6.5: Solution Concentrations and Dilutions
- Page ID
- 431298
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Define concentration and use the terms concentrated and dilute to describe the relative concentration of a solution.
- Calculate (w/v) percent concentration.
- Determine concentration in molarity.
- Complete dilution calculations.
Solution Concentration
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.
Weight/Volume Percent
One way to describe the concentration of a solution is by the percent of the solute in a solution. Percent means "parts per hundred". It means the number of parts of solute present every 100 parts of solution. The most commonly used percent measurement is weight/volume percent. The weight/volume percent unit (%w/v), is commonly used for intravenous (IV) fluids. It is calculated by dividing the grams of the solute by the milliliters of the solution and expressing the result as a percent.
For example, if a solution is prepared from \(10. g \: \ce{NaCl}\) in enough water to make a \(150 \: \text{mL}\) solution, the weight/volume percent is
\[\begin{align} \text{weight/volume percent =} & \frac{\text{grams solute}}{\text{volume solution}} \times 100\% \\ &= \frac{10. \: \text{g} \: \ce{NaCl}}{150 \: \text{mL solution}} \times 100\% \\ &= 6.7\% (w/v) \end{align}\]
Molarity
A concentration unit based on moles of solute is molarity. 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".
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
- Mass of \(\ce{NH_4Cl} = 42.23 \: \text{g}\)
- 1 mole of \(\ce{NH_4Cl} = 53.50 \: \text{g}\)
- Volume of solution \(= 500.0 \: \text{mL} = 0.5000 \: \text{L}\)
Unknown
- Molarity \(= ? \: \text{M}\)
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.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. Dilution is adding more solvent (water) to a solution. Dilution is important in health care because some drugs must be diluted to proper concentration before being administered. The number of moles of the solute does not change, but the total volume of the solution increases.
An useful equation to use when doing dilutions is
\[C_1 \times V_1 = C_2 \times V_2\]
The concentration can be in any other unit as long as \(C_1\) and \(C_2\) 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 \(C_2\).
\[C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl}\]
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.
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
- Stock \(\ce{HNO_3} \: \left( C_1 \right) = 16 \: \text{M}\)
- \(V_2 = 8.00 \: \text{L}\)
- \(C_2 = 0.50 \: \text{M}\)
Unknown
- Volume of stock \(\ce{HNO_3} \: \left( V_1 \right) = ? \: \text{L}\)
The unknown in the equation is \(V_1\), the necessary volume of the concentrated stock solution.
Step 2: Solve.
\[V_1 = \frac{C_2 \times V_2}{V_1} = \frac{0.50 \: \text{M} \times 8.00 \: \text{L}}{16 \: \text{M}} = 0.25 \: \text{L} = 250 \: \text{mL}\]
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.
Contributors and Attributions
- Deboleena Roy, Ph.D. (Deparment of Science and Engineering @ American River College)
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)