3.2: Units of Concentration
- Page ID
- 284474
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objective
- Learn to determine specific concentrations with several common units.
Rather than qualitative terms (Section 11.2 - Definitions), we need quantitative ways to express the amount of solute in a solution; that is, we need specific units of concentration. In this section, we will introduce several common and useful units of concentration.
Molarity (M) is defined as the number of moles of solute divided by the number of liters of solution:
\[molarity \: =\: \frac{moles\: of\: solute}{liters\: of\: solution}\nonumber \]
which can be simplified as
\[M\: =\: \frac{mol}{L},\; or\; mol/L\nonumber \]
As with any mathematical equation, if you know any two quantities, you can calculate the third, unknown, quantity.
For example, suppose you have 0.500 L of solution that has 0.24 mol of NaOH dissolved in it. The concentration of the solution can be calculated as follows:
\[molarity \: =\: \frac{0.24\: mol\: NaOH}{0.500L}=0.48\, M\; NaOH\nonumber \]
The concentration of the solution is 0.48 M, which is spoken as "zero point forty-eight molarity" or "zero point forty-eight molar." 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):
\[22.4\cancel{gHCl}\times \frac{1\: mol\: Hcl}{36.5\cancel{gHCl}}=0.614\, M\; HCl\nonumber \]
Now we can use the definition of molarity to determine a concentration:
\[M \: =\: \frac{0.614\: mol\: HCl}{1.56L}=0.394\, M\nonumber \]
Example \(\PageIndex{1}\):
What is the molarity of a solution made when 32.7 g of NaOH are dissolved to make 445 mL of solution?
Solution
To use the definition of molarity, both quantities must be converted to the proper units. First, convert the volume units from milliliters to liters:
\[445\cancel{mL}\times \frac{1\: L}{1000\cancel{mL}}=0.445\, L\nonumber \]
Now we convert the amount of solute to moles, using the molar mass of NaOH, which is 40.0 g/mol:
\[32.7\cancel{gNaOH}\times \frac{1\: mol\: NaOH}{40.0\cancel{gNaOH}}=0.818\, mol\: NaOH\nonumber \]
Now we can use the definition of molarity to determine the molar concentration:
\[M \: =\: \frac{0.818\: mol\: NaOH}{0.445L}=1.84\, M\: NaOH\nonumber \]
Exercise \(\PageIndex{1}\)
What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution?
- Answer
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1.57 M
Another way to specify an amount is percentage composition by mass (or mass percentage, % m/m). It is defined as follows:
\[\%m/m\: =\: \frac{mass\: of\: solute}{mass\: of\: entire\: sample}\times 100\%\nonumber \]
It is not uncommon to see this unit used on commercial products (Figure \(\PageIndex{1}\) - Concentration in Commercial Applications)
![e06d00c8a941e1fdde0005935083a5d7.jpg](https://chem.libretexts.org/@api/deki/files/91930/e06d00c8a941e1fdde0005935083a5d7.jpg?revision=1&size=bestfit&width=368&height=550)
The percentage of urea in this package is 5% m/m, meaning that there are 5 g of urea per 100 g of product.
Example \(\PageIndex{1}\):
What is the mass percentage of Fe in a piece of metal with 87.9 g of Fe in a 113 g sample?
Solution
Using the definition of mass percentage, we have
\[\%m/m\: =\: \frac{87.9\, g\, Fe}{113\, g\, sample}\times 100\%=77.8\%\, Fe\nonumber \]
Exercise \(\PageIndex{1}\)
What is the mass percentage of H2O2 in a solution with 1.67 g of H2O2 in a 55.5 g sample?
- Answer
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3.01%
Related concentration units are parts per thousand (ppth), parts per million (ppm) and parts per billion (ppb). Parts per thousand is defined as follows:
\[ppth\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1000\nonumber \]
There are similar definitions for parts per million and parts per billion:
\[ppm\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000\nonumber \]
and
\[ppb\: =\: \frac{mass\: of\: solute}{mass\: of\: sample}\times 1,000,000,000\nonumber \]
Each unit is used for progressively lower and lower concentrations. The two masses must be expressed in the same unit of mass, so conversions may be necessary.
Example \(\PageIndex{1}\):
If there is 0.6 g of Pb present in 277 g of solution, what is the Pb concentration in parts per thousand?
Solution
Use the definition of parts per thousand to determine the concentration. Substituting
\[\frac{0.6g\, Pb}{277g\, solution}\times 1000=2.17\, ppth\nonumber \]
Exercise \(\PageIndex{1}\)
If there is 0.551 mg of As in 348 g of solution, what is the As concentration in ppm?
- Answer
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1.58 ppm
For ionic solutions, we need to differentiate between the concentration of the salt versus the concentration of each individual ion. Because the ions in ionic compounds go their own way when a compound is dissolved in a solution, the resulting concentration of the ion may be different from the concentration of the complete salt. For example, if 1 M NaCl were prepared, the solution could also be described as a solution of 1 M Na+(aq) and 1 M Cl−(aq) because there is one Na+ ion and one Cl− ion per formula unit of the salt. However, if the solution were 1 M CaCl2, there are two Cl−(aq) ions for every formula unit dissolved, so the concentration of Cl−(aq) would be 2 M, not 1 M.
In addition, the total ion concentration is the sum of the individual ion concentrations. Thus for the 1 M NaCl, the total ion concentration is 2 M; for the 1 M CaCl2, the total ion concentration is 3 M.
Key Takeaway
- Quantitative units of concentration include molarity, mass percentage, parts per thousand (ppth), parts per million (ppm), and parts per billion (ppb).
Exercise \(\PageIndex{1}\)
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What is the molarity of a solution made by dissolving 13.4 g of NaNO3 in 345 mL of solution?
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What is the molarity of a solution made by dissolving 332 g of C6H12O6 in 4.66 L of solution?
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What are the individual ion concentrations and the total ion concentration in 0.66 M Mg(NO3)2?
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What are the individual ion concentrations and the total ion concentration in 1.04 M Al2(SO4)3?
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If the C2H3O2– ion concentration in a solution is 0.554 M, what is the concentration of Ca(C2H3O2)2?
6. If the Cl− ion concentration in a solution is 2.61 M, what is the concentration of FeCl3?
Answers
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0.457 M
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Mg2+ = 0.66 M; NO3− = 1.32 M; total: 1.98 M
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0.277 M