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2.2: Concentration

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  • Concentration is a general measurement unit stating the amount of solute present in a known amount of solution

    \[\textrm{Concentration} = \dfrac{\text{amount of solute}}{\text{amount of solution}} \tag{2.1}\]

    Although we associate the terms “solute” and “solution” with liquid samples, we can extend their use to gas-phase and solid-phase samples as well. Table \(\PageIndex{1}\) lists the most common units of concentration.

    Molarity and Formality

    Both molarity and formality express concentration as moles of solute per liter of solution. There is, however, a subtle difference between molarity and formality. Molarity is the concentration of a particular chemical species. Formality, on the other hand, is a substance’s total concentration without regard to its specific chemical form. There is no difference between a compound’s molarity and formality if it dissolves without dissociating into ions. The formal concentration of a solution of glucose, for example, is the same as its molarity. (A solution that is 0.0259 M in glucose is 0.0259 F in glucose as well.)

    For a compound that ionizes in solution, such as \(\ce{NaCl}\), molarity and formality are different. Dissolving 0.1 moles of \(\ce{CaCl_2}\) in 1 L of water gives a solution containing 0.1 moles of \(\ce{Ca_{2+}}\) and 0.2 moles of \(\ce{Cl^{-}}\). The molarity of \(\ce{NaCl}\), therefore, is zero since there is essentially no undissociated \(\ce{NaCl}\). The solution, instead, is 0.1 M in \(\ce{Ca^{2+}}\) and 0.2 M in \(\ce{Cl^{-}}\). The formality of \(\ce{NaCl}\), however, is 0.1 F since it represents the total amount of \(\ce{NaCl}\) in solution. The rigorous definition of molarity, for better or worse, is largely ignored in the current literature, as it is in this textbook. When we state that a solution is 0.1 M \(\ce{NaCl}\) we understand it to consist of \(\ce{Na^{+}}\) and \(\ce{Cl^{-}}\) ions. The unit of formality is used only when it provides a clearer description of solution chemistry.

    Molarity is used so frequently that we use a symbolic notation to simplify its expression in equations and in writing. Square brackets around a species indicate that we are referring to that species’ molarity. Thus, \(\ce{[Na^+]}\) is read as “the molarity of sodium ions.”

    Table \(\PageIndex{1}\): Common Units for Reporting Concentration
    Name Units Symbol
    molarity \(\mathrm{\dfrac{moles\: solute}{liters\: solution}}\) M
    formality \(\mathrm{\dfrac{moles\: solute}{liters\: solution}}\) F
    normality \(\mathrm{\dfrac{equivalents\: solute}{liters\: solution}}\) N
    molality \(\mathrm{\dfrac{moles\: solute}{kilograms\: solvent}}\) m
    weight percent \(\mathrm{\dfrac{grams\: solute}{100\: grams\: solution}}\) % w/w
    volume percent \(\mathrm{\dfrac{mL\: solute}{100\: mL\: solution}}\) % v/v
    weight-to-volume percent \(\mathrm{\dfrac{grams\: solute}{100\: mL\: solution}}\) % w/v
    parts per million \(\mathrm{\dfrac{grams\: solute}{10^6\: grams\: solution}}\) ppm
    parts per billion \(\mathrm{\dfrac{grams\: solute}{10^9\: grams\: solution}}\) ppb

    An alternative expression for weight percent is

    \[\mathrm{\dfrac{grams\: solute}{grams\: solution}\times 100}\]

    You can use similar alternative expressions for volume percent and for weight-to-volume percent.


    Normality is a concentration unit that is no longer in common use. Because you may encounter normality in older handbooks of analytical methods, it can be helpful to understand its meaning. Normality defines concentration in terms of an equivalent, which is the amount of one chemical species reacting stoichiometrically with another chemical species. Note that this definition makes an equivalent, and thus normality, a function of the chemical reaction in which the species participates. Although a solution of \(\ce{H_2SO_4}\) has a fixed molarity, its normality depends on how it reacts. You will find a more detailed treatment of normality in Appendix 1.

    One handbook that still uses normality is Standard Methods for the Examination of Water and Wastewater, a joint publication of the American Public Health Association, the American Water Works Association, and the Water Environment Federation. This handbook is one of the primary resources for the environmental analysis of water and wastewater.


    Molality is used in thermodynamic calculations where a temperature independent unit of concentration is needed. Molarity is based on the volume of solution containing the solute. Since density is a temperature dependent property a solution’s volume, and thus its molar concentration, changes with temperature. By using the solvent’s mass in place of the solution’s volume, the resulting concentration becomes independent of temperature.

    Weight, Volume, and Weight-to-Volume Ratios

    Weight percent (% w/w), volume percent (% v/v) and weight-to-volume percent (% w/v) express concentration as the units of solute present in 100 units of solution. A solution of 1.5% w/v \(\ce{NH_4NO_3}\), for example, contains 1.5 gram of \(\ce{NH_4NO_3}\) in 100 mL of solution.

    Parts Per Million and Parts Per Billion

    Parts per million (ppm) and parts per billion (ppb) are ratios giving the grams of solute to, respectively, one million or one billion grams of sample. For example, a steel that is 450 ppm in Mn contains 450 μg of Mn for every gram of steel. If we approximate the density of an aqueous solution as 1.00 g/mL, then solution concentrations can be express in ppm or ppb using the following relationships.

    \[\mathrm{ppm = \dfrac{mg}{L} = \dfrac{g}{g}}\]

    \[\mathrm{ppb = \dfrac{g}{L} = \dfrac{ng}{mL}}\]

    For gases a part per million usually is a volume ratio. Thus, a helium concentration of 6.3 ppm means that one liter of air contains 6.3 μL of \(\ce{He}\).

    ppm: mg/L vs. mg/g?

    You should be careful when using parts per million and parts per billion to express the concentration of an aqueous solute. The difference between a solute’s concentration in mg/L and mg/g, for example, is significant if the solution’s density is not 1.00 g/mL. For this reason many organizations advise against using the abbreviation ppm and ppb (see If in doubt, include the exact units, such as 0.53 mg \(\ce{Pb^{2+}/L}\) for the concentration of lead in a sample of seawater.

    Converting Between Concentration Units

    The most common ways to express concentration in analytical chemistry are molarity, weight percent, volume percent, weight-to-volume percent, parts per million and parts per billion. By recognizing the general definition of concentration given in equation 2.1, it is easy to convert between concentration units.

    Example \(\PageIndex{1}\)

    A concentrated solution of ammonia is 28.0% w/w \(\ce{NH_3}\) and has a density of 0.899 g/mL. What is the molar concentration of \(\ce{NH_3}\) in this solution?


    \[\mathrm{\dfrac{28.0\: g\: NH_3} {100\: g\: solution} \times \dfrac{0.899\: g\: solution}{mL\: solution} \times \dfrac{1\: mol\: NH_3} {17.04\: g\: NH_3} \times \dfrac{1000\: mL}{L} = 14.8\: M}\]

    Example \(\PageIndex{2}\)

    The maximum permissible concentration of chloride in a municipal drinking water supply is 2.50 × 102 ppm \(\ce{Cl^{-}}\). When the supply of water exceeds this limit it often has a distinctive salty taste. What is the equivalent molar concentration of \(\ce{Cl^{-}}\)?


    \[\mathrm{\dfrac{2.50 \times 10^2\: mg\: Cl^-}{L} \times \dfrac{1\: g}{1000\: mg}\times \dfrac{1\: mol\: Cl^-}{35.453\: g\: Cl^-} = 7.05 \times 10^{-3}\: M}\]

    Exercise \(\PageIndex{1}\)

    Which solution—0.50 M \(\ce{NaCl}\) or 0.25 M \(\ce{SrCl_2}\)—has the larger concentration when expressed in μg/mL?
    Click here to review your answer to this exercise.


    Figure \(\PageIndex{1}\): Graph showing the progress for the titration of 50.0 mL of 0.10 M \(\ce{HCl}\) with 0.10 M \(\ce{NaOH}\). The \(\ce{[H^{+}]}\) is shown on the left y-axis and the pH on the right y-axis.


    Sometimes it is inconvenient to use the concentration units in Table \(\PageIndex{1}\). For example, during a reaction a species’ concentration may change by many orders of magnitude. If we want to display the reaction’s progress graphically we might plot the reactant’s concentration as a function of time or as a function of the volume of a reagent being added to the reaction. Such is the case in Figure \(\PageIndex{1}\) for the titration of \(\ce{HCl}\) with \(\ce{NaOH}\). (Acid–base titrations, as well as several other types of titrations, are covered in Chapter 9.) The y-axis on the left-side of the figure displays the \(\ce{[H^+]}\) as a function of the volume of \(\ce{NaOH}\). The initial \(\ce{[H^+]}\) is 0.10 M and its concentration after adding 80 mL of \(\ce{NaOH}\) is 4.3 × 10-13 M. We can easily follow the change in \(\ce{[H^+]}\) for the first 14 additions of \(\ce{NaOH}\). For the remaining additions of \(\ce{NaOH}\), however, the change in \(\ce{[H^+]}\) is too small to see.

    When working with concentrations spanning many orders of magnitude, it is often more convenient to express concentration using a p-function. The p-function of X is written as pX and is defined as


    The pH of a solution that is 0.10 M \(\ce{H^{+}}\) is

    \[\mathrm{pH = -\log[H^+] = -\log(0.10) = 1.00}\]

    and the pH of 4.3 × 10-13 M \(\ce{H^+}\) is

    \[\mathrm{pH = −\log[H^+] = -\log(4.3 \times 10^{-13}) = 12.37}\]

    Figure \(\PageIndex{1}\) shows that plotting pH as a function of the volume of \(\ce{NaOH}\) provides more detail about how the concentration of \(\ce{H^+}\) changes during the titration.

    A more appropriate equation for pH is

    \[\mathrm{pH = -\log_{10}(a_{H^+})}\]

    where aH+ is the activity of the hydrogen ion. See Chapter 6 for more details. For now the approximate equation is sufficient.

    \[\mathrm{pH \approx -\log[H^+]}\]

    Example \(\PageIndex{3}\)

    What is \(\ce{pNa}\) for a solution of 1.76 × 10-3 M \(\ce{Na_3PO_4}\)?


    Since each mole of \(\ce{Na_3PO_4}\) contains three moles of \(\ce{Na^+}\), the concentration of \(\ce{Na^+}\) is

    \[\mathrm{[Na^+] = (1.76 \times 10^{-3}) \left (\dfrac{3\; mol\; Na^+}{1\; mol\; Na_3PO_4}\right) = 5.28 \times 10^{-3}\; M}\]

    and \(\ce{pNa}\) is

    \[\mathrm{pNa = -\log_{10}[Na^+] = -\log_{10}(5.28 \times 10^{-3}) = 2.277}\]

    Example \(\PageIndex{4}\)

    What is the \(\ce{[H^+]}\) in a solution that has a pH of 5.16?


    The concentration of \(\ce{H^+}\) is

    \[\mathrm{pH \approx -\log[H^+] = 5.16}\]

    \[\mathrm{\log[H^+] = -5.16}\]

    \[\mathrm{[H^+] = antilog(-5.16) = 10^{-5.16} = 6.9 \times 10^{-6}\:M}\]

    Note: If \(X = 10^a\), then \(\log_{10}(X) = a\).

    Exercise \(\PageIndex{2}\)

    What is \(\ce{pK}\) and \(\ce{pSO_4}\) for a solution containing 1.5 g \(\ce{K_2SO_4}\) in a total volume of 500.0 mL?

    Click here to review your answer to this exercise.