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Lecture #2Intro: Expressions of Concentrations and Natural
Abundance(Stumm & Morgan, Chapt.1 & 3.4 )
(Pg. 4-11; 97-105)(Pankow, Chapt. 2.8)
David Reckhow CEE 680 #2 1
(Benjamin, 1.2-1.5)
Updated: 25 January 2019 Print version
Natural Water Environment
Chemistry of: Water column Sediments Soil Groundwater Atmosphere
David Reckhow CEE 680 #2 2S&M: Fig. 1.1; Pg. 2
Model Complexity: Phases Single Phase to multi-phase
David Reckhow CEE 680 #2 3
AqueousSolution
AqueousSolution
SolidPhase
GasAqueousSolution
SolidPhase
AqueousSolution
AqueousSolution
Gas
Solidα
Solidβ
Solidγ
AqueousSolution
Solidα
Solidβ
Solidγ
Gas
Based on S&M: Fig. 1.2 Pg. 4
Abundances?
s
David Reckhow CEE 680 #2 4
Elemental abundance in crust O Si Al Fe Ca Na Mg K Ti H P Mn F
David Reckhow CEE 680 #2 5
Elemental abundance in fresh water
David Reckhow CEE 680 #2 6
From: Stumm & Morgan, 1996; Benjamin, 2002; fig 1.1
Rock-WQ Connection
David Reckhow CEE 680 #2 7
Water Solutes reflect rock mineralogy; e.g.
Limestone CaCO3, mostly
Dolomite CaMg(CO3)2,
mostly Gypsum
CaSO4
“Stiff diagram”From Hounslow, 1995Water Quality Data; Analysis and Interpretation
Review Units
Mass based Molarity Molality Normality Mole fraction Atmospheres
Chemical Stoichiometry mass balance balancing equations
Thermodynamics law of mass action types of equilibria
David Reckhow CEE 680 #2 8
SI Unit prefixesFactor Prefix Symbol10-1 deci d10-2 centi c10-3 milli m10-6 micro µ10-9 nano n10-12 pico p10-15 femto f10-18 atto a
David Reckhow CEE 680 #2 9
Factor Prefix Symbol101 deka da102 hecto d103 kilo k106 mega M109 giga G1012 tera T1015 peta P1018 exa E
Mass Based Concentration Units Solid samples
David Reckhow CEE 680 #2 10
soilin Pb ppm 5.17soil 10
Pb 5.17soil g 1x10
Pb g 105.17soil 1
Pb 5.17
m
63
3
=
==−
ggx
kgmg
m
m
ppbkggppmkgmg
1/11/1
==
µ
Liquid samples
David Reckhow CEE 680 #2 11
in water Fe ppm 35.0 water10
Fe 35.0 waterg 10
Fe g 1035.0 waterg 10
Fe 35.0 waterg 10
water1 water1
Fe 35.0
m
63
3
3
3
=
===−
ggxmg
LxL
mg
Density of Water at 5ºC
David Reckhow CEE 680 #2 12
Mass/Volume Units Mass/Mass Units Typical Applications g/L (grams/liter) (parts per thousand) Stock solutions mg/L (milligrams/liter)
10-3g/L ppm (parts per million) Conventional pollutants
(DO, nitrate, chloride) µg/L (micrograms/liter)
10-6g/L ppb (parts per billion) Trihalomethanes, Phenols.
ng/L (nanograms/liter) 10-9g/L
ppt (parts per trillion) PCBs, Dioxins
pg/L (picograms/liter) 10-12g/L
Pheromones
Gas phase concentration Gas samples (compressible)
Could be converted to a ppmm basis But this would change as we compress the air
sample Could also be converted to a ppmv basis Independent of degree of compression But now we need to convert mass of ozone to
volume of ozone
David Reckhow CEE 680 #2 13
air 1Ozone 056.0
3mmg
Ideal Gas Law An ideal gas
Will occupy a certain fixed volume as determined by:
regardless of the nature of the gas Where:
P=pressure V=volume n=number of moles T=temp R=universal gas constant=0.08205 L-atm/mole-ºK GFW=gram formula weight
David Reckhow CEE 680 #2 14
PRT
GFWgmass
PRTnV
nRTPV)(
==
=
=22.4 Lat 1 atm, 273.15ºK
GFWgmassn )(
=
By definition:
Convert mass to moles Now we know that ozone’s formula is O3
Which means it contains 3 oxygen atoms Therefore the GFW = 3x atomic weight of oxygen in
grams or 48 g/mole
n=mass(g)/GFW n=0.056x10-3g/(48 g/mole) n=0.00117x10-3 moles
David Reckhow CEE 680 #2 15
air 1Ozone 056.0
3mmg
air 1Ozone 1000117.0
3
3
mmolesx −
Now determine ppmv
David Reckhow CEE 680 #2 16
airin O ppb 26airin O ppm 026.0
air L 1x10Ozone L 10026.0
air 1/4.22 1000117.0
3v
3v
3
3
3
3
==
=
=
=
−
−
xm
moleLxmolesxvolume
volumeppmair
ozonev
Mole & volume fractions Based on the ideal gas law:
The volume fraction (ratio of a component gas volume to the total volume) is the same as the mole fraction of that component
Therefore:
And since the fraction of the total is one-millionth of the number of ppm:
David Reckhow CEE 680 #2 17
total
i
total
i
nn
VV
=
PRTnV
PRTnV
totaltotal
ii
=
=
total
i
total
iv n
nVVppm =≡−610
Defined as:mole fraction
Defined as:Volume fraction
Partial pressures Based on the ideal gas law:
And defining the partial pressure (Pi) as the pressure a component gas (i) would exert if all of the other component gases were removed.
We can write:
Which leads to:
And:
David Reckhow CEE 680 #2 18
RTVnP
nRTPV
=
=
RTV
nPtotal
ii = RT
VnP
total
totaltotal =
total
i
total
i
nn
PP
=
vtotaltotal
itotali ppmP
nnPP 610−==
and
Earth’s Atmosphere
David Reckhow CEE 680 #2 19
Divide by 100 and you get the partial pressure for a total pressure of 1 atm.
Molarity One mole of any substance contains 6.02 x 1023 (Avogadro’s
number) elementary chemical units (e.g., molecules). It is very convenient to measure concentrations in moles, since
reactions conform to the law of definite proportions where integer ratios of reactants are consumed (e.g., 1:1, 1:2, etc.) on both a molecular and molar basis.
It is calculated by:
Often use M, mM, µM (molar, millimolar, micromolar) To represent: moles/L, 10-3 moles/L, 10-6 moles/L
David Reckhow CEE 680 #2 20
GFWL
massMolarity =
Normality Like molarity, but takes into account the stoichiometric
ratios of reactants and products Measured in equivalents per liter
And Z is an integer related to the number of exchangeable hydrogen ions, or electrons the chemical has, or its overall charge
David Reckhow CEE 680 #2 21
GEWL
massNormality =
ZGFWGEW =
“Complete” water analysisSpecies mg/L meq/LBicarbonate 153 2.5Chloride 53 1.5Sulfate 19.2 0.4Calcium 44 2.2Magnesium 10.9 0.9Sodium 25.3 1.1Potassium 7.8 0.2
David Reckhow CEE 680 #2 22
Anion-Cation Balance
David Reckhow CEE 680 #2 23
0 1 2 3 4 5
Cations
Anions
Conc. (mequiv./L)
HCO3- Cl-
SO4-2
Ca+2 Mg+2
K+
Na+
Total Hardness
Carbonate Hardness
Non-carbonate Hardness
Common Constituents N, P, and S containing compounds are often
expressed in terms of their elemental concentration Examples
66 mg of (NH4)2SO4 added to 1 L of water 85 mg of NaNO3 added to 1 L of water
David Reckhow CEE 680 #2 24
Example: element/group conc. Consider a solution of Ammonium Sulfate
prepared by dissolving 66 g of the anhydrous compound in water and diluting to 1 liter. What is the concentration of this solution in: a) g/L? b) moles/L? c) equivalents/L? d) g/L as sulfate? e) g/L as N?
David Reckhow CEE 680 #2 25
Example (cont.) a) 66 g/L b) The gram formula weight of ammonium sulfate is 132 g/mole. So,
using equation 2.7, on gets: Molarity = (66 g/L)/(132 g/mole) = 0.5 moles/L or 0.5 M.
c) Without any specific information regarding the use of this solution,one might simply presume that either the sulfate group or the ammoniumgroup will be the reacting species. In either case, Z should be equal totwo (product of the oxidation state times the number of groups). So: Normality = 0.5 moles/L * 2 equivalents/mole = 1 equivalent/L or 1.0 N or N/1.
David Reckhow CEE 680 #2 26
Example (cont.) d) The GFW for sulfate is:
GFW = 32 + 4*16 = 96.
The molarity of sulfate is: Molarity = 0.5 moles-(NH4)2SO4/L * 1 mole-
SO4/mole-(NH4)2SO4
= 0.5 moles-SO4/L
Then, one :gets mass/L = Molarity * GFW
= 0.5 moles-SO4/L *96 g-SO4/mole-SO4
= 48 g-SO4/L
David Reckhow CEE 680 #2 27
Example (cont.) e) The GFW for nitrogen is simply 14:
The molarity of nitrogen is: Molarity = 0.5 moles-(NH4)2SO4/L *
2 moles-N/mole-(NH4)2SO4
= 1 mole-N/L
Again, one gets: mass/L = Molarity * GFW = 1 mole-N/L
* 14 g-N/mole-N = 14 g-N/L or 14 g NH3-N/L
David Reckhow CEE 680 #2 28
Calcium carbonate units Used for major ion concentrations in drinking
waters Alkalinity Hardness
Since CaCO3 is divalent (Z=2) and its GFW is 100 g, its GEW is 50 g 50 g/equivalent or 50 mg/meq 50,000 mg/equivalent
David Reckhow CEE 680 #2 29
To next lecture
David Reckhow CEE 680 #2 30
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