DILUTION: CONCEPT, HOW TO DO IT, EXAMPLES, EXERCISES - CHEMISTRY

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

This last procedure, although it sounds strange, 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.

As an aqueous solution of phenolphthalein is diluted, its color fades. Source: Pxhere.

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.

How is a dilution done?

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 reaching the volume calculated for the diluted solution.

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

Where m _{i is} the mass of solute in the concentrated solution used to make the dilution, and m _{f is} the mass of solute in 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.

Application

If you want to prepare a dilution, you must know how many times the concentrated solution has to be diluted 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 must be taken to make the dilution? If the final volume of the diluted solution (v _f) and the dilution factor are known, 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

The volume of the calculated concentrated solution (v _i) is measured by means of a graduated pipette or cylinder, and poured into a volumetric flask. Then, diluent is added until the flask's capacity is reached, which indicates the volume of the diluted solution (v _f).

Serial dilution

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

A 1/5 serum dilution 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 diluent to the paint to reduce the intensity of the coloring or to decrease its viscosity.

-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 reduce its acidity concentration.

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

Exercises

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 be the volume of the 0.5 M NaCl solution needed to prepare this 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 isolate v _i and solve:

v _i = 1 L / 20

= 0.05 L

= 50 mL

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

Exercise 2

How many mL of a sulfuric acid (H ₂ SO ₄) reagent with a concentration of 95% (m / m) and a density of 1.84 g / mL are required to prepare 250 mL of a solution of sulfuric acid to the 0.5 M? Sulfuric acid molecular weight: 98 g / mol.

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

m = v d

We determine the mass of H ₂ SO ₄ 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 a liter of 95% H ₂ SO ₄ solution was assumed, 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 ₂ SO ₄ 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 is added and made up to 250 mL with water.

Exercise 3

How many mL of water must be added to 50 mL to a 0.25 M calcium chloride (CaCl ₂) solution to prepare a 0.0125 M CaCl ₂ solution ?

Again, there is no mass loss of CaCl ₂ once it is diluted:

v _i c _i = v _f c _f

We isolate 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

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

DILUTION: CONCEPT, HOW TO DO IT, EXAMPLES, EXERCISES - CHEMISTRY - 2025