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Units of Concentration

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Title: Units of Concentration


1
Units of Concentration
  • Chemical concentration is an important unit of
    measure in all aspects of chemical fate,
    transport and treatment in environmental and
    engineered systems.
  • The unit of expression for concentration depends
    upon the chemical and what media it is being
    measured in. The following table lists the most
    common ways to express chemical concentrations.

2
Units of Concentration
3
Mass/Mass Units
  • Mass/mass units are commonly expressed in
    percent, parts per million, parts per billion,
    parts per trillion, and so on.
  • For example 1 mg of toluene is placed in 1 kg of
    water equals 1 ppmm. Parts per million (ppm or
    ppmm) is defined as the number of units of mass
    of chemical per million units of total mass.

4
Mass/Mass Units
  • The general formula for ppm or ppmm is expressed
    as

Where, mi is the mass of chemical,i, (g) and
mtotal is the total mass in the sample. Note
that 106 is really a conversion factor and it has
the implicit units of ppmm/mass fraction (mass
fraction mi/mtotal)
5
Mass/Mass Units
  • In the previous equation mi/mtotal is defined as
    the mass fraction, and the conversion factor of
    106 is similar to the conversion factor of 102
    that is used to convert fractions to percentages.
    For example,

Similar definitions are used for the units ppbm
and pptm and by mass. That is 1 ppbm equals 1
part per billion or 1 g chemical per billion
(109) g total, so that the number of ppbm in a
sample is equal to mi/mtotalx109. Also, 1 pptm
equals 1 part per trillion or 1 g chemical per
trillion (1012).
6
Mass/Mass Units
  • Mass/mass concentrations can also be reported
    with the units explicitly shown (e.g. mg/kg,
    ?g/kg). In soils and sediments, 1 ppmm equals 1
    mg of chemical per kg of solid (mg/kg) and 1 ppbm
    equals 1 ?g/kg. Percent by mass is equal to the
    number of g chemical per 100 g total.

7
Mass/Mass Units
  • Example A one-kg sample of soil is analyzed for
    the chemical trichloroethene (TCE). The analysis
    indicates that the sample contains 5.0 mg of TCE.
    What is the TCE concentration in ppmm and ppbm?

Note that in soil and sediments, mg/kg equals
ppmm and ?g/kg equals ppbm.
8
Mass/Volume Units
  • It is common to use concentration units in terms
    of mass per volume because it is easy to use
    these units in mass balance calculations. For
    gas-phase concentrations common usage is in terms
    of mass per volume of air or mg/m3 or ?g/m3. In
    aqueous phase applications ppmm is equivalent to
    mg/L. This is because the density of pure water
    is approximately 1,000 g/L at 5oC. This is
    demonstrated in the following example. At 20oC,
    the density has decreased slightly to 998.2 g/L.
    In addition, this equality is strictly true for
    dilute solutions, in which any dissolved material
    does not contribute significantly to the mass of
    water, and the total density remains
    approximately 1,000 g/L. Most wastewaters and
    natural waters can be considered dilute, except
    perhaps sea water and brines.

9
Mass/Volume Units
  • Example Concentration in Water
  • One liter of water is analyzed and found to
    contain 5.0 mg TCE. What is the TCE
    concentration in mg/L and ppmm?

To convert to ppmm, which is a mass/mass unit, it
is necessary to convert volume of water to mass
of water, by dividing by the density of water,
which is approximately 1,000 g/L.
10
Mass/Volume Units
  • Example Concentration in Air
  • What is the carbon monoxide (CO) concentration
    expressed in ?g/m3 of a 10-L gas mixture that
    contains 10-6 mole of CO?

Solution In this case the measured quantities
are presented in units of moles chemical/total
volume. To convert to mass chemical/total
volume, convert the moles of chemical to mass of
chemical by multiplying by COs molecular weight.
Note that the molecular weight of CO (28 g/mole)
is equal to 12 (atomic weight of C) plus 16
(atomic weight of O).
11
Mass/Volume Units
  • Example Concentration in Air

12
Volume/Volume Mole/Mole Units
  • Units of volume fraction or mole fraction are
    frequently used for gas concentrations. The most
    common volume fraction units are ppmv (parts per
    million by volume) which is defined as

Where Vi/Vtotal is the volume fraction and the
factor 106 is a conversion factor, with units of
106 ppmv/(volume fraction).
13
Volume/Volume Mole/Mole Units
  • Other common units for gaseous chemicals are ppbv
    (parts per 109 by volume).
  • The advantage of volume/volume units is that
    gaseous concentrations reported in these units do
    not change as a gas is expanded or compressed.
  • Atmospheric concentrations expressed as
    mass/volume (e.g. ?g/m3) decrease as the gas
    expands, since the chemical mass remains constant
    but the volume increases.

14
Volume/Volume Mole/Mole Units
  • Using the Ideal Gas Law to Convert ppmv to ?g/m3.
  • The Ideal Gas Law is used to convert gaseous
    concentrations between mass/volume and
    volume/volume. The Gas Law states that the
    pressure (P) times the Volume occupied by the gas
    (V) is equal to the number of moles (n) times the
    gas constant (R) times the absolute temperature
    (T) in degrees Kelvin or Rankine. This is
    written as

15
Volume/Volume Mole/Mole Units
  • Using the Ideal Gas Law to Convert ppmv to ?g/m3.
    Here , R, the universal gas constant, may be
    expressed in many different sets of units. Some
    of the most common are

16
Volume/Volume Mole/Mole Units
  • Using the Ideal Gas Law to Convert ppmv to ?g/m3.
    Since R is expressed in several different units,
    always be careful of the units and cancel them
    out to ensure the use of the correct value of R.
  • The Gas Law states that the volume occupied by a
    given number of molecules of any gas is the same,
    no matter what the molecular weight or
    composition of the gas, as long as the pressure
    and temperature are constant. The Ideal Gas Law
    can be rearranged to show that the volume
    occupied by n moles of gas is equal to

17
Volume/Volume Mole/Mole Units
  • At standard conditions (P 1 atm, T 273.15 K),
    one mole of any pure gas will occupy a volume of
    22.4 L. This result can be derived by using the
    corresponding value of R (0.08205 Latm/moleK)
    and the above form of the Ideal Gas Law. At
    other temperatures and pressures, this volume
    varies as determined by the equation form.

18
Volume/Volume Mole/Mole Units
  • Example Gas Concentration in Volume Fraction.
  • A gas mixture contains 0.001 mole sulfur dioxide
    (SO2) and 0.999 mole of air. What is the SO2
    concentration, expressed in units of ppmv?

To solve, convert the number of moles of SO2 to
volume using the ideal gas law and the total
number of moles to volume. Then divide by the
two expressions.
19
Volume/Volume Mole/Mole Units
  • Example Gas Concentration in Volume Fraction.

Substitute these volume terms in the above
equation and solve for ppmv.
20
Volume/Volume Mole/Mole Units
  • The last example shows that the terms RT/P cancel
    out. This demonstrates an important point that
    is useful in calculating volume fraction or mole
    fraction concentrations. For gases, the volume
    ratios and mole ratios are equivalent. This is
    clear from the Gas Law because at constant
    temperature and pressure the volume occupied by a
    gas is proportional to the number of moles.
    Therefore ppmv can be expressed as

21
Volume/Volume Mole/Mole Units
  • Therefore, in any given problem, either units of
    volume or units of moles can be used to calculate
    ppmv. Being aware of this fact will save
    unnecessary conversions between moles and volume.
    The mole ratio (moles i/total moles) is
    sometimes referred to as the mole fraction X.
  • The following example shows how the Ideal Gas Law
    can be used to convert concentrations between
    ?g/m3 and ppmv.

22
Volume/Volume Mole/Mole Units
  • Example Convert Gas Concentration between ?g/m3
    and ppmv. The concentration of SO2 is measured
    in air to be 100 ppbv. What is this
    concentration in units of ?g/m3? Assume the
    temperature is 28oC and pressure is 1 atm.
    Remember that T(oK) is equal to T(oC) plus
    273.15.
  • Solution To accomplish this conversion, use the
    Gas Law to convert the volume of SO2 to moles of
    SO2, resulting in units of moles/L. This can be
    converted to ?g/m3 using the molecular weight of
    SO2 (MW 64). This method will be used to
    develop a general formula for converting between
    ppmv and ?g/m3.

23
Volume/Volume Mole/Mole Units
  • Example Convert Gas Concentration between ?g/m3
    and ppmv. First, use the definition of ppbv to
    obtain a volume ratio for SO2.

Now convert the volume of SO2 to units of mass.
First, convert the volume to a number of moles,
using a rearranged format of the Gas Law
(n/VP/RT) and the given temperature and
pressure
24
Volume/Volume Mole/Mole Units
  • Example Convert Gas Concentration between ?g/m3
    and ppmv.

25
Volume/Volume Mole/Mole Units
  • Example Convert Gas Concentration between ?g/m3
    and ppmv. Now convert the moles to mass units

The above example demonstrates that there is a
useful conversion for converting air
concentrations between units of ?g/m3 and ppmv
can be written as
MW molecular weight of chemical species, R
0.08205 Latm/moleK, T temperature (K), P
pressure (atm) and 1,000 conversion factor
(1000 L 1 m3).
26
Volume/Volume Mole/Mole Units
  • Note that for 0.0oC, RT has the value of 22.4
    Latm/mole, while at 20C RT has a value of 24.2
    Latm/mole.

27
Partial-Pressure Units
  • In the atmosphere, concentrations of chemicals in
    the gas and particulate phases may be determined
    separately. A substance will exist in the
    gas-phase if the atmospheric temperature is above
    the chemicals boiling (or sublimation) point or
    if its concentration is below the saturated vapor
    pressure of the chemical at a specified
    temperature. The major and minor gaseous
    constituents of the atmosphere all have boiling
    points well below atmospheric temperatures.
  • Concentrations of these species typically are
    expressed either as volume fractions (e.g., ,
    ppmv, or ppbv) or as partial pressures (units of
    atmospheres, atm). The following table
    summarizes the concentrations of the most
    abundant atmospheric constituents.

28
Partial-Pressure Units
  • Composition of the Atmosphere (dry basis)

29
Partial-Pressure Units
  • The total pressure exerted by a gaseous mixture
    may be considered as the sum of the partial
    pressures exerted by each component of the
    mixture. The partial pressure of each component
    is equal to the pressure that would be exerted if
    all the of the other components of the mixture
    were suddenly removed. Partial pressure is
    commonly written as Pi, where i refers to the
    particular gas being considered.
  • For example, the partial pressure of O2 in the
    atmosphere is PO2 is 0.21 atm.

30
Partial-Pressure Units
  • Remember, the Ideal Gas Law states that, at a
    given temperature and volume, pressure is
    directly proportional to the number of moles of
    gas present therefore, pressure fractions or
    partial pressures are identical to mole fractions
    (and volume fractions).
  • For this reason, partial pressure can be
    calculated as the product of the mole and volume
    fraction and the total pressure. For example,

31
Partial-Pressure Units
  • If the previous equation is rearranged, ppmv can
    be calculated from the partial pressures as

Consequently, volume, moles, and partial
pressures can be used to calculate ppmv.
32
Partial-Pressure Units
  • Example. Concentration as Partial Pressure.
  • The concentration of gas-phase polychlorinated
    biphenyls (PCBs) in the air above Lake Superior
    was measured to be 450 picograms per cubic meter
    (pg/m3). What is the partial pressure (in atm)
    of PCBs? Assume the temperature is 0C, the
    atmospheric pressure is 1 atm, and the average
    molecular weight of PCBs is 325.

33
Partial-Pressure Units
  • Example. Concentration as Partial Pressure.
    Solution
  • The partial pressure is defined as the mole or
    volume fraction times the total gas pressure.
    First, the number of moles of PCB in air can be
    calculated. Then the ideal gas law can be used
    to calculate that one mole of gas at 0C and 1 atm
    occupies 22.4 L.

Multiplying the mole fraction by the total
pressure (1 atm) yields the PCB partial pressure
of 3.1x10-14 atm.
34
Partial-Pressure Units
  • Example. Concentration as Partial Pressure
    Corrected For Moisture. What is the partial
    pressure (atm) of carbon dioxide (CO2) when the
    barometer reads 29.0 inches of Hg, the relative
    humidity is 80 , and the temperature is 70F?
    Use the table to obtain the concentration of CO2
    in dry air.
  • Solution The partial-pressure concentration
    units in the table are for dry air, so the
    partial pressure must first be corrected for the
    moisture present in the air.
  • From the table the partial pressure of CO2 in dry
    air is 350 ppmv. The partial pressure will be
    this volume fraction times the total pressure of
    dry air. The total pressure of dry air is the
    total atmospheric pressure (29.0 in Hg) minus the
    contribution of water vapor.

35
Partial-Pressure Units
  • Example. Concentration as Partial Pressure
    Corrected For Moisture. Solution The vapor
    pressure of water can be obtained from the
    physical properties of water and is 0.36 lb/in2
    at 70F. Thus the total pressure of dry air is

The partial pressure of CO2 would be
36
Mole/Volume Units
  • Units of moles per liter (molarity, M) are often
    used to report concentrations of compounds
    dissolved in water. Molarity is defined as the
    number of moles of compound per liter of
    solution. Concentrations expressed in these
    units read as molar. Example Concentration as
    Molarity
  • Convert the 5 ppm concentration of TCE to units
    of molarity. The molecular weight of TCE is
    131.4 g/mole.
  • Solution Remember, in water, ppmm is equivalent
    to mg/L, so the concentration of TCE is 5.0 mg/L.
    Conversion to molarity units requires only the
    molecular weight.

37
Mole/Volume Units
  • Often, concentrations below 1 M are expressed in
    units of millimoles per liter, or millimolar
  • (1mM 10-3 moles/L)
  • or
  • micromolar (1 ?M 10-6 moles/L).
  • Thus the TCE concentration can be expressed as
    0.038 mM or 38 ?M.

38
Normality Units
  • Normality (equivalents/L) is typically used in
    defining the chemistry of water, especially in
    instances where acid/base and oxidation/reduction
    reactions taking place. It is also used in
    determining the accuracy of a water analysis such
    as calculating the chemical dosages in water
    treatment.
  • Reporting concentration on an equivalent basis is
    useful if two chemical species reacting have the
    same strength on an equivalent basis, a 1-mL
    volume of reactant number 1 will react with a
    1-mL volume of reactant number 2. In acid/base
    chemistry the number of equivalents per mole of
    acid equals the number of moles of H the acid
    can potentially donate.

39
Normality Units
  • For example, HCl has one equivalent/mole, H2SO4
    has two equivalents/mole, and H3PO4 has three
    equivalents per mole.
  • Likewise, the number of equivalents per mole of a
    base equals the number of moles of H that will
    react with one mole of base. Thus NaOH has 1
    equivalents/mole, CaCO3 has two equivalents/mole,
    and PO4-3 has 3 equivalents/mole.

40
Normality Units
  • Example Calculation of Equivalent Weight
  • What is the equivalent weight of HCl, H2SO4,
    NaOH, CaCO3, and Aqueous CO2?
  • SolutionThe equivalent weight is found by
    dividing the molecular weight by the number of
    equivalents.

41
Normality Units
  • Example Calculation of Equivalent Weight
  • SolutionDetermining the equivalent weight of
    aqueous CO2 requires a bit of thinking and some
    new information. Aqueous CO2 is not an acid
    until it hydrates in water and forms carbonic
    acid (CO2H2O?H2CO3). So aqueous CO2 really has
    two eqv/mole. Thus the equivalent weight of
    aqueous CO2 is

42
Example - Equivalents for Mineral Analysis
  • Solution
  • Component mg/L Equiv. Mol. Wt.
    Eq.Wt. Meq/L
  • CO2 8.8 2 44.0
    22.0 0.40
  • Ca 40.0 2 40.0 20.0
    2.00
  • Mg 14.7 2 24.4 12.2
    1.21
  • Na 13.7 1 23.0
    23.0 0.60
  • Alk. 135.0 2 100.0 50.0
    2.70
  • Cl- 17.8 1 35.0
    35.0 0.51
  • SO4 29.0 2 96.0
    48.0 0.60
  • as CaCO3

43
Example - Mineral Analysis for Two Stage Lime
Soda Ash Softening
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