4.2: Solution Concentration
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- To describe the concentrations of solutions quantitatively
Many people have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations. In this section, we will focus on a unit of concentration called molarity which is a useful unit for many applications in chemistry. In addition, concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in parts per million (ppm), which is grams of solute per 106 g of solution, or in parts per billion (ppb), which is grams of solute per 109 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in Table \(\PageIndex{1}\) and will be discussed in detail in chapter 13.
Table \(\PageIndex{1}\): Common Units of Concentration
| Concentration | Units |
|---|---|
| m/m | g of solute/g of solution |
| m/v | g of solute/mL of solution |
| ppm | g of solute/106 g of solution |
| μg/mL | |
| ppb | g of solute/109 g of solution |
| ng/mL |
Molarity
The most common unit of concentration is molarity (M). It is defined as the number of moles of solute present in exactly 1 L of solution. It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution:
\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.2}\]
A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?
Solution
Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:
\[\begin{align*} M &=\dfrac{mol\: solute}{L\: solution} \\[4pt] &=\dfrac{0.133\:mol}{355\:mL\times \dfrac{1\:L}{1000\:mL}} \\[4pt] &= 0.375\:M \label{3.4.1} \end{align*}\]
A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL?
- Answer
-
0.05 M
How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example \(\PageIndex{1}\)?
Solution
In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 3.4.2, 0.375 M:
\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.3} \nonumber\]
\[ \begin{align*} \mathrm{mol\: solute} &= \mathrm{ M\times L\: solution} \label{3.4.4} \\[4pt] \mathrm{mol\: solute} &= \mathrm{0.375\:\dfrac{mol\: sugar}{L}\times \left(10\:mL\times \dfrac{1\:L}{1000\:mL}\right)} &= \mathrm{0.004\:mol\: sugar} \label{3.4.5} \end{align*} \]
What volume (mL) of the sweetened tea described in Example \(\PageIndex{1}\) contains the same amount of sugar (mol) as 10 mL of the soft drink in this example?
- Answer
-
80 mL
Distilled white vinegar (Figure \(\PageIndex{1}\)) is a solution of acetic acid, \(CH_3CO_2H\), in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?
Solution
As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:
\[\mathrm{\mathit M=\dfrac{mol\: solute}{L\: solution}=\dfrac{25.2\: g\: \ce{CH3CO2H}\times \dfrac{1\:mol\: \ce{CH3CO2H}}{60.052\: g\: \ce{CH3CO2H}}}{0.500\: L\: solution}=0.839\: \mathit M} \label{3.4.6} \nonumber\]
\[M=\mathrm{\dfrac{0.839\:mol\: solute}{1.00\:L\: solution}} \nonumber\]
Calculate the molarity of 6.52 g of \(CoCl_2\) (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.
- Answer
-
0.674 M
How many grams of NaCl are contained in 0.250 L of a 5.30-M solution?
Solution
The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example \(\PageIndex{3}\):
\[M=\mathrm{\dfrac{mol\: solute}{L\: solution}} \label{3.4.9}\]
\[\mathrm{mol\: solute= \mathit M\times L\: solution} \label{3.4.10}\]
\[\mathrm{mol\: solute=5.30\:\dfrac{mol\: NaCl}{L}\times 0.250\:L=1.325\:mol\: NaCl} \label{3.4.11}\]
Finally, this molar amount is used to derive the mass of NaCl:
\[\mathrm{1.325\: mol\: NaCl\times\dfrac{58.44\:g\: NaCl}{mol\: NaCl}=77.4\:g\: NaCl} \label{3.4.12}\]
How many grams of \(CaCl_2\) (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride?
- Answer
-
5.55 g \(CaCl_2\)
When performing calculations stepwise, as in Example \(\PageIndex{3}\), it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example \(\PageIndex{4}\), the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.
In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (Example \(\PageIndex{5}\)). This eliminates intermediate steps so that only the final result is rounded.
In Example \(\PageIndex{3}\), we found the typical concentration of vinegar to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid?
Solution
First, use the molar mass to calculate moles of acetic acid from the given mass:
\[\mathrm{g\: solute\times\dfrac{mol\: solute}{g\: solute}=mol\: solute} \label{3.4.13}\]
Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute:
\[\mathrm{mol\: solute\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.14}\]
Combining these two steps into one yields:
\[\mathrm{g\: solute\times \dfrac{mol\: solute}{g\: solute}\times \dfrac{L\: solution}{mol\: solute}=L\: solution} \label{3.4.15}\]
\[\mathrm{75.6\:g\:\ce{CH3CO2H}\left(\dfrac{mol\:\ce{CH3CO2H}}{60.05\:g}\right)\left(\dfrac{L\: solution}{0.839\:mol\:\ce{CH3CO2H}}\right)=1.50\:L\: solution} \label{3.4.16}\]
What volume of a 1.50-M KBr solution contains 66.0 g KBr?
- Answer
-
0.370 L
The Preparation of Solutions
To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure \(\PageIndex{1}\) illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always less than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure \(\PageIndex{2}\), for some substances this effect can be significant, especially for concentrated solutions.
The solution contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2•2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of \(\ce{CoCl2•2H2O}\)?
Given: mass of solute and volume of solution
Asked for: concentration (M)
Strategy:
To find the number of moles of \(\ce{CoCl2•2H2O}\), divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.
Solution:
The molar mass of CoCl2•2H2O is 165.87 g/mol. Therefore,
\[ moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \nonumber \]
The volume of the solution in liters is
\[ volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \nonumber \]
Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is
\[ molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \nonumber \]
The solution shown in Figure \(\PageIndex{2}\) contains 90.0 g of (NH4)2Cr2O7 in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?
- Answer
-
\[(NH_4)_2Cr_2O_7 = 1.43\: M \nonumber\]
To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation \(\ref{4.5.2}\). We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example \(\PageIndex{3}\).
The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.
Given: molarity, volume, and molar mass of solute
Asked for: mass of solute
Strategy:
- Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.
- Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.
Solution:
A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:
\( V_L M_{mol/L} = moles \)
\( 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose \)
B We then convert the number of moles of glucose to the required mass of glucose:
\( mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose \)
Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.
- Answer
-
2.3 g NaCl
Summary
The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution.
Key Equations
- \(M=\mathrm{\dfrac{mol\: solute}{L\: solution}}\)
Glossary
- molarity (M)
- unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution
Contributors and Attributions
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).