Accomplish calculations involving solution dilution is to check the amount of solvent that has been added or removed from them, which results in the following possible changes to their concentrations:
Removal of solvent part: When part of the solvent is removed from a solution, the amount of solute approaches or becomes greater than the amount of solvent, making the solution concentrated.
Addition of more solvent: When the solution receives an extra amount of solvent, its mass becomes even greater than that of solute, making the solution dilute.
the formulas commonly used to perform these calculations are:
a) To molarity:
Mi. Vi = VF.VF
Mi = initial molarity of the solution
vi = initial volume
MF = final molarity of the solution
VF = final volume of solution
NOTE: Final volume is the sum of the initial volume and the added volume (Vf = Vi + Va) or the subtraction of the initial volume by the volume of solvent that was removed (Vf = Vi - Ve).
b) To common concentration:
Ci.Vi = CF.VF
Ci = initial molarity of the solution
ÇF = final molarity of the solution
c) To mass title:
Ti.mi = Tf.mf
Ti = Initial title of the solution
mi = mass of initial solution
Tf = Final title of the solution
mf = mass of final solution
See some examples of using the formulas above in calculations involving solution dilution:
Example 1: 50 g of a solution of H2ONLY4 63% by mass is added to 400 g of water. The mass percentage of H2ONLY4 in the solution obtained is:
a) 7%.
b) 9%.
c) 10%.
d) 12%.
e) 16%.
Exercise data:
mi = 50g
Pi = 63%
NOTE: Transforming the given percentage for initial title by dividing by 100, we have:
Ti = 0.63
TF =?
PF = ?
Before finding the percentage, we must initially determine the value of the final security (TF) through the following expression:
mi. Ti = mF.TF
50.0.63 = 450.TF
31.5 = 450.TF
31,5 = TF
450
TF = 0,07
After we find the value of the final title, just multiply it by 100 to get the mass percentage of H2ONLY4 required:
P = TF.100
P = 0.07,100
P = 7%, Letter a).
Example 2: To a 100 mL sample of NaOH at a concentration of 20 g/L was added enough water to make up to 500 mL. The concentration, in g/L, of this new solution is equal to:
a) 2.
b) 3.
c) 4.
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d) 5.
e) 8.
Exercise data:
Ci = 20g/L
Vi = 100 mL
VF= 500 ml
ÇF = ?
To determine the final concentration value (CF), just use the expression below:
Ci.Vi = CF.VF
20,100 = CF.500
2000 = CF.500
2000 = CF
500
ÇF = 4g/L letter C).
Example 3: Dilution is a very common operation in our daily lives. An example is when we prepare a soft drink from a concentrated juice. Consider 100 mL of a given juice where the solute concentration is 0.4 mol. L-1. The volume of water, in mL, that should be added so that the concentration of the solute drops to 0.04 mol. L-1 will be from:
a) 1000.
b) 900.
c) 500.
d) 400.
Exercise data:
Mi = 0.4 mol/L
Vi = 100 mL
NOTE: it is not necessary to transform the unit mL of the initial volume because the exercise asks for the added volume also in mL.
Va = ?
VF= ?
MF = 0.004 mol/L
The statement does not provide the final volume (VF) and asks to calculate the added volume (Va). For this, we must first remember that the final volume is the sum of the initial volume (Vi) plus the added volume:
VF = Vi + Va
Adding the initial volume value in the expression above, we will have:
VF = 100 + Va
So if we replace the VF above in the expression for calculations in dilution, we can find the value of the added volume:
Mi. Vi = MF.VF
0.4,100 = 0.04. (100 + Va)
NOTE: We must multiply the 0.04 for each data inside the parentheses:
40 = 4 + 0.04.Va
40 - 4 = 0.04.Va
36 = 0.04.Va
36 = go
0,04
Va = 900 mL, letter B).
Example 4: When 100 cm are diluted3 of a 0.5 mol/dm solution3 to 0.2 mol/dm3, what will be the volume of the new solution obtained?
a) 2500 cm3
b) 250 cm3
c) 200 cm3
d) 2000 cm3
Exercise data:
Mi = 0.5 mol/dm3
vi = 100 cm3
NOTE: it is not necessary to transform the unit cm3 because all the alternatives bring volume results in cm3.
VF= ?
MF = 0.2 mol/dm3
To find the final volume value, just add the values provided in the expression below:
Mi. Vi = MF.VF
0.5,100 = 0.2.VF
50 = 0.2.VF
50 = VF
0,2
VF = 250 cm3 - letter B).
By Me. Diogo Lopes Dias
Would you like to reference this text in a school or academic work? Look:
DAYS, Diogo Lopes. "Calculations involving dilution of solutions"; Brazil School. Available in: https://brasilescola.uol.com.br/quimica/calculos-envolvendo-diluicao-solucoes.htm. Accessed on June 28, 2021.
