7.19: Concentrations: Dilution

Learning Objectives

• State whether the concentration of a solution is directly or indirectly proportional to its volume.
• Write the dilution equation.
• Define dilution.
• Apply the dilution equation to calculate the final concentration, or the final volume, of a diluted solution.

The previous section of this chapter discussed how the terms "concentrated" and "dilute" could be applied to describe the concentration of a solution, compared to another homogenous mixture.  The following paragraphs will describe the processes through which the relative concentration of a solution can be altered and how those changes impact quantitative solution measurements.

Dilution Equation and Indicator Words

As stated previously, a concentration quantitatively ratios the amount of solute that is contained in a solution to the amount of solution that is present, overall.  This generic definition of concentration is represented in the equation that is shown below. 

\(\text{Concentration (C)}\) = \( \dfrac{ \rm{Amount \; of \; Solute}}{\rm{Amount \; of \; Solution}}\)

Because most of the solutions that are of interest to chemists are liquid-phase solutions, which are prepared using liquid solvents, scientists usually quantify the amount of solution that is present using a volumetric measurement.  Replacing the "Amount of Solution" in the generic concentration equation that is shown above with a volumetric quantity results in the equation that is shown below.

\(\text{C}\) = \( \dfrac{ \rm{Amount \; of \; Solute}}{\rm{V_{solution}}}\)

In order to alter the concentration of a solution, either the amount of solute or the volume of the overall solution must be changed.  Because solutes are usually more expensive than solvents, chemists generally do not alter the quantity of solute that is utilized to prepare a solution and, instead, modify the amount of solvent, and, therefore, the overall volume of the solution, in order to change the concentration of that solution.  As stated in the previous section, when the amount of solute and, therefore, the numerator in the equation that is shown above, remains constant, increasing the volume of the solution raises the value in the denominator of this ratio, which, in turn, decreases the concentration of the solution.  Therefore, the volume and concentration of a solution are indirectly, or inversely, proportional to one another.

As no denominator is explicitly-written on the left side of the equal sign in the equation that is shown above, its value is understood to be an unwritten "1."  Inserting this value generates the following equation

\( \dfrac{ \rm{C}}{1}\) = \( \dfrac{ \rm{Amount \; of \; Solute}}{\rm{V_{solution}}}\)

which can be subsequently rearranged through cross-multiplication, as shown below.

(\({\rm{C}}\))(\({\rm{V_{solution}}}\)) = (\(\rm{Amount \; of \; Solute}\))(\({1}\))

Finally, removal of the "1" yields the following mathematical relationship.  

(\({\rm{C}}\))(\({\rm{V_{solution}}}\)) = \(\rm{Amount \; of \; Solute}\)

As discussed in Chapter 6, multiplying two quantities, such as volume and concentration, that are indirectly proportional, results in the calculation of a constant.  Based on the final equation that is shown above, the "constant" value corresponds to the amount of solute that is present in the solution, which is consistent with the information that is described above.  Therefore, the value of the corresponding concentration-volume product after the completion of an experiment must be identical to its value at the beginning of that experiment, as shown in the following equation sequence.

(\({\rm{C_{initial}}}\))(\({\rm{V_{solution, \; initial}}}\)) = \(\rm{Amount \; of \; Solute}\) = (\({\rm{C_{final}}}\))(\({\rm{V_{solution, \; final}}}\))

Because all of the values in an equation sequence are equal to one another, the left- and right-most quantities can be directly related to one another, as shown below.

(\({\rm{C_{initial}}}\))(\({\rm{V_{solution, \; initial}}}\)) = (\({\rm{C_{final}}}\))(\({\rm{V_{solution, \; final}}}\))

This equation, which, due to the experimental process in which it is applied, is referred to as the dilution equation, can also be stated using variables that have abbreviated or modified subscripts, as shown below.  As stated above, chemists generally modify the amount of solvent that is present in a solution, in order to change the concentration of that solution.  However, since the solute and solvent in a solution must, by definition, be uniformly-distributed, removing a specified amount of solvent without also altering the amount of solute that is present in a solution is prohibitively challenging.  Therefore, in order to reliably control the process of changing the concentration of a homogeneous mixture, more solvent must be added to the solution.  This process is known as dilution, because, relative to the solution from which it was prepared, the final solution contains the same amount of solute in a larger volume of solution, and, therefore, is more dilute because of the solvent that has been added.

\({\rm{C_{i}}}\)\({\rm{V_{i}}}\) = \({\rm{C_{f}}}\)\({\rm{V_{f}}}\)
\({\rm{C_{1}}}\)\({\rm{V_{1}}}\) = \({\rm{C_{2}}}\)\({\rm{V_{2}}}\)

Finally, because the values that are incorporated into the mathematical statements that are shown above correspond to solution volumes and concentrations that are measured before and after a dilution, words such as "dilutes," "diluted," and "diluting" indicate that one of the variations of the dilution equation should be applied to solve a given problem.

Dilution Calculations

Before the dilution equation can be applied, the validity of the units that are associated with the given numerical values must be confirmed.  Four types of concentrations, mass percent, volume percent, mass/volume percent, and molarity, which are expressed using units of % m/m, % v/v, % m/v, and mol/L or M, respectively, have been presented and discussed in this chapter.  Furthermore, the volume of a solution must be reported in milliliters (mL) when calculating a volume percent or a mass/volume percent and must be given in liters (L), in order to calculate a molarity.  As shown above, the dilution equation contains both an "initial" and a "final" variable for each of these quantities.  Therefore, in order to achieve unit cancelation, the units in which the initial and final values are expressed must be identical for a given variable.  

For example, calculate the volume of a 99.5% v/v solution of acetic acid, HC₂H₃O₂, that must be diluted to prepare 5.6 liters of vinegar, which has a concentration of 5.0% v/v acetic acid, HC₂H₃O₂.

The word "diluted" indicates that the dilution equation should be applied to solve this problem.  Before this equation can be applied, each numerical quantity that is given in the problem must be assigned to a variable, and the validity of the units that are associated with these numerical values must be confirmed.  As stated above, the concentration of a solution must be expressed as a mass percent (% m/m), a volume percent (% v/v), a mass/volume percent (% m/v), or a molarity (mol/L or M), and the volume of the solution must be given in milliliters (mL) or liters (L).  Finally, in order to achieve unit cancelation, the units in which the initial and final values are expressed must be identical for a given variable.

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.

Numerical Quantity	Variable	Unit Validity
99.5% v/v	C1
V1	V1
5.0% v/v	C2
5.6 L	V2

Since V₁ is the only variable that cannot be assigned to a numerical value in the given problem, the initial volume of the solution is the unknown quantity that will be calculated upon solving the dilution equation.  The problem does not specify whether the final answer should be expressed in liters or milliliters.  Therefore, the given unit for the final volume, "liters," is acceptable for this problem.  Finally, the given concentrations are both reported using identical acceptable units and, therefore, can be utilized to solve the given problem.

The quantities that are shown in the table above can be incorporated into the dilution equation, as shown below.  In order to solve for V₁, the quantities on both sides of this equation must be divided by the value of C₁, "99.5% v/v," which causes the cancelation of the concentration unit, "% v/v," as this unit is present in the numerator and the denominator in the resultant fraction.  The unit that remains after this cancelation is "liters," which is an acceptable unit for expressing the volume of a solution.  The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantity in its denominator.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below. 

\({\rm{C_{1}}}\)\({\rm{V_{1}}}\) = \({\rm{C_{2}}}\)\({\rm{V_{2}}}\)

(\({99.5 \%\ \rm{v/v}}\))(\({\rm{V_{1}}}\)) = (\(5.0 \%\ \rm{v/v}\))(\({5.6 \; \rm{L}}\))

\(\rm{V_{1}}\) = \({\dfrac{({5.0 \%\ \cancel{\rm{v/v}}}) ({5.6 \; \rm{L}})} {({99.5 \%\ \cancel{\rm{v/v}}})}}\) 

\(\rm{V_{1}}\) = \({0.281407... \; \rm{L}} ≈ {0.28 \; \rm{L}}\)

Finally, recall that the quantities that are related by the dilution equation are indirectly proportional.  Therefore, since the concentration of the solution decreased, its volume must increase.  Because the final volume of the solution, V₂, is larger than the calculated initial volume, V₁, the value of the final answer is reasonable.

Exercise \(\PageIndex{1}\)

Calculate the concentration that results upon diluting 250 milliliters of a 3.9 M solution to a final volume of 1.55 liters.

Answer

The word "diluting" indicates that the dilution equation should be applied to solve this problem.  Before this equation can be applied, each numerical quantity that is given in the problem must be assigned to a variable, and the validity of the units that are associated with these numerical values must be confirmed.  As stated above, the concentration of a solution must be expressed as a mass percent (% m/m), a volume percent (% v/v), a mass/volume percent (% m/v), or a molarity (mol/L or M), and the volume of the solution must be given in milliliters (mL) or liters (L).  Finally, in order to achieve unit cancelation, the units in which the initial and final values are expressed must be identical for a given variable.

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.
 

Numerical Quantity	Variable	Unit Validity
3.9 M	C1
250 mL	V1
C2	C2
1.55 L	V2

Since C₂ is the only variable that cannot be assigned to a numerical value in the given problem, the final concentration of the solution is the unknown quantity that will be calculated upon solving the dilution equation.  The problem does not specify whether the final answer should be expressed as a mass percent (% m/m), a volume percent (% v/v), a mass/volume percent (% m/v), or a molarity (mol/L or M).  Therefore, the given unit for the initial concentration, "M," is acceptable for this problem.  However, the units that are associated with the given values for the initial volume, V₁, and the final volume, V₂, of the solution are not consistent with one another and, therefore, will not cancel when incorporated into the dilution equation.  In order to remedy this discrepancy, one of these units must be converted to match the other.  Because both of the given units, "milliliters" and "liters," are valid units for reporting the volume of a solution, altering either unit is acceptable.  The conversion of V₂ to a corresponding quantity in milliliters is shown below.

\( {\text {1.55}} {\cancel{\rm{L} }} \times\) \( \dfrac{1,000 \; \rm{mL} }{\cancel{\rm{L} }}\) = \( {\text {1,550}} \; \rm{mL} \)

The updated numerical values that are summarized in the following table are all expressed in appropriate units and, therefore, can be utilized to solve the given problem.
 

Numerical Quantity	Variable	Unit Validity
3.9 M	C1
250 mL	V1
C2	C2
1,550 mL	V2

The quantities that are shown in the table above can be incorporated into the dilution equation, as shown below.  In order to solve for C₂, the quantities on both sides of this equation must be divided by the value of V₂, "1,550 mL," which causes the cancelation of the volume unit, "mL," as this unit is present in the numerator and the denominator in the resultant fraction.  The unit that remains after this cancelation is "M," which is an acceptable unit for expressing the concentration of a solution.  The numerical solution is calculated by multiplying the values that are present in the numerator of this fraction and then dividing by the quantity in its denominator.  Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below. 

\({\rm{C_{1}}}\)\({\rm{V_{1}}}\) = \({\rm{C_{2}}}\)\({\rm{V_{2}}}\)

(\({3.9 \; M}\))(\(250 \; {\rm{mL}}\)) = \({\rm{C_{2}}}\)(\({1,550 \; \rm{mL}}\))

\(\rm{C_{2}}\) = \({\dfrac{({3.9 \; M}) ({250 \; \cancel{\rm{mL}}})} {({1,550 \cancel{\rm{mL}}})}}\) 

\(\rm{C_{2}}\) = \({0.629032... \; M} ≈ {0.63 \; M}\)

Finally, recall that the quantities that are related by the dilution equation are indirectly proportional.  Therefore, since the volume of the solution increased, its concentration must decrease.  Because the calculated final concentration of the solution, C₂, is smaller than the initial concentration, C₁, the value of the final answer is reasonable.