CEE 680: Water Chemistry

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

http://www.ecs.umass.edu/cee/reckhow/courses/680/slides/680l02p.pdf

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


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


Elemental abundance in crust O Si Al Fe Ca Na Mg K Ti H P Mn F


Elemental abundance in fresh water


From: Stumm & Morgan, 1996; Benjamin, 2002; fig 1.1

Rock-WQ Connection


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


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


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


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


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


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


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


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


air 1Ozone 056.0

3mmg

air 1Ozone 1000117.0

3

3

mmolesx −

Now determine ppmv


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:


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:


RTVnP

nRTPV

=

=

RTV

nPtotal

ii = RT

VnP

total

totaltotal =

total

i

total

i

nn

PP

=

vtotaltotal

itotali ppmP

nnPP 610−==

and

Earth’s Atmosphere


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


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


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


Anion-Cation Balance


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


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?


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.


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


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


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


To next lecture


http://www.ecs.umass.edu/cee/reckhow/courses/680/slides/680l03.pdf

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CEE 680: Water Chemistry