5.5: Solutions
- Page ID
- 475568
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Learn some terminology involving solutions.
- Express the amount of solute in a solution in various concentration units.
When one substance dissolves into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the substance that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with many different types and forms of solutes and solvents. In this chapter, we will focus on solution where the solvent is water. An aqueous solution is water that contains one or more dissolved substance. The dissolved substances in an aqueous solution may be solids, gases, or other liquids.
In order to be a true solution, a mixture must be stable. When sugar is fully dissolved into water, it can stand for an indefinite amount of time, and the sugar will not settle out of the solution. Further, if the sugar-water solution is passed through a filter, it will remain with the water. This is because the dissolved particles in a solution are very small, usually less than \(1 \: \text{nm}\) in diameter. Solute particles can be atoms, ions, or molecules, depending on the type of substance that has been dissolved. Water typically dissolves most ionic compounds and polar molecules. Nonpolar molecules, such as those found in grease or oil, do not dissolve in water.
Solution Concentrations
to the most concentrated (on the right).
A concentrated solution is one in which there is a large amount of solute in a given amount of solvent. A dilute solution is one in which there is a small amount of solute in a given amount of solvent. A dilute solution is a concentrated solution that has been, in essence, watered down. Think of the frozen juice containers you buy in the grocery store. What you have to do is take the frozen juice from inside these containers and usually empty it into 3 or 4 times the container size full of water to mix with the juice concentrate and make your container of juice. Therefore, you are diluting the concentrated juice. When we talk about solute and solvent, the concentrated solution
has a lot of solute versus the dilute solution that would have a smaller amount of solute.
The terms "concentrated" and "dilute" provide qualitative methods of describing concentration. Although qualitative observations are necessary and have their place in every part of science, including chemistry, we have seen throughout our study of science that there is a definite need for quantitative measurements in science. This is particularly true in solution chemistry. In this section, we will explore some quantitative methods of expressing solution concentration.
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as dilute or concentrated are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors. When we say that vinegar is \(5\%\) acetic acid in water, we are giving the concentration. If we said the mixture was \(10\%\) acetic acid, this would be more concentrated than the vinegar solution and we know exactly by how much.
Molarity
One way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution.
\[\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \label{defMolarity} \]
The symbol for molarity is \(\text{M}\) or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. For example, the expression \(\left[ \ce{Ag^+} \right]\) refers to the molarity of the silver ion in solution. Solution concentrations expressed in molarity are the easiest to calculate with but the most difficult to make in the lab. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.
It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore
\[\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber \]
Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl (aq).” You would read as "a 3.00 molar sodium chloride solution," meaning that there are 3.00 moles of NaOH dissolved per one liter of solution.
Be sure to note that molarity is calculated as the total volume of the entire solution, not just volume of solvent! The solute contributes to total volume.
If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L?
First, convert the mass of solute to moles using the molar mass of HCl (36.5 g/mol):
\[22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.5\cancel{gHCl}}=0.614\, mol\; HCl \nonumber \]
Now we can use the definition of molarity to determine a concentration:
\[M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber \]
Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.
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
Steps for Problem Solving |
Example |
|---|---|
| Identify the "given"information and what the problem is asking you to "find." |
Given: Mass \(= 42.23 \: \text{g} \: \ce{NH_4Cl}\) Volume solution \(= 500.0 \: \text{mL} = 0.5000 \: \text{L}\) Find: Molarity = ? M |
| List other known quantities. | Molar mass \(\ce{NH_4Cl} = 53.50 \: \text{g/mol}\) |
| Plan the problem. |
1. The mass of the ammonium chloride is first converted to moles.
2. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters. \(\mathrm{M=\frac{mol\: NH_4Cl}{L\: solution}} \) |
| Cancel units and calculate. |
Now substitute the known quantities into the equation and solve. \(\begin{align} 42.23 \ \cancel{\text{g} \: \ce{NH_4Cl}} \times \dfrac{1 \: \text{mol} \: \cancel{\ce{NH_4Cl}}}{53.50 \: \text{g} \: \ce{NH_4Cl}} &= 0.7893 \: \text{mol} \: \ce{NH_4Cl} \\ \dfrac{0.7893 \: \text{mol} \: \ce{NH_4Cl}}{0.5000 \: \text{L solution}} &= 1.579 \: \text{M} \end{align}\) |
| Think about your result. | The molarity is \(1.579 \: \text{M}\), meaning that a liter of the solution would contain \(1.579 \: \text{mol} \: \ce{NH_4Cl}\). Four significant figures are appropriate. |
What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?
- Answer
-
1.57 M C6H12O6
What is the concentration, in \(\text{mol/L}\), where \(137 \: \text{g}\) of \(\ce{NaCl}\) has been dissolved in enough water to make \(500 \: \text{mL}\) of solution?
- Answer
-
4.69 M NaCl
Figure \(\PageIndex{1}\) illustrates how a solution is prepared with the measured amount of a solid solute and the desired final volume of the solution.
Percent Concentrations
There are several ways of expressing the concentration of a solution by using a percentage. The percent can further be determined in one of two ways: (1) the ratio of the mass of the solute divided by the mass of the solution or (2) the ratio of the volume of the solute divided by the volume of the solution.
Percent by Volume
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 \(\left( \frac{\text{volume}}{\text{volume}} \right)\) of the solution. If a solution is made by adding \(40 \: \text{mL}\) of ethanol to \(20 \: \text{mL}\) of water, the percent by volume is:
\[\begin{align} \text{Percent by volume} &= \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\ &= \frac{40 \: \text{mL ethanol}}{240 \: \text{mL solution}} \times 100\% \\ &= 16.7\% \: \text{ethanol} \end{align} \nonumber \]
Frequently, ingredient labels on food products and medicines have amounts listed as percentages (see figure below).
Percent by Mass
The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100:
\[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber \]
mass of solution = mass of solute + mass solvent
If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration.
Suppose that a solution was prepared by dissolving \(25.0 \: \text{g}\) of sugar into \(100.0 \: \text{g}\) of water.
The mass of the solution is
mass of solution = 25.0g sugar + 100.0g water = 125.0 g
The percent by mass would be calculated by:
\[\text{Percent by mass} = \frac{25.0 \: \text{g sugar}}{125.0 \: \text{g solution}} \times 100\% = 20.0\% \: \text{sugar} \nonumber \]
A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution?
Solution
We can substitute the quantities given in the equation for mass/mass percent:
\(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\)
A dextrose (also called D-glucose, C6H12O6) solution with a mass of 2.00 × 102 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution?
- Answer
-
7.90 %
Using Mass Percent in Calculations
Sometimes you may want to make up a particular mass of solution of a given percent by mass and need to calculate what mass of the solute to use. Using mass percent as a conversion can be useful in this type of problem. The mass percent can be expressed as a conversion factor in the form \(\frac{g \; \rm{solute}}{100 \; \rm{g solution}}\) or \(\frac{100 \; \rm g solution}{g\; \rm{solute}}\)
For example, if you need to make \(3000.0 \: \text{g}\) of a \(5.00\%\) solution of sodium chloride, the mass of solute needs to be determined.
Solution
Given: 3000.0 g NaCl solution
5.00% NaCl solution
Find: mass of solute = ? g NaCl
Other known quantities: 5.00 g NaCl is to 100 g solution
The appropriate conversion factor (based on the given mass percent ) can be used follows:
To solve for the mass of NaCl, the given mass of solution is multiplied by the conversion factor.
\[g NaCl = 3,000.0 \cancel{g \: NaCl \:solution} \times \frac{5.00 \:g \: NaCl}{100\cancel{g \: NaCl \: solution}} = 150.0g \: NaCl \nonumber \]
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 \(\left( 3000 \: \text{g} \right)\) to calculate the mass of the water that would need to be added.
What is the amount (in g) of hydrogen peroxide (H2O2) needed to make a 6.00 kg , 3.00 % (by mass) H2O2 solution?
Answer
180. g H2O2
Summary
- Solutions are composed of a solvent (major component) and a solute (minor component).
- Concentration of solutions can be more precisely expressed in terms of molarity, percent by volume, and percent by mass.
Contributors and Attributions
Allison Soult, Ph.D. (Department of Chemistry, University of Kentucky)
Henry Agnew (UC Davis)