to obtain a 0.5 M NaCl solution.

-The addition of water or another thinner to the paint to reduce the intensity of the coloring or to decrease the viscosity of the same.

-The addition of milk to the coffee to reduce the concentration of the coffee and give it a smoother and sweeter taste.

-Dilute a lemonade with water to lower its acidity concentration.

-Make the dilution of a serum to make the titration of any antibody present in it.  

Training

Exercise 1

How many times must a 0.5 M NaCl solution be diluted to obtain 1 liter of the 0.025 M solution, and what will also be the volume of the 0.5 M NaCl solution needed to prepare said diluted solution?

We start from the dilution factor:

FD = c_i / c_F

We have all the data:

c_i = initial concentration (0.5 M)

c_F  = final concentrate (0.025 M)

And we therefore calculate FD:

FD = 0.5M / 0.025M

= 20

The 0.5 M NaCl solution must be diluted 20 times to obtain a 0.025 M NaCl solution.

With this value of DF we can now calculate the initial volume to be taken from the concentrated solution for this dilution:

FD = v_F / v_i

We solve for v_i and we solve:

v_i = 1 L / 20

= 0.05 L

= 50 mL

Therefore, 50 mL of the 0.5 M NaCl solution are required to prepare one liter of the 0.025 M NaCl solution..

Exercise 2

How many mL of a sulfuric acid reagent (H_twoSW₄) with a concentration of 95% (m / m) and a density of 1.84 g / mL, are required to prepare 250 mL of a 0.5 M sulfuric acid solution? Sulfuric acid molecular weight: 98 g / mol.

The first step is to calculate the molarity of the concentrated sulfuric acid:

m = vd

We determine the mass of H_twoSW₄ corresponding to the solution with the given density:

m = 1,000 mL · 1.84 g / mL

= 1,840 g

Since sulfuric acid is 95% pure, its actual mass must be calculated:

m = 1,840 g (95/100)

= 1,748 g

Because one liter of solution H_twoSW₄ at 95%, the moles present in these grams will directly give us the molarity:

M = (1,748 g / L) / (98 g / mol)

= 17.83

We know that the mass of H_twoSW₄ that is diluted is the same before and after dilution:

m_i = m_F

c_i V_i = c_F V_F

v_i = c_F V_F  / c_i

And we solve for v_i:

v_i = 0.5 M 250 mL / 17.83 M

= 7.010 mL

Then, to prepare 250 mL of a 0.5 M sulfuric acid solution, a portion of water is placed in the volumetric flask to avoid splashing, and 7.010 mL of concentrated sulfuric acid are added and made up to 250 mL with water..

Exercise 3

How many mL of water should be added to 50 mL to a solution of calcium chloride (CaCl_two) 0.25 M, to prepare a CaCl solution_two at 0.0125 M?

Again, there is no loss of mass of CaCl_two once it is going to be diluted:

v_i C_i = v_F C_F

Solve for and solve for v_F:

v_F  = v_i  C_i / c_F

= 50 mL 0.25 M / 0.0125 M

= 1,000 mL

Volume of water to be added = v_{F -} v_i

1,000 mL - 50 mL = 950 mL

It is therefore necessary to add 950 mL of water to 50 mL to the 0.5 M calcium chloride solution. In this way, 1,000 mL of 0.0125 M calcium chloride solution will be prepared..

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
2. Wikipedia. (2020). Dilution (equation). Recovered from: en.wikipedia.org
3. Jessie A. Key. (s.f.). Dilutions and Concentrations. Recovered from: opentextbc.ca
4. ChemTeam. (s.f.). Dilution: Definition and Calculations. Recovered from: chemteam.info
5. David R. Caprette. (2012). Making Dilutions. Recovered from: ruf.rice.edu