Learning Outcomes
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another (see figure below). Also, be aware that the terms "concentrate" and "dilute" can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, you would be concentrating it, because the ratio of solute to solvent would be increasing. If you were to add more water to an aqueous solution, you would be diluting it because the ratio of solute to solvent would be decreasing.
Figure \(\PageIndex{1}\): Solutions of a red dye in water from the most dilute (on the left) to the most concentrated (on the right).One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute. This percentage can be determined in one of three ways: (1) the mass of the solute divided by the mass of solution, (2) the volume of the solute divided by the volume of the solution, or (3) the mass of the solute divided by the volume of the solution. Because these methods generally result in slightly different vales, it is important to always indicate how a given percentage was calculated.
When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent (mass/mass), which is the grams of solute per \(100 \: \text{g}\) of solution.
\[\text{Percent by mass} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\]
Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100 \: \text{g}\) of water. The percent by mass would be calculated as follows:
\[\text{Percent by mass} = \frac{25 \: \text{g sugar}}{125 \: \text{g solution}} \times 100\% = 20\% \: \text{sugar}\]
Sometimes, you may want to make a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed. For example, let's say you need to make \(3.00 \times 10^3 \: \text{g}\) of a sodium chloride solution that is \(5.00\%\) by mass. You can rearrange and solve for the mass of solute.
\[\begin{align} \% \: \text{by mass} &= \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\% \\ 5.00\% &= \frac{\text{mass of solute}}{3.00 \times 10^3 \: \text{g solution}} \times 100\% \\ \text{mass of solute} &= 150. \: \text{g} \end{align}\]
You would need to weigh out \(150 \: \text{g}\) of \(\ce{NaCl}\) and add it to \(2850 \: \text{g}\) of water. Notice that it was necessary to subtract the mass of the \(\ce{NaCl}\) \(\left( 150 \: \text{g} \right)\) from the mass of solution \(\left( 3.00 \times 10^3 \: \text{g} \right)\) to calculate the mass of the water that would need to be added.
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of the solution expressed as a percent, yields the percent by volume (volume/volume) of the solution. If a solution is made by taking \(40. \: \text{mL}\) of ethanol and adding enough water to make \(240. \: \text{mL}\) of solution, the percent by volume is:
\[\begin{align} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \\ &= 16.7\% \: \text{ethanol} \end{align}\]
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
Figure \(\PageIndex{2}\): Hydrogen peroxide is commonly sold as a \(3\%\) by volume solution for use as a disinfectant.It should be noted that, unlike in the case of mass, you cannot simply add together the volumes of solute and solvent to get the final solution volume. When adding a solute and solvent together, mass is conserved, but volume is not. In the example above, a solution was made by starting with \(40 \: \text{mL}\) of ethanol and adding enough water to make \(240 \: \text{mL}\) of solution. Simply mixing \(40 \: \text{mL}\) of ethanol and \(200 \: \text{mL}\) of water would not give you the same result, as the final volume would probably not be exactly \(240 \: \text{mL}\).
The mass-volume percent is also used in some cases and is calculated in a similar way to the previous two percentages. The mass/volume percent is calculated by dividing the mass of the solute by the volume of the solution and expressing the result as a percent.
For example, if a solution is prepared from \(10 \: \ce{NaCl}\) in enough water to make a \(150 \: \text{mL}\) solution, the mass-volume concentration is
\[\begin{align} \text{Mass-volume concentration} & \frac{\text{mass solute}}{\text{volume solution}} \times 100\% \\ &= \frac{10 \: \text{g} \: \ce{NaCl}}{150 \: \text{mL solution}} \times 100\% \\ &= 6.7\% \end{align}\]
Two other concentration units are parts per million and parts per billion. These units are used for very small concentrations of solute such as the amount of lead in drinking water. Understanding these two units is much easier if you consider a percentage as parts per hundred. Remember that \(85\%\) is the equivalent of 85 out of a hundred. A solution that is \(15 \: \text{ppm}\) is 15 parts solute per 1 million parts solution. A \(22 \: \text{ppb}\) solution is 22 parts solute per billion parts solution. While there are several ways of expressing two units of \(\text{ppm}\) and \(\text{ppb}\), we will treat them as \(\text{mg}\) or \(\mu \text{g}\) of solutes per \(\text{L}\) solution, respectively.
For example, \(32 \: \text{ppm}\) could be written as \(\frac{32 \: \text{mg solute}}{1 \: \text{L solution}}\) while \(59 \: \text{ppb}\) can be written as \(\frac{59 \: \mu \text{g solute}}{1 \: \text{L solution}}\).
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity \(\left( \text{M} \right)\) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.
\[\text{Molarity} \: \left( \text{M} \right) = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{\text{mol}}{\text{L}}\]
Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol \(\text{M}\), which is read as "molar". For example, a solution labeled as \(1.5 \: \text{M} \: \ce{NH_3}\) is a "1.5 molar solution of ammonia".
Example \(\PageIndex{1}\)
A solution is prepared by dissolving \(42.23 \: \text{g}\) of \(\ce{NH_4Cl}\) into enough water to make \(500.0 \: \text{mL}\) of solution. Calculate its molarity.
Solution
Step 1: List the known quantities and plan the problem.
Known
Unknown
The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.
Step 2: Solve.
\[42.23 \: \text{g} \: \ce{NH_4Cl} \times \frac{1 \: \text{mol} \: \ce{NH_4Cl}}{53.50 \: \text{g} \: \ce{NH_4Cl}} = 0.7893 \: \text{mol} \: \ce{NH_4Cl}\]
\[\frac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L}} = 1.579 \: \text{M}\]
Step 3: Think about your result.
The molarity is \(1.579 \: \text{M}\), meaning that a liter of the solution would contain 1.579 moles of \(\ce{NH_4Cl}\). Having four significant figures is appropriate.
Figure \(\PageIndex{3}\): Volumetric flasks come in many sizes, each designed to prepare a different volume of solution.When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).
\[\text{mol}_1 = \text{mol}_2\]
Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, we can set those equal.
\[M_1 \times L_1 = M_2 \times L_2\]
Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:
\[M_1 \times V_1 = M_2 \times V_2\]
Additionally, the concentration can be in any other unit as long as \(M_1\) and \(M_2\) are in the same unit.
Suppose that you have \(100. \: \text{mL}\) of a \(2.0 \: \text{M}\) solution of \(\ce{HCl}\). You dilute the solution by adding enough water to make the solution volume \(500. \: \text{mL}\). The new molarity can easily be calculated by using the above equation and solving for \(M_2\).
\[M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}\]
The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Another common dilution problem involves deciding how much a highly concentrated solution is required to make a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.
Example \(\PageIndex{2}\)
Nitric acid \(\left( \ce{HNO_3} \right)\) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is \(16 \: \text{M}\). How much of the stock solution of nitric acid needs to be used to make \(8.00 \: \text{L}\) of a \(0.50 \: \text{M}\) solution?
Solution
Step 1: List the known quantities and plan the problem.
Known
Unknown
The unknown in the equation is \(V_1\), the necessary volume of the concentrated stock solution.
Step 2: Solve.
\[V_1 = \frac{M_2 \times V_2}{V_1} = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}\]
Step 3: Think about your result.
\(250 \: \text{mL}\) of the stock \(\ce{HNO_3}\) solution needs to be diluted with water to a final volume of \(8.00 \: \text{L}\). The dilution from \(16 \: \text{M}\) to \(0.5 \: \text{M}\) is a factor of 32.
Concentration is important in healthcare because it is used in so many ways. It's also critical to use units with any values to ensure the correct dosage of medications or report levels of substances in blood, to name just two.
Another way of looking at concentration such as in IV solutions and blood is in terms of equivalents. One equivalent is equal to one mole of charge in an ion. The value of the equivalents is always positive regardless of the charge. For example, \(\ce{Na^+}\) and \(\ce{Cl^-}\) both have 1 equivalent per mole.
\[\begin{array}{ll} \textbf{Ion} & \textbf{Equivalents} \\ \ce{Na^+} & 1 \\ \ce{Mg^{2+}} & 2 \\ \ce{Al^{3+}} & 3 \\ \ce{Cl^-} & 1 \\ \ce{NO_3^-} & 1 \\ \ce{SO_4^{2-}} & 2 \end{array}\]
Equivalents are used because the concentration of the charges is important than the identity of the solutes. For example, a standard IV solution does not contain the same solutes as blood but the concentration of charges is the same.
Sometimes, the concentration is lower in which case milliequivalents \(\left( \text{mEq} \right)\) is a more appropriate unit. Just like metric prefixes used with base units, milli is used to modify equivalents so \(1 \: \text{Eq} = 1000 \: \text{mEq}\).
Example \(\PageIndex{3}\)
How many equivalents of \(\ce{Ca^{2+}}\) are present in a solution that contains 3.5 moles of \(\ce{Ca^{2+}}\)?
Solution
Use the relationship between moles and equivalents of \(\ce{Ca^{2+}}\) to find the answer.
\[3.5 \: \text{mol} \cdot \frac{2 \: \text{Eq}}{1 \: \text{mol} \: \ce{Ca^{2+}}} = 7.0 \: \text{Eq} \: \ce{Ca^{2+}}\]
Example \(\PageIndex{4}\)
A patient received \(1.50 \: \text{L}\) of saline solution which has a concentration of \(154 \: \text{mEq/L} \: \ce{Na^+}\). What mass of sodium did the patient receive?
Solution
Use dimensional analysis to set up the problem based on the values given in the problem, the relationship for \(\ce{Na^+}\) and equivalents and the molar mass of sodium. Note that if this problem had a different ion with a different charge, that would need to be accounted for in the calculation.
\[1.50 \: \text{L} \cdot \frac{154 \: \text{mEq}}{1 \: \text{L}} \cdot \frac{1 \: \text{Eq}}{1000 \: \text{mEq}} \cdot \frac{1 \: \text{mol} \: \ce{Na^+}}{1 \: \text{Eq}} \cdot \frac{22.99 \: \text{g}}{1 \: \text{mol} \: \ce{Na^+}} = 5.31 \: \text{g} \: \ce{Na^+}\]
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
The concentration of a solution tells the amount of solute dissolved in a given amount of solution.
A measured amount of solute is dissolved in enough solvent to make the desired volume of the solution, as illustrated in Fig. 5.4.1. The concentration of the solution can be expressed in different ways using mass, volume, or mole units, as explained in the following.
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
Given: 0.015 mg in 1L solution, Desired: ? ppb m/v of Lead in water
EPA’s action limit for lead is 0.015 mg/L in drinking water. What is this limit in ppb of lead m/v in the drinking water?
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
Given: 1.3 mg copper in 1L solution, Desired: ? ppm m/v of Copper in water
EPA’s action limit for copper is 1.3 mg/L in drinking water. What is this limit in ppm of copper m/v in the drinking water?
Like percentage concentration, the ppm and ppb can be mass/mass (m/m), volume/volume (v/v) or mass/volume (m/v).
Parts per billion (ppb) is a number or ratio expressed as a fraction of billion (10 9 ).
For example, 2 ppm means \(\frac{2}{1,000,000}\) or 2:1,000,000, where 2 is the part, and 1,000,000 is the total. The concentration in ppm is calculated as a million times of part by total, i.e.: \begin{equation} \text { Concentration in ppm }=\frac{p a r t}{\text { Total }} \times 10^{6}\nonumber \end{equation}
Parts per million (ppm) is a number or ratio expressed as a fraction of a million (10 6 ).
How many grams of clindamycin antibiotics are in a 45 mL capsule of the 1.0% (m/v) clindamycin?
The two conversion factors from ms/v % concentration are: \begin{equation} \frac{\text { given } g \text { solute }}{100 \mathrm{~mL} \text { solution }}\quad \text { and }\quad \frac{100 \mathrm{~mL} \text { solution }}{\text { given } \mathrm{g} \text { solute }}\nonumber \end{equation}
What is the mass/volume % of glucose solution prepared by dissolving 50 g glucose in enough water to make 1000 mL of solution?
The mass/volume percent concentration expresses the mass units of solute in a hundred volume units of solution.
What is the volume of bromine (Br 2 ) in 250 mL of 4.8% v/v of Br 2 solution in carbon tetrachloride?
The two conversion factors for v/v % concentration are: \begin{equation} \frac{\text { given mL solute }}{100 \mathrm{~mL} \text { solution }} \quad\text {, and }\quad \frac{100 \mathrm{~mL} \text { solution }}{\text { given } \mathrm{~mL} \text { solute }}\nonumber \end{equation}
What is the volume % of rose extract in a solution prepared by dissolving 14.0 mL rose extract in a solvent to make 200 mL of solution?
Fig. 5.4.2 shows the volume percent concentration ranges of different classes of fragrances.
The mathematical form of the v/v % is:
The volume percent concentration expresses the volume units of solute in a hundred volume units of the solution.
Neosporin antibiotic is a 3.5% m/m neomycin solution. How many grams of neomycin are in 50 g of ointment?
Note that the mass% concentration and its reciprocal are two conversion factor: \(\frac{\text { given } g \text { solute }}{100 g \text { solution }}\) and \(\frac{100 g \text { solution }}{\text { given g solute }}\)
What is the mass % of NaOH in a solution prepared by dissolving 10.0 g NaOH in 100 g water?
Note that the total is solute and solvent added together, i.e., solution.
\begin{equation} \operatorname{Mass}(\%)=\frac{\text { mass of solute }(g)}{\text { mass of solute }(g)+\text { mass of solvent }(g)} \times 100=\frac{\text { mass of solute }(g)}{\text { mass of solution }(g)} \times 100\nonumber \end{equation}
The mass percent concentration expresses the mass units of solute in a hundred mass units of the solution.
The units cancel in the fraction calculation part, and a % sign is added to the answer to tell that it is a fraction out of a hundred.
Given part = 50.0 g NaCl, and total = 500 g solution. Desired: %NaCl in the solution?
A 50.0 g NaCl is dissolved in water to make a 500 g solution. What is the percentage of NaCl in the solution?
For example, 5% means 5:100, where 5 is the part, and 100 is the total. The percentage is calculated as a hundred times of part by total, i.e.,
Percentage (%) is a number or ratio that represent a fraction of 100.
Given: Volume of solution = 2.50 L solution, concentration of solution = \(1.12 \mathrm{~M} \mathrm{~NaOH}=\frac{1.12 \mathrm{~mol} \mathrm{~NaOH}}{1 \mathrm{~L} \text { solution }}\), Desired: ? moles of NaOH?
A 2.50 L of 1.12 M NaOH solution contains how many moles of NaOH?
How many litters of 0.211 M HCl solution are needed to provide 0.400 mol of HCl?
\[\frac{n \text { (moles of solute) }}{V \text { (Litters of solution) }} \quad\text { and }\quad \frac{V \text { (Litters of solution) }}{n \text { (moles of solute })}\nonumber\]
The two conversion factor from molarity are the following:
Calculations: First, convert the given units of mass and volume into the corresponding units that the formula takes, then plug the values in the formula and calculate.
What is the molarity (M) of a solution prepared by dissolving 50.0 g NaOH in enough water to make 250 mL solution?
The most common solution concentration unit used in chemistry is molarity (M):
Molarity (M) expresses the moles of solute in a liter of solution.
Dilution of a solution is the addition of a solvent to decrease the solute concentration of the solute in the solution.
The product of concentration (C) and volume (V) is the amount of solute, i.e.,
\begin{equation}
\text { Amount of solute }=C \text { in } \frac{\text { amount of solute }}{\text { volume of solution }} \times V \text { in volume of solution. }\nonumber
\end{equation}
The amount of solute does not change by adding solvent. Therefore, the product of concentration and volume, i.e., CV, which is the amount of solute, is a constant, i.e.,
\begin{equation}
C_{1} V_{1}=C_{2} V_{2}=\text { amount of solute }\nonumber
\end{equation}
Fig. 5.4.3 shows that if the initial concentration is C1, the initial volume is V1, and after dilution, the final concentration is C2, the final volume is V2, then C1V1 = C2V2 = amount of solute that is constant. If three of the four variables in this equation are known, the missing one can be calculated, as explained in the following example.
Keep in mind that the concentrations and volumes should be in the same units on both sides of the equation: C1V1 = C2V2. If they are not in the same units, convert them to the same units before plunging them in the formula.
how much volume of 11.3 M HCl is needed to prepare 250 mL of 2.00 M HCl?
Solution
Given: C1 = 11.3 M HCl, C2 = 2.00 M HCl, V2 = 250 mL solution, Desired V1 = ?
Formula: C1V1 = C2V2, rearrange it to isolate the desired parameter: \(V_{1}=\frac{C_{2} V_{2}}{C_{1}}\)
Calculations: \[V_{1}=\frac{C_{2} V_{2}}{C_{1}}=\frac{2.00 \mathrm{~M~HCl} \times 250 \mathrm{~mL} \text { solution }}{11.3 \mathrm{~M} \mathrm{~HCl}}=44.2 \mathrm{~mL} \text { solution }\nonumber\]
what is the molarity of the NaOH solution prepared by diluting 100 mL of 0.521 M NaOH solution to 500 mL?
Solution
Given C1 = 0.521 M NaOH, V1 = 100 mL solution, V2 = 500 mL solution, Desired: Concentration of the final solution C2 = ? M NaOH
Formula: C1V1 = C2V2, rearrange to isolate the desired parameter: \(C_{2}=\frac{C_{1} V_{1}}{V_{2}}\)
Calculations: \[C_{2}=\frac{C_{1} V_{1}}{V_{2}} = \frac{0.521 \mathrm{~M~NaOH} \times 100 \mathrm{~mL} \mathrm{~NaOH}}{500 \mathrm{~mL} \mathrm{~NaOH}}=0.104 \mathrm{~M~NaOH}\nonumber\]
Dopamine is administered intravenously to a patient to increase blood pressure. How many milliliters (mL) of a 4.0% (m/v) dopamine solution is needed to prepare 250 mL of a 0.030% m/v) solution?
Solution
Given: C1 = 4.0% (m/v), C2 = 0.030% (m/v), V2 = 250 mL solution, Desired V1= ?
Formula: C1V1 = C2V2, rearrange to isolate the desired parameter: \(V_{1}=\frac{C_{2} V_{2}}{C_{1}}\)
Calculations: \[V_{1}=\frac{C_{2} V_{2}}{C_{1}}=\frac{0.030 \% \times 250 \mathrm{~mL} \text { solution }}{4.0 \%}=5.0 \mathrm{~mL} \text { of solution }\nonumber\]
Take a unit volume of a given solution and add enough solvent to increase the volume of the solution 10 times for a logarithmic diluiton.
A logarithmic dilution is ten times dilution, i.e., proven by the following formula: \[C_{2}=\frac{C_{1} V_{1}}{V_{2}}=\frac{1 \mathrm{~mL} \times C_{1}}{10 \mathrm{~mL}}=0.1 \times C_{1}\nonumber\]
Repeating the above step with the diluted solution results in 10x10 = 100-time dilution, and repeating third-time results in 10x10x10 = 1000-time dilution. This dilution of 10 times in each step is called logarithmic dilution. Fig. 5.4.4 shows that five steps of logarithmic dilution on a 10% initial solution results in a concentration of 10 ppm in the final solution.
Figure \(\PageIndex{4}\): Logarithmic dilution: five steps of logarithmic dilution on a 10% initial solution results in a concentration of 10 ppm of the final solution. Source: Grasso Luigi / CC BY-SA (https://creativecommons.org/licenses/by-sa/4.0)