-
Lecture #5: Water Chemistry
(Chapter 9 in Faure, 1998) Activity vs. Concentration with
Minimal Mathematics Most textbooks define the concentration and
activity ("effective concentration") of a chemical species in a
water sample with the following equation: a = m where: a =
"activity" or "effective concentration" of the dissolved species m
= molar or molal concentration of the dissolved species = "activity
coefficient" But, what does this equation really mean on an atomic
level? Lets look at the significance of concentration and activity
by using KCl (sylvite) dissolved in water as an example. Potassium
chloride (KCl) is very soluble in water. In very dilute solutions,
abundant water molecules almost completely separate KCl into
distinctive K+ and Cl- ions (see Figure 1 on the next sheet,
Nebergall et al., 1976, p. 316). The positive partial charges (+)
on the hydrogens of the water molecules surround the Cl- and the
partial negative charges (-) on the oxygens of other water
molecules are attracted towards the K+. The presence of abundant
water molecules around the K+ and Cl- essentially makes the ions
independent of each other. The potassium concentration of the
solution refers to the total number of potassium atoms in the
solution or the mass of the potassium in the solution. Typically,
the concentration is given in milligrams of potassium per liter of
water (mg/L). Because a liter of water has an approximate mass of l
kilogram (compare Table 9.1, p. 114), we can also express mg/L
mg/kg, where: 1 mg/kg = 1 part per million (ppm). If we're more
interested in the number of potassium atoms in the solution rather
than the mass of the potassium in the solution, we can express the
concentration as moles of solute/liter of solvent = molar = M.
Solute = dissolved material. Solvent = the dissolving medium,
often water. A mole of a substance is the atomic weight of the
element or the formula weight of the compound expressed in grams.
For example, the atomic weight of mercury is 200.59. Therefore, 1
mole of pure liquid mercury is 200.59 grams. One mole contains
Avogadro's number of molecules or atoms, or 6.022 x 1023.
Therefore, one mole of
-
mercury contains 6.022 x 1023 atoms of mercury. The formula
weight of KCl is 39.0983 (K) + 35.453 (Cl) = 74.5513 g/mole. One
mole of solid KCl is 74.5513 grams and contains 6.022 x 1023
molecules of KCl and 6.022 x 1023 atoms of potassium and 6.022 x
1023 atoms of chloride. If the concentration of KCl in an aqueous
solution is very low, it is easier to use millimoles of KCl/liter
of water (mM/L). Recall that m = milli = 1/1000, micro = =
1/millionth. The activity of potassium in a solution refers to the
total number of independent K+ in the solution. Because the
potassiums are well dissolved and dissociated from the chlorides in
dilute KCl solutions, the K+ activity and potassium concentration
of a dilute (let's say 0.0001 M) KCl solution are essentially
equal. That is, essentially 100% of the potassiums are K+ that are
unbonded to Cl-. In more concentrated KCl solutions, there are less
water molecules to shield all of the K+ and Cl- from each other. In
concentrated solutions, the unshielded ions begin to attract each
other and form K+Cl- pairs or complexes (Figure 2, Nebergall et
al., 1976, p. 316). That is, some of the K+ are no longer "active"
or independent of the Cl-. Therefore, in a concentrated solution
(such as 1 M KCl), the concentration of the potassium is 1 M, but
the activity, the number of independent K+, may be much less. The
K+Cl- pairs and groups of associated K+Cl- won't affect the boiling
or freezing points of the solutions as much as if they were well
dissociated and active K+ and Cl-. Specifically, independent or
active K+ and Cl- attract the partial charges of water molecules
and hinder the water molecules from boiling off or arranging
themselves into ice. Therefore, K+ and Cl- lower the freezing point
of water and raise the boiling point. However, when K+ and Cl-
combine into K+Cl- pairs and groups in more concentrated KCl
solutions, the K+Cl- partially neutralize each others charges, less
water molecules are then attracted to the potassiums and chlorides,
and the water molecules are not as hindered from boiling off or
freezing into ice as they are in the presence of strongly charged
and independent K+ and Cl-. Although concentrated KCl solutions
with their K+Cl- pairs and groups still suppress freezing points
and enhance boiling points, the freezing and boiling points are not
as affected as they would be if the potassium and chloride ions
were entirely independent and active. Once an evaporating KCl
solution reaches 4.65 M or 347 g/L at 20oC, the solution becomes
saturated. That is, the water molecules can no longer separate the
high concentrations of potassiums and chlorides. The K+, Cl-, and
K+Cl- readily attract each other and begin to grow into KCl
crystals, which precipitate (Figure 3 on the attached sheet).
Concentration = what's there. Activity = how much of what's there
really affects the chemical and physical properties of the
solution. Activities are used with liquids and solids. Fugacities
describe the activities of gases.
-
We use activity coefficients, , to mathematically relate the
concentration and activity of a species in a solution. a = m again
where: a = activity m = molar or molal concentration = activity
coefficient In very dilute solutions, = 1 and a = m. Example: Let's
say that we have 0.100 moles of NaCl in 1.000 kilogram of pure
water (0.100 molal NaCl).
molal = number of moles of a substance dissolved in 1.000
kilogram of solvent (often water).
For simplicity, let's convert moles to grams: NaCl = 22.98977 g
Na /mole + 35.453 g Cl / mole = 58.443 g NaCl / mole. Therefore,
5.8443 g of NaCl are dissolved in 1 kg of water to form a 0.100
NaCl molal solution. From the literature, = 0.778 for a 0.100 molal
NaCl solution at 25oC. (Later, in Chapter 10, well discuss the
equations that are used to calculate [activity coefficients]).
Therefore: a = 0.100 x 0.778 = 0.0778 molal of Na+ independent of
Cl- and 0.0778 molal of Cl- independent of Na+. This means that
0.1000 - 0.0778 = 0.0222 molal of Na+ and Cl- have formed ion
groups with each other and are "inactive" or not part of the Na+
and Cl- activity. pH pH = "the negative log of the H+ activity in a
water sample." The pH values do not include hydrogens covalently
bonded with oxygen as H2O, but only include "H+." In reality,
however, H+ does not exist in water. Instead, the H+ attaches onto
the - of individual water molecules as H3O+ or, less commonly,
groups of
-
hydrogen bonded water molecules to form H5O2+, H7O3+, and larger
complexes. Scientists, being lazy and practical, prefer to write H+
as an abbreviation for H3O+,
+ +H5O2 , H7O3 , etc.
ketches of H3O+ and hydrogen-bonded H5O2+ and H7O3 :
as undissociated H2O. Only about 1 out of 1014 H2O molecules
issociates as follows:
H2O + H2O H3O + OH r using simplified abbreviations:
H2O H+ + OH-
f 1014 H2O molecules into H+ + OH at 25oC:
--- [H2O] 1014
here:
+], [OH-] = activities of H+ and OH-, respectively, in
water.
In water at 25oC, the + and OH- activities and concentrations
are approximately equal.
he preferred format for the equation is:
= constant
+S
Almost all water exists d
+ - O We can write the following mathematical equation to
represent the dissociation of 1 out
-o [H+] [OH-] 1 ------------- = --------- W [H (H+), (OH-) =
concentrations of H+ and OH-, respectively, in water. H T
Activities of the Products -------------------------------
-
Activities of the Reactants r:
- = KW = 10-14 [H2O]
iates) into H+ + OH , athematically [H+] x [OH-] = 10-14 and
[H2O] becomes 1.
so 14 H2O molecules
reaks up into H+ + OH-, then from the above equation, KW =
10-14.
xample Problem: What is the concentration of H+ and OH- in pure
water?
e know:
---- = KW = 10-14
2O]
ince [H2O] = 1:
+] [OH-] = KW = 10-14
pure water, the following reaction is balanced:
2O = H+ + OH
mounts (moles) of H+ and OH-, and therefore the activities f the
two species are equal:
+] = [OH-]
ecause [H+] = [OH-], we can use algebra and substitute [H+] for
[OH-]:
+] = [H+]2 = 10-14
[H+]2 = 10-14 +] = 10-7 = [OH-]
he activities are expressed in moles/liter (molar, M).
Therefore: [H+] = [OH-] = 10-7 M.
O [H+] [OH-] ------------ Because only 1 out of 1014 H2O
molecules breaks up (dissoc -m KW = a constant, an equilibrium
constant for the reaction: H2O H+ + OH-, KW is alcalled the
ion-product constant of water. Because only 1 out of 10b E W [H+]
[OH-] ---------[H S [H In
-H The reaction produces equal ao [H B [H+] [OH-] = [H+] [H
_____ ____
[H T
-
Because we're dealing with dilute solutions, activities
approximately equal oncentrations. Therefore: (H+) = (OH-) 10-7 M
at 25oC.
ater has activities oncentrations) of [H+] = [OH-] = 10-7 M and
a pH of 7.
OH is the -log of [OH-].
OH = 14 - pH.
solution with a pH of 7 has a pOH of 7, and a solution with a pH
of 4 has a OH of 10.
e can design the following table:
er /liter pH pOH KW at 25oC Comment
c pH = -log of [H+] = -log of 10-7 = 7. Theoretically, pure w(c
p p Therefore, a p W [H+], moles/lit
[OH-], moles
100 = 1 10-14 0 14 1 x 10-14 Strong acid 10-1 = 0.1 10-13 1 13 1
x 10-14 10-2 = 0.01 10-12 2 12 1 x 10-14 10-6 10-8 6 8 1 x 10-14
Weak acid 10-7 10-7 7 7 1 x 10-14 Neutral 10-8 10-6 8 6 1 x 10-14
Weak base 10-12 10-2 = 0.01 12 2 1 x 10-14 10-13 10-1 = 0.1 13 1 1
x 10-14 10-14 100 = 1 14 0 1 x 10-14 Strong base The pH values of
some common materials:
emon juice = 2
hite vinegar = 3
eer, coffee = 4
lean" rainwater (contains carbonic acid from contact with CO2 in
air) = 5.7
ilk = 6.8
gram of baking soda (NaHCO3) in 100 ml H2O = 8
eawater = 8.1 to 8.3
ilk of magnesia = 10.5
L W B "C M 1 S M
-
Household ammonia = 11 Clam shells are made of calcite or
aragonite (both are calcium carbonate, CaCO3). CaCO3 is fairly
soluble in acidic freshwater. Therefore, freshwater clams are
coveredwith organic layers, which protect their CaCO
marine clams do not require protective organic overings on the
outside of their shells.
hell Chemistry of Freshwater and Marine Clams:
ainwater Chemistry
3 shells from being dissolved in acidic freshwater. Seawater has
pH values of 8.1 to 8.3. CaCO3 does not appreciably dissolve in
alkaline (basic) seawater. Therefore, c S R Uncontaminated
rainwater and purified (distilled) laboratory water does not have a
pH7. Carbon dioxide from the air readily reacts with the water to
produce carbonic acid (H
of
O3), which gives distilled water and "clean" rainwater a
slightly acidic pH of about .7:
H2O + CO2 = H2CO3
arbonic acid readily decomposes into bicarbonate (HCO3 ) and
carbonate (CO32-):
2O = HCO + H3O
H2CO3 + 2H2O = CO32- + 2 H3O
bonic acid also reacts with calcite (CaCO3) in limestones and
produces caves! That :
H2CO3 + H2O = H3O + HCO3-
H3O+ + CaCO3 = Ca2+ + H2O + HCO3
ater, we'll demonstrate that the pH of "pure" rain water is
about 5.7.
2C5
-C
- + H2CO3 + H 3
+ Caris
+
- L
-
Rainwater chemistry is also affected by other sources:
reacts with water and oxygen in the atmosphere to produce
sulfuric acid rain:
2SO2 + 2H2O + O2 = 2H2SO4
and other nitrogen xides into the atmosphere, which produce
nitric acid rain:
4NO + 2H2O + 3O2 4HNO3 4NO2 + O2 + 2H2O 4HNO3
ow itrates into the atmosphere, all of which may be converted
into nitric acid.
y during storms, and send NaCl and ther salts as dusts into the
atmosphere.
ts, unicipal waste incinerators, ocean water,
aper mills, and chloralkali plants.
H Buffers
Coal-fired power plants release SO2 gases into the atmosphere.
The SO2
Cars, lightning, and coal-fired power plants release NO, NO2o
Organisms and fertilizers also release ammonia into the air and
winds may bln Ocean waves break on beaches, especiallo Mercury is
released into the atmosphere as Hg0 vapor by coal-fired power
planvolcanoes, crematoria, forest fires, mp
p
well
e H2CO3 and HCO3 , which are capable of
eutralizing (buffering) the acid or base:
o neutralize an acid (H+) that has been added to seawater:
H + HCO3 H CO3 o neutralize a base (OH-) that has been added to
seawater:
OH- + H2CO3 HCO3 + H2O
Buffering is the ability of a solution to resist a change in pH
when an acid or base is added. "Pure" distilled water is poorly
buffered. Just a little acid or base will greatly change the pH of
distilled water. For example, a small drop of nitric acid may cause
the pH of a distilled water sample to drop from 5.7 to below 2.0.
In contrast, seawater isbuffered. A small addition of acid or base
only leads to a small change in pH. For example, a small drop of
nitric acid to a sample of seawater may cause no noticeablchange in
the pH. The seawater contains -n T
+ - 2 T
-
-
Another example of a buffer (a weak acid and a salt of the
acid):
cetic acid (vinegar) + sodium acetate
n acid, such as HCl, e H is removed or neutralized by the
following buffering reaction:
H3CO2 + H+ CH3CO2H ch as NaOH, the OH- is removed or neutralized
by the following
uffering reaction:
H3CO2H + OH CH3CO2 + H2O H near 7.35.
ike seawater, blood neutralizes acids and bases with HCO3 and
H2CO3.
he buffering reactions in human blood are:
o neutralize an acid in human blood:
H + HCO3 H2CO3 o neutralize a base in human blood:
OH- + H2CO3 HCO3 + H2O e produced by mixing a weak base and a
salt of the weak base (such as
H3 and NH4+).
nearly constant pH values may be destroyed too much acid or base
is added to them.
easily
neutralize the acid. gain, seawater also contains abundant HCO3
that neutralizes acids.
eferences
a The solution contains both CH3CO2- and CH3CO2Ho. If we add
a
+th
- oC If we add a base, sub
o - -C Human blood is another good buffer and typically attempts
to maintain a p
-L T T
+ - T
- Buffers may also bN Obviously, the ability of buffers to
maintainif Buffering has profound effects on water pollution. For
example, acid rain mayacidify a poorly buffered lake and kill its
fish, since the lake contains no large concentrations of dissolved
anions to neutralize the acid. Acid-sensitive lakes are located in
Minnesota and New England, where granites, sandstones and other
water-insoluble silicate-rich rocks are present. In contrast, lakes
that contain limestones tend to be better buffered because the
calcite in the limestones can dissolve to
-A R
-
Nebergall, W.H.; F.C. Schmidt and H.F. Holtzclaw, Jr., 1976,
College Chemistry, 5th ed., D.C. Heath & Co., Lexington,
Massachusetts.
-
Lecture #6: More Water Chemistry
(Chapter 20 in Faure, 1998) Total Dissolved Solids (TDS) Total
dissolved solids (TDS) refer to the total amount of solids
dissolved in an aqueous solution, usually given in mg/L. Solids are
ASSUMED to be "dissolved" if they can pass through a 0.45 micron
(m) filter. Filtering removes any suspended materials from a water
sample. TDS is usually determined through one of the three
following methods:
Evaporate a known volume of a filtered water sample to dryness
and weigh the residue.
(+): easy and cheap to do.
(-): Some residues are hydrated, such as gypsum (CaSO42H2O). The
water in the residues throws off the results. Heating the residue
to drive off the water may also evaporate some of the desired
solids.
Measure the major components (Na, Ca, K, Si, Al, sulfate,
carbonate species, Mg, chloride, Fe, etc.) in the filtered water
and add up their concentrations.
(+): Usually fairly accurate.
(-): Costly and locally important ions may be overlooked in the
analyses, such as boron or fluoride.
Measure the conductivity of the solution and use an
equation:
The conductivity of the solution is easily measured with a
portable meter.
TDS AC C = conductivity in microsiemens or micromhos. A = a
constant taken from the literature, usually 0.55 to 0.75, depending
on SO42-/Cl- of the water sample.
-
(+): Simple method. The meter is easy to use.
(-): Errors associated with the value of A.
(-): The equation doesn't work with seawater and other high TDS
solutions.
Waters are typically classified into several "quality classes"
by their TDS values (p. 372): Fresh: 0 to 1000 mg/L TDS. Usually
suitable for drinking by humans. Brackish: 1000 to 10,000 mg/L TDS.
Unfit for human consumption, but lower TDS brackish waters may be
used for industrial purposes, crop irrigation or watering
livestock. Saline: 10,000 to 100,000 mg/L TDS. Seawater is about
35,000 mg/L. These waters are generally unfit for industrial or
agricultural uses. Brine: >100,000 mg/L TDS. Great Salt Lake,
Utah and some deep groundwaters. Used for underground disposal of
liquid wastes. Underground Injection of Liquid Wastes into the
Brines: Water Hardness Water hardness is determined by the amount
of Mg2+ and Ca2+ in the water: Hardness = 2.5(Ca2+) + 4.1(Mg2+) The
concentrations of Mg2+ and Ca2+ are in mg/L.
-
"Hard water": Hardness > 150 mg/L "Intermediate water":
Hardness of 60 - 150 mg/L "Soft water": Hardness < 60 mg/L Hard
water leaves calcium and magnesium carbonate deposits in water
heaters, boilers, shower heads, pipes and drains. In boilers and
water heaters, the scale acts as an insulator, which requires more
fuel to heat the water. Hard water keeps soaps from lathering and
produces bathtub rings. Hard water leaves scum on clothes rather
than cleaning them in washing machines. Therefore, more soap is
needed to clean yourself and your clothes with hard water. If thick
hard water deposits are present on the inner walls of a water
heater tank, the tanks must become very hot to boil the insulated
water. If the scale cracks, hot water may flow into the cracks and
become exposed to hot elemental iron in the boiler wall. Explosive
hydrogen gas may then be produced: 4 H2O + 3 Fe0 (hot) Fe3O4 + 4 H2
(BOOM!!) Note: Fe3O4 = (Fe3+)2Fe2+O4 Water Softening Water
"softening" refers to the removal of calcium from water with sodium
zeolites. In the softening process, sodium is released into the
water. Removal of calcium from water with sodium zeolites: 2
NaAlSi2O6nH2O + Ca2+ Ca(AlSi2O6)2nH2O + 2 Na+
The reaction is then reversed with concentrated NaCl solutions
and the zeolite is regenerated for reuse: Ca(AlSi2O6)2nH2O + 2 Na+
2 NaAlSi2O6nH2O + Ca2+ Dissolved Oxygen (DO) Oceans, lakes, ponds
and streams must contain enough dissolved oxygen (DO) to support
aquatic life. DO is usually measured with a meter and may be done
in the field. The DO concentration of a water sample is affected by
the life forms in the water, temperature, pressure, and water
chemistry, including the presence of pollutants. Organic matter,
FeS2, other sulfides, and many life forms consume DO. Typically,
fresh water at
-
5oC has a maximum DO concentration of about 12 mg/L (see the
attached sheet with Table 8-1 from Drever, 1997, p. 167). At 30oC,
the DO concentration drops to 7.5 mg/L. Biological Oxygen Demand
(BOD) and Chemical Oxygen Demand (COD) Some organic pollutants are
nutrients for organisms. Therefore, the release of sewage and other
organic pollutants into surface waters may cause dramatic increases
in the growth of bacteria, algae, and other organisms. The
organisms obtain energy from organic compounds by breaking down the
organics into simpler compounds and CO2. Overpopulating bacteria
and algae may lower DO concentrations and cause fish kills and
other ecological damage in lakes, ponds, and streams. Not every
organic compound is biodegradable and readily available as food for
organisms. As examples, tars, most plastics, and other organics
consisting of large and complex molecules tend to be
non-biodegradable. Organisms prefer to eat small organic molecules
and small molecules are also more susceptible to destruction from
oxygen or other chemical reactions. Biological (or biochemical)
oxygen demand (BOD) estimates the amount of biodegradable carbon in
a water sample, which is a potential food supply for bacteria and
other organisms. High BOD surface waters may cause organisms to
overpopulate and kill fish. Chemical oxygen demand (COD) measures
the amount of organic matter in a water sample that may decompose
through chemical oxidation. Both BOD and COD concentrations in mg/L
provide some indication of the health or level of pollution in a
ground or surface water. Biological (or Biochemical) Oxygen Demand
(BOD) While pH and dissolved oxygen may be measured in the field
with meters, BOD must be measured in an analytical laboratory. The
sample is diluted with oxygenated water. The sample is then stored
in the dark at 20oC and the dissolved oxygen concentration is
measured 5 days later. The amount of dissolved oxygen that is
consumed over the five days gives some idea of the amount of
readily biodegradable (reactive) organic contaminants that are
present in the water sample. That is, the bacteria and other
organisms in the water consume the organic matter and oxygen in the
water during the 5-day analysis. Example: A sewage treatment plant
is being constructed along a river. How does the amount of
biodegradable organic matter vary in the river before the
construction of the plant, during plant construction, and once the
plant is operating? Take BOD measurements downstream from the plant
construction site before, during and after the plant is built.
-
Chemical Oxygen Demand (COD) Chemical oxygen demand (COD)
analyses are generally more rigorous and faster than BODs. COD is a
measure of the amount of chemically destructible organic matter in
a water sample, which includes the biodegradable organic materials
detected by BOD analyses and organics that are more resistant to
biodegradation and not detected by BODs. Specifically, COD is a
measure of the amount of organic materials that will react with
strong oxidizing agents, either potassium permanganate (KMnO4) or
potassium dichromate (K2Cr2O7). (We'll discuss the oxidation of
organic matter later.) BOD is probably a better estimate of the
amount of carbon that is likely to react in nature, but COD
analyses don't require 5 days to analyze. You also have to subtract
out the carbonate (inorganic carbon) concentrations (CO32-, HCO3-,
H2CO30) from the CODresults. The amounts of CO
32-, HCO3-, H2CO30 in a water sample will vary with pH and
other conditions. Professional analytical chemists can usually
determine the concentrations of various carbonate species in water
samples. Total Organic Carbon (TOC) Total Organic Carbon (TOC)
refers to the entire concentration of all organic materials in an
unfiltered water sample, including: microorganisms, plant and
animal debris, suspended materials and dissolved materials. If the
sample is filtered at 0.45 microns (m) before the carbon analysis,
the results are called "dissolved" organic carbon (DOC). Filtering
at 0.45 m is arbitrary and was once considered the size boundary
between "dissolved" and "suspended" particles. That is, particles
smaller than 0.45 m were once considered "dissolved" and
potentially able to readily move through ground and surface waters.
Particles larger than 0.45 m were considered "suspended" and were
expected to settle out of surface waters in short periods of time
or be filtered out of ground water as the ground water passed
through the subsurface. As we will see later when we discuss
colloids, these assumptions are now known to be often incorrect and
may lead to profound errors in predicting the movement of
contaminants through natural waters. TOC and DOC analyses are
performed in analytical laboratories. The carbon in the dried
residue of a water sample is burned, the resulting CO2 is measured,
and the TOC or DOC is then calculated from the CO2 analysis after
subtracting out the inorganic carbon (carbonate). Rainwater,
seawater and most groundwater typically have about 0.5 mg/L of DOC.
DOC values of rivers and lakes usually are 2-10 mg/L (Drever, 1997,
p. 107). Waters from soils, marshes and bogs may contain more than
50 mg/L DOC. "Black waters" are brownish, high DOC waters that are
common in tropical wetlands.
-
How Good are Your Water Analyses? Did You Analyze All of the
Important Constituents? Possible Answers Using Water Ion Balances
If the chemical analyses of a water sample are reliable and
complete, the number of anion charges should equal the number of
cation charges. If the measured cation and anion charges are not
similar, then:
significant errors may be present in the analyses. For example,
if the sulfate (SO42-) analysis is too low, then the total measured
amount of cations may exceed the total measured amount of
anions.
some ions that are usually not in significant concentrations in
water samples may
be present in significant concentrations in the sample and are
being overlooked (such as: iron, aluminum, boron or fluoride).
The following equation is used in the ion balance analysis: E =
[( /z/mc - /z/ma) / ( /z/mc + /z/ma)] x 100 where: E = charge
balance error in %. For acceptable results, E < 5%. mc = moles
of each cation ma = moles of each anion /z/ = absolute value of the
charge of the ion Of course, it's possible that errors in the
cation analyses may cancel out errors in the anion analyses.
Therefore, a low E value does not guarantee the accuracy of the
chemical analyses of a water analysis. Furthermore, the sum of the
concentrations in mg/L of the anions, cations and dissolved neutral
species (such as H2CO3o and SiO2o) should equal the total dissolved
solids (TDS) of the sample in mg/L. For the charge balance equation
to work, we need to balance the charges on all of the ions in the
water sample. This means that we need to work with moles and not
mg/L (weight per volume) in the equation. We also need to account
for charge differences (z) among the different ions.
-
For example, a charge balance between Ca2+ and HCO3- in water:
Ca2+ HCO3- atomic mass units = 40.08 atomic mass units = 61.02 1
cation +2 charge 1 anion -1 charge
We need two HCO3- to charge balance every Ca2+. We're interested
in the charges of each ion and their numbers (moles), and not their
weights. Therefore, we'll work with moles (m) and charges (z), and
not milligrams.
Example: A water sample has the following concentrations:
Species Except pH, concentrations in ppm Na+ 120 K+ 15 Ca2+ 380
Mg2+ 22 Sr2+ 0.8 SO42- 1115 Cl- 15 HCO3- 150 SiO20 21 pH 7.4
Perform a charge balance. (Watch significant digits!!): SiO2o and
any other neutral species have z = 0, and are not involved in the
charge balance. They are ignored. The pH of the water = 7.4 or H+ =
1 x 10-7.4 M is insignificant and will be ignored in the charge
balance analysis. At pH 7, the OH- concentration is also
insignificant. parts per million = ppm = mg/kg
-
Cations Cation ppm =
mg/kg mg/kg 1000 mg/g = g/kg
atomic mass (g/mole)
= moles/kg (molal, m)
Absolute value of charge on ion, /z/
mcz
Na+ 120 0.120 22.99 0.00522 1 0.00522 K+ 15 0.015 39.10 0.00038
1 0.00038 Ca2+ 380 0.380 40.08 0.00948 2 0.01896 Mg2+ 22 0.022
24.31 0.000905 2 0.00181 Sr2+ 0.8 0.0008 87.62 0.000009 2 0.000018
Cation Sum
0.0264
Anions
Anion ppm = mg/kg
mg/kg 1000 mg/g = g/kg
mass (g/mole)
= moles/kg (molal, m)
Absolute value of charge on ion, /z/
maz
SO42- 1115 1.115 96.06 0.01161 2 0.02322 Cl- 15 0.015 35.45
0.00042 1 0.00042 HCO3- 150 0.150 61.02 0.00246 1 0.00246 Anion
Sum
0.0261
Significant digits: DO NOT round off the results until you get
the final answer!! E = [0.0264 - 0.0261)/(0.0264 + 0.0261)] x 100 E
= 0.0003/0.0525 x 100 E = 0.6%, which is very good. E < 5% is
acceptable. Chemical Compatibility in a Water Sample: A Few
Examples Some chemical species are incompatible with each other in
water. For example, if the sulfate concentration of a water sample
is much greater than about 1000 mg/L, the barium concentration
should be less than 1 mg/L. In the presence of sulfate, barium
precipitates as water-insoluble barium sulfate. Unless the pH is
low, barium concentrations should also be low in the presence of
more than 1000 mg/L of carbonate. Barium carbonate is also sparsely
soluble in unacidified water. Silver is insoluble in the presence
of chloride in water. These relationships may be used to detect
questionable analytical results. More examples will be given
later.
-
Brief Comments on Field and Laboratory Blanks and Standards
Field and laboratory blanks and standards are important for
verifying the quality of analytical data from laboratories. For
example, if you believe that your surface or ground water samples
contain about 100 mg/L of chloride and 100 mg/L of sodium, dissolve
ultrapure sodium hydroxide (NaOH) and hydrochloric acid (HCl) into
clean (distilled and deionized) laboratory water to produce a
standard that contains only 100 mg/L of chloride, 100 mg/L of
sodium and nothing else. Anomalously submit the standard to the
analytical laboratory along with your water samples. Also submit,
as "blanks," samples of distilled and deionized water and any acids
that are used to preserve the ground or surface water samples. For
the blanks and standards, use the same types of containers that
were used for your water samples. If the water and acids blanks and
their containers are properly clean and if the laboratory is using
appropriate procedures, nothing significant should be detected in
the blanks. The analytical laboratory should only detect sodium and
chloride in the standard and they should obtain values that are
within 5-10% of the known concentrations of sodium and chloride
(that is, in the case of a 100 mg/L chloride and sodium standard,
the lab should report between 90 to 110 mg/L of sodium and
chloride). If the values are off, find another laboratory. It is
always a good idea, if you can afford it, to submit a few samples
to a second laboratory to confirm the analytical results of the
primary laboratory. Furthermore, if your water samples contain more
than 5,000 mg/L TDS, it is important to find a laboratory that
routinely analyzes seawater and other high TDS samples. You may
have to ship your water samples off to New Orleans, New York or
other coastal cities to find an analytical laboratory that
routinely analyzes high TDS samples. There have been cases of
laboratories in the Lexington area finding toxic concentrations of
silver and other contaminants in water samples. In reality, the
metals were not in significant concentrations. High concentrations
of sodium, chloride, sulfate and other ions in water samples often
produce interferences in analytical equipment that appear as
falsely high concentrations of silver, cadmium, arsenic and other
toxic metals. For example, I've seen an arsenic-free 1000 mg/L
aluminum standard produce a false measurement of about "3 mg/L" of
arsenic. Many analytical chemists are not trained to spot and
correct for these interferences and simply report the imaginary
results as being real. Obviously, bad data can lead to erroneous
interpretations and, without proper confirmation, could lead to
unnecessary remediation (site clean up) efforts that waste a lot of
valuable time and money. In other words, DONT ALWAYS BELIEVE YOUR
DATA! More later... Piper and Stiff Diagrams "Pictures and Graphs
are Worth a Thousand Numbers." Water chemical data are better seen
and understood in Piper and Stiff diagrams rather than in long
tables and lists of numbers. Piper diagrams are discussed on p.
372, 374 in Faure (1998). Stiff diagrams are explained in Appendix
I of Drever, p. 409-412, which are attached to this handout.
-
Piper Diagrams Piper diagrams show the major anion and cation
chemistry of a water sample as a point on a trilinear diagram, as
shown in Figure 20.2 on p. 374. Before the concentrations of the
major cations and anions can be plotted on the Piper diagram, the
concentrations must be in the form of "equivalents of ion per
kilogram of water." Here are the conversion steps from ppm to
equivalents/kg, which closely resemble the steps used with the
charge balance equation: ppm = mg/kg mg/kg 1000 mg/g = g/kg g/kg
atomic mass units in g/mole = mole/kg x /z/ = equivalents/kg
Example: Convert 40.00 ppm Ca2+ to equivalents/kg: 40.00 ppm =
40.00 mg/kg 40.00 mg/kg 1000 mg/g = 0.0400 g/kg 0.0400 g/kg 40.08
g/mole = 9.980 x 10-4 moles/kg 9.980 x 10-4 moles/kg /+2/ = 1.996 x
10-3 equivalents/kg Let's say that we have a water sample with the
following concentrations: Ca2+ = 40.00 ppm, 1.996 x 10-3
equivalents/kg Mg2+ = 15.00 ppm, 1.234 x 10-3 equivalents/kg Na+ =
120 ppm, 5.22 x 10-3 equivalents/kg K+ = 20.0 ppm, 0.51 x 10-3
equivalents/kg Sum cations = 8.96 x 10-3 equivalents/kg Next
convert the concentrations of the cations to percentages of the
total equivalents/kg. The percentages of potassium and sodium are
combined: Ca2+ 22.3 % Mg2+ 13.7 % Na+ + K+ 64.0% ______________
Sum: 100.0 % They should sum to 100.0% and they do.
-
For the anions:
The alkalinity of the water sample (essentially CO32- + HCO3- in
most waters) was measured as 2.8 meq/kg.
2.8 meq/kg = milliequivalents/kg = 2.8 x 10-3 equivalents/kg
Sulfate = 234 ppm = 4.88 x 10-3 equivalents/kg Chloride = 45.0 ppm
= 1.27 x 10-3 equivalents/kg The anions' equivalents/kg are also
summed and converted to percentages: Alkalinity (CO32- + HCO3-)
31.3 % SO42- 54.5 % Cl- 14.2 % ___________ Sum anions: 100.0% The
percentages are then plotted on the Piper plot, as shown in Figure
1, Drever (1997, p. 410) on the attached sheets. On the lower left
side of the Piper plot is a triangle with Mg on the top, Ca on the
lower left and Na + K on the lower right. Each corner of the
triangle represents 100% of the element on the corner. In other
words, a water sample with 100% Mg and no Ca, Na, or K would plot
as a point on the very top of the triangle. Figure 1 on p. 410
shows the location of a point with 64.0% Na + K, 13.7 % Mg and 22.3
% Ca. As expected, the point is located nearest to the Na + K
corner. The anions are plotted on a triangle on the lower right
corner of the Piper plot (Figure 1, p. 410 on the attached sheet).
100% Alkalinity is located on the lower left corner of the
triangle, 100% sulfate on the top corner of the triangle and 100%
chloride on the lower right corner. Finally, a line is drawn from
the point in the cation triangle parallel to the triangular graph
lines and into the diamond in the center of the diagram (Figure 1,
p. 410). Another line is drawn from the anion point parallel to the
graph lines in the anion triangle and into the diamond. The point
in the diamond of the Piper plot, where the two lines cross,
represents the major anion and cation chemistry of the water
sample. Notice that the points represent relative concentrations
between the major ions rather than absolute concentrations.
Therefore, the chemistries of a 100,000 mg/L TDS brine and a 500
mg/L TDS fresh water sample could plot in the same location on the
diagram. Piper diagrams are very useful for plotting the
chemistries of groundwaters or natural waters from a given area and
comparing them. Also, the chemistry of a particular river, lake or
groundwater may be plotted on a Piper diagram over time to track
chemical
-
changes in the water because of increased pollution, droughts,
increased crop irrigation, urban development, etc. The data are
easier to interpret on Piper diagrams than in long lists of tables.
Stiff Diagrams A Stiff diagram is another convenient way of
plotting the major ion chemistry of a water sample (Drever, 1997,
p. 411-412 on the attached sheets). Stiff diagrams are polygons.
Stiff diagrams for particular rivers, lakes or well waters may be
easily and conveniently plotted on a map of an area. Typically,
only the Na + K, Mg, Ca, carbonate, sulfate, and chloride
concentrations are plotted on Stiff diagrams. The ions are usually
plotted in milliequivalents/liter. The larger the polygon is, the
higher the TDS of the water. Furthermore, if the major cations and
anions are balanced, the right side of the polygon should have
roughly the same area as the left side. If not, as with the ion
balance equation, the following problems may be present:
significant errors may be present in the analyses. For example,
if the sulfate analysis is too low, then the total area represented
by the cations may exceed the area of the polygon represented by
the anions.
some ions that are usually not in significant concentrations in
water samples may
be present in this sample and are being overlooked (such as:
iron, aluminum, boron or fluoride).
Figure 3 in Drever, 1997 (p. 412 on the attached sheets) shows
how Stiff diagrams are plotted. Notice the shape and size
differences of the high TDS (Figure 3b) and low TDS (Figure 3c)
calcium carbonate-rich waters. The cation and anion concentrations
in milliequivalents/liter of the water analysis on page 409 in
Drever (1997, see attachments) are:
Ca2+ 1.996 x 10-3 equivalents/kg 2.0 milliequivalents/liter Mg2+
1.234 x 10-3 equivalents/kg 1.2 milliequivalents/liter Na+ + K+
5.73 x 10-3 equivalents/kg 5.7 milliequivalents/liter Alkalinity
(CO32- + HCO3-) 2.8 x 10-3 equivalents/kg 2.8
milliequivalents/liter SO42- 4.88 x 10-3 equivalents/kg 4.9
milliequivalents/liter Cl- 1.27 x 10-3 equivalents/kg 1.3
milliequivalents/liter Because it's difficult to plot Stiff
diagrams to an accuracy of greater than 0.1 milliequivalents/liter,
only two significant digits are listed above.
-
Using a ruler and the guidelines in Figure 3, the Stiff diagram
may be easily plotted. A Stiff diagram for the water analysis
plotted on the Piper diagram in Figure 1, p. 410 is shown in Figure
3a, p. 412 (see attached sheets). Estimating Relative TDS values
and Charge Balances for Water Samples Using Stiff Diagrams on Maps.
Which water sample has the lowest TDS? Which water sample shows a
charge imbalance? References:
Drever, J.I., 1997, The Geochemistry of Natural Waters, 3 ed.,
Prentice Hall, Upper Saddle River, NJ 07458.
rd
-
Lecture #9: Systems and Chemical Equilibrium
(Chapters 9 - 11 in Faure, 1998) Systems A system is a part of
the Universe that is under consideration or study. A garbage
landfill is an example of a system. Lake Superior or the Atlantic
Ocean also may be identified as systems. Our home galaxy, the Milky
Way, may be considered as another system. The surroundings are the
rest of the Universe outside of the system. Systems may be
classified as open, closed, or isolated. An open system allows for
the exchange of matter and energy with its surroundings. For
example, the biosphere is an open system. A tree is another open
system. It takes energy from the sun, absorbs water with its roots,
and releases oxygen and water to soils and the atmosphere. Closed
systems exchange energy with their surroundings, but not matter.
Isolated systems do not exchange energy or matter with their
surroundings. Closed systems are rare in nature, but may exist
under laboratory conditions. Isolated systems are essentially
absent from nature and can only be approximated in the laboratory.
A garbage landfill is an attempt to develop a closed system.
Supposedly, the waste is indefinitely locked up in the landfill.
The landfill is designed to keep the wastes from contaminating
soils, sediments and water outside of the landfill (the immediate
surroundings). At the same time, landfill engineers usually dont
care if seasonal temperatures within the landfill vary somewhat
because of the exchange of heat energy between the landfill and its
surroundings. That is, the engineers have no interest or budget in
making the landfill an isolated system. Open, Closed and Isolated
Systems: matter, energy
Open System Surroundings energy only
Closed System Surroundings
[ Isolated System ] Surroundings
1
-
A Landfill as a Closed System (the Ideal) and as an Open System
(Reality): Systems may also be described as dynamic (changing) or
static (not changing) over time. Most systems on Earth are dynamic.
Trees grow and die. Continental elevations build up through
tectonic uplift and volcanism, and wear down through erosion. The
Moon, in contrast, is essentially static and only changes slightly
from the impacts of cosmic dust and occasional meteorites. Chemical
Equilibrium If we mix an aqueous solution of 1 M barium chloride
with another aqueous solution of 1 M sodium sulfate, the mixture
will immediately turn cloudy white. The cloudy precipitate is
barium sulfate. We can write an equation for the reaction.
Initially, the reaction will move to the right to produce the
barium sulfate precipitate: Ba2+ + 2 Cl- + 2 Na+ + SO42- + H2O
BaSO4 + H2O + 2 Cl- + 2 Na+ Since many of the chemical species in
the mixture don't readily react, we can rewrite the reaction to
only show the relevant reactants and products: Ba2+ + SO42- BaSO4
We can graph the progress of the reaction between 1 M Ba2+ and 1 M
SO42- (compare Figure 9.1, p. 111): 1 | | | | Molar | Conc. | | | 0
|________________________________ Time ------ SO42- concentration
***** Ba2+ concentration +++++ BaSO4 concentration
2
-
Sometime later, the reaction will reach equilibrium: Ba2+ +
SO42- = BaSO4 At chemical equilibrium, the concentrations appear
constant or parallel to each other on the above graph. At
equilibrium, both sides of the reaction occur at the same rate
(that is, the rate of Ba2+ + SO42- BaSO4 is the same as Ba2+ +
SO42- BaSO4). However,this DOES NOT mean that the species on each
side of the reaction are in equal concentrations. At equilibrium,
chemical analyses indicate that almost all of the barium and
sulfate are located in the BaSO
That
ts of
4. However, a small amount of the barium and the sulfate, only
about 1.37 mg/L of barium and about 0.96 mg/L of sulfate (1 x 10-5
M of each), will exist as Ba2+ and SO42-. At equilibrium, these
trace amounts of Ba2+ and SO42- are constantly reacting to form
small amounts of BaSO4 and small amounts of BaSO4 are constantly
dissolving to replenish the trace amounts of Ba2+ and SO42-. is, at
equilibrium, the formation of barium sulfate from the remaining
trace amounts of Ba2+ and SO42- (the forward reaction) is balanced
by the dissolution of small amounbarium sulfate to produce Ba2+ and
SO42- (the reverse reaction). Equilibrium Constants In Lecture #5,
when we defined pH, we derived an ion-product constant (KW) for
water, which is a type of equilibrium constant: H2O = H+ + OH-
Activities of the Products ------------------------------- =
equilibrium constant Activities of the Reactants [H+] [OH-]
------------- = KW = 10-14 [H2O] [H2O] = 1 Therefore: [H+] [OH-] =
10-14 Similarly, we can derive equilibrium constants for other
chemical reactions. That is, when a reaction, such as Ba2+ + SO42-
= BaSO4, reaches equilibrium, the ratio of the activities of the
products to the reactants yields a constant, an equilibrium
constant, Keq. [BaSO4] ----------------- = Keq [Ba2+] [SO42-]
3
-
The general format for the Keq of a reaction at equilibrium is:
aA + bB = cC + dD where: A and B = the reactants C and D = the
products a, b, c, and d = the number of moles of A, B, C and D,
respectively. [C]c [D]d ----------- = Keq [A]a [B]b where: [C] and
[D] = activities of the products [A] and [B] = activities of the
reactants Keq = equilibrium constant. Each chemical reaction has
its own Keq.
If [A], [B], [C], or [D] are undissolved solids or liquids in
solutions, their activities are considered to be 1 and they are
ignored in the equilibrium constant calculations. However if the
solids or liquids are dissolved neutral molecules (such as SiO20 or
H2CO30), then they are considered part of the calculations.
Similarly, undissolved gases are not considered in the Keq
calculations. However, dissolved gases (such as O20, Hg0, and
CO20), along with ions and other dissolved neutral molecules, are
considered in the Keq calculations. As examples:
Ba2+ + SO42- + H2O = BaSO4
H2O and BaSO4 are distinctively separate, undissolved and
electrically neutral compounds, therefore for this reaction: 1
------------- = Keq [Ba2+] [SO42-] With the following reaction,
notice the zero charge on the carbonic acid: H2CO30 = H2O + CO2
4
-
The zero charge symbol indicates that the carbonic acid exists
as dissolved molecules rather than as a distinct liquid. Therefore,
the carbonic acid molecules are part of the equilibrium constant
calculations. Also, CO2 gas may dissolve in the water along with
the carbonic acid molecules, H2CO30. Therefore, the dissolved
portion of the CO2 is also considered in the Keq calculations: Keq
= [CO20] / [H2CO30] Later, we'll show how to distinguish dissolved
from non-dissolved CO2 for the Keq calculations. So, if you see a
zero superscript on a neutral species (including if the species is
a dissolved gas), they are relevant to the equilibrium constant
calculations. However, water, undissolved gases and solid
precipitates are ignored in the calculations.
As mentioned earlier, concentrations are usually easier to
measure than activities. As long as the concentrations are low and
the temperatures of the solutions are around 25oC, we can use
concentrations instead of activities to calculate the equilibrium
constants. Like Faure (1998, p. 113), we'll use brackets [ ] to
denote activities and parentheses ( ) to denote concentrations in
equilibrium constant calculations. Note: some textbooks use the
opposite convention, that is, [ ] denote concentrations and ( ) to
denote activities. LeChateliers Principle The addition or removal
of reactants or products from a reaction readily disrupts the
equilibrium of the reaction. LeChatelier's Principle says that if
the concentrations of reactants or products, temperature, or
pressure are changed for a reaction in equilibrium, the equilibrium
will shift in a way to undo the effect of the change. Let's look at
the barium sulfate reaction again:
Ba2+ + SO42- = BaSO4 What would happen if we add more Ba2+ to
the mixture? The excess barium would react with much of the 0.96
mg/L of SO42- to produce more barium sulfate. In other words,
LeChatelier's Principle states that the reaction would move to the
right and the 0.96 mg/L of SO42- would decrease to even lower
concentrations. What would happen if we were to filter out (remove)
the barium sulfate from the mixture? The small amounts of Ba2+ and
SO42- would tend to react and precipitate new barium sulfate.
However, with the solid barium sulfate essentially removed, there
is no abundant barium sulfate to dissolve and replace the small
amounts of Ba2+ and SO42-. Therefore, according to LeChatelier's
Principle, the removal of barium sulfate from the mixture through
filtering would cause the reaction to move towards the right.
5
-
The effects of LeChatelier's Principle may be seen when
sulfate-rich groundwater reacts with spilled water-soluble barium
nitrate from a fireworks factory. The presence of SO42- >>
Ba2+ in the mixing of barium nitrate and sulfate-rich groundwater
will cause the reaction to move to the right according to
LeChatelier's Principle: Ba2+ + excess SO42- BaSO4 Because of the
excess sulfate from the groundwater, the concentration of the
remaining Ba2+may be much lower than 1.37 mg/L and not a serious
environmental problem. On the other hand, in rare circumstances, if
Ba2+ >> SO42- + CO32- + HCO3-, not all of the Ba2+ will
precipitate as BaCO3 or BaSO4, and the Ba2+ may be a groundwater
contamination problem. LeChateliers Principle: Barium Nitrate
Pollution from Fireworks Factories into Low and High Sulfate
Groundwaters: LeChatelier's Principle is also affected by the
addition or removal of heat. Let's look at the following
reaction:
CaO + H2O = Ca(OH)2 + Heat!! When lime (CaO) is placed in water,
it violently reacts with the water to produce Ca(OH)2 and a lot of
heat! Sometimes the reaction is so exothermic (heat-releasing) that
the water boils! Now, imagine what would happen if you were to add
considerably more heat to a container with reacting lime and water.
LeChatelier's Principle says that the reaction would move to the
left. The excess heat would cause Ca(OH)2 to dehydrate to CaO and
steam:
CaO + H2O (as steam) Ca(OH)2 + More heat It will take a lot of
heat to get the reaction to noticeably move to the left because
substantial amounts of Ca(OH)2 are not dehydrated until
temperatures reach about 580oC. On the other hand, the addition of
ice to the reaction would remove the produced heat and force the
reaction to the right.
6
-
The effect of temperature on LeChatelier's Principle may also be
seen with the following reaction between H2 and Cl2 gases to
produce HCl gas: H2 + Cl2 = 2 HCl + Heat At higher temperatures,
the reaction goes to the left and HCl decomposes to H2 and Cl2. If
the heat is removed by lowering the temperature, the reaction moves
to the right and more of the H2 and Cl2 gases react to produce HCl.
Solubility Product Constant A solubility product constant (Ksp) is
a special type of equilibrium constant that is used to describe the
dissolution of a slightly soluble compound in water and calculate
the activities or concentrations of the dissolution products in a
saturated solution. When studying the dissolution of a solid in
water with Ksp values, the dissolving solids are considered
reactants. As an example: BaSO4 = Ba2+ + SO42- [Ba2+] [SO42-] = Ksp
From Faure (1998, p. 132): Ksp = 10-10. Often Ksp is written as
pKsp = 10.0 or the negative log of Ksp. By knowing the Ksp, we can
calculate the solubility of Ba2+ and SO42- in a saturated solution
of BaSO4:
[Ba2+] [SO42-] = 10-10
Because BaSO4 dissolves into an equal number of moles of Ba2+
and SO42-: [Ba2+] = [SO42-]
Therefore, to find [Ba2+], we can substitute [Ba2+] for [SO42-]:
[Ba2+] [Ba2+] = 10-10 [Ba2+]2 = 10-10 _____ [Ba2+] = 10-10 = 10-5 M
= [SO42-] 10-5 moles/L SO42- x 96.06 g/mole SO42- = 0.0009606 g/L
SO42-
0.0009606 g/L SO42- = 0.96 mg/L SO42-
7
-
10-5 moles/L Ba2+ x 137.34 g/mole Ba2+ = 0.00137 g/L Ba2+
0.00137 g/L Ba2+ = 1.37 mg/L Ba2+ How many moles of BaSO4
dissolved to produce the 10-5 M SO42- and 10-5 M Ba2+? BaSO4 = Ba2+
+ SO42- Therefore 10-5 M of BaSO4 dissolves to produce 10-5 M SO42-
and the 10-5 M
2+Ba .
10-5 M of BaSO4 x 233.4 g/mole BaSO4 = 0.002334 g/L = 2.33 mg/L
BaSO4
As a check, we note that:
.37 mg/L Ba + 0.96 mg/L SO42- = 2.33 mg/L dissolving BaSO4
e
o proper treatment and disposal. Using e pKsp of Ag2SO4 from
Faure (1998, p. 132):
Ag2SO4 = 2 Ag+ + SO42-
[Ag+]2 [SO42-] = Ksp = 10-4.81 = 1.55 x 10-5
te. be doubled to
qual the silver activity. Note: this is hard to visualize!!
Therefore:
[Ag+] = 2 x [SO42-]
e unknowns, e activities of the silver and sulfate in the
saturated silver sulfate solution.
[Ag+]2 [SO42-] = 1.55 x 10-5
[Ag+] = 2 x [SO42-]
ebra, let's substitute the second equation into the first to
solve for the sulfate ctivity:
[2 x SO42-]2 [SO42-] = 1.55 x 10-5
2+ 1 A more complex example: Is it illegal to dump a saturated
solution of Ag2SO4 down the drain? Note: Federal regulations state
that it is illegal to dump silver solutions down thdrain if they
contain more than 5 mg/L of silver. Solutions with more than 5 mg/L
of silver are considered hazardous and must undergth Notice that
the dissolution of Ag2SO4 produces twice as many moles of silver as
sulfaWe have two unknowns: [Ag+] and [SO42-]. The sulfate activity
has toe We now have two equations for the two unknowns. We can now
solve for thth Using alga
8
-
2- 4 [SO4 ]2 [SO42-] = 1.55 x 10-5
4 [SO ]3 = 1.55 x 10-5
[SO42-]3 = 3.88 x 10-6
Take the cube root of both sides.
[SO42-] = 0.0157 M
[Ag+] = 2 x [SO42-]
[Ag+] = 2 x 0.0157 M = 0.0314 M = 0.0314 moles/L
Ag+ = 107.8682 g/mole
[Ag+] = 3.39 g/L = 3390 mg/L, It's illegal.
isequilibrium
2- 4
Since:
D
on in ome supersaturated with some
ecies. A reaction that is NOT at equilibrium has:
Keq ]a [B]b
l calculate an activity product (AP) from the activities of the
reactants nd products:
= AP Keq ]a [B]b
the reaction involves ions, the activity product is an ion
activity product (IAP).
nd B (reactants) so that equilibrium may be eventually achieved
aure, 1998, p. 131):
aA + bB cC + dD
Some reactions are so slow that they may take years to reach
equilibrium. If a reactia solution is not at equilibrium, the
solution may becsp [C]c [D]d ----------- [A But we can stila [C]c
[D]d ----------- [A If If AP/Keq > 1, the activities or
concentrations of the products (C and D) are higher than expected
and LeChatelier's Principle dictates that the reaction should
proceed to the left to regenerate more A a(F
9
-
10
um roceed to the right to
roduce more products and eventually achieve equilibrium:
aA + bB cC + dD t equilibrium:
aA + bB = cC + dD
AP/Keq = 1 or AP = Keq
If AP/Keq < 1, A and B (the reactants) are more abundant than
expected at equilibriand LeChatelier's Principle dictates that the
reaction should pp
A
-
Lecture #12: Ionic Strength of Aqueous Solutions
(Chapter 10 in Faure, 1998) In previous lectures, we examined
the solubilities of substances in pure water at various
temperatures. However, pure water doesn't exist in nature and, for
environmental geochemistry, we need to examine the solubilities of
toxic substances in natural waters, which may have substantial TDS
values. Earlier, we described how excess SO42-, CO32-, or Ba2+ may
influence the solubility of BaSO4 in water via LeChateliers
Principle or through the precipitation of BaCO3. Na+, Cl-, Mg2+,
and other ions that normally don't react with Ba2+ or SO42- in
water may actually increase the solubility of BaSO4, if they are in
great abundance. As mentioned earlier, in dilute solutions, water
molecules may surround the Ba2+ or SO42- and hinder them from
reacting. As Ba2+ and SO42- concentrations substantially increase
in pure water, the Ba2+ and SO42- begin to find each other and
precipitate as BaSO4. However, the presence of abundant Na+ and Cl-
may increase the solubility of BaSO4 by more effectively shielding
the Ba2+ and SO42- from each other than water molecules.
Furthermore, +2 and -2 ions have greater abilities to shield Ba2+
and SO42- from each other than -1 or +1 ions. Ionic strength
describes the concentrations of ions in water. The attached figure
from Bird and Krauskopf (1995, p. 51) shows the solubility of BaSO4
in moles/liter (molarity, M) with increasing ionic strength. The
following equation is often used to calculate ionic strength from
the concentrations of the major cations and anions in a water
sample: I = 1/2 (mi (z2)) where: I = ionic strength mi = molal
concentrations of each ion (not mg/L) z = the charge of the ion
Unlike ionic strength, TDS (total dissolved solids) ignores the
charges on the ions and includes the concentrations of neutral
species, such as SiO20 and H2CO30. Ions in aqueous solutions are
electrolytes because the conduct electricity. The more ions in a
water sample, the greater the ability of the sample to conduct
electricity and the higher the ionic strength. Because the number
of ions in a water sample affects both its conductivity and ionic
strength, it's not surprising that equations may be derived to
relate ionic strength (in molality) and electrical conductivity
(Langmuir, 1997, p. 124):
-
Used with NaCl waters: I 0.8 x 10-5 Conductivity in mhos/cm Used
with Ca-Mg sulfate waters: I 1.7 x 10-5 Conductivity in mhos/cm
Used with calcium carbonate waters: I 1.9 x 10-5 Conductivity in
mhos/cm These equations are cheaper alternatives for estimating
ionic strength than measuring the major ions for $$$ and using I =
1/2 mi (z2). TDS and ionic strength are also closely related.
Langmuir (1997, p. 124) provides several equations that relate the
two parameters (TDS is in mg/L and I in molality): I 2.5 x 10-5 TDS
for an average water I 2 x 10-5 TDS for NaCl waters. I 2.8 x 10-5
TDS for calcium carbonate waters. Some typical ionic strengths (in
molal) for natural waters: Most rivers and freshwater lakes are
about 0.001 Seawater = 0.7 Oilfield brines = 5 Ionic strength is
often unitless in books, but if you work out the units, it should
be molality, m. As mentioned earlier, concentrations may be readily
measured in a laboratory, but activities are more difficult to
determine. The ionic strength of an aqueous solution may be used to
calculate the activity coefficients () for various chemical species
in a sample. The activities are then calculated with: activity of
the species = concentration of species (a = m). Two equations are
most often used to determine activity coefficients in low TDS
solutions. One is the Debye-Hckel equation (Faure, 1998 equation
10.46, p. 140), which works for solutions with I < 0.1 m: __ __
log i = (-Azi2I ) / (1 + BaI ) where:
log i = log of the activity coefficient for a particular ionic
species, such as Na+ or SO42-. zi = charge on the ion.
-
I = ionic strength. A, B = different variables related to
temperature and the dielectric constant of the solvent, see Table
10.3 in Faure (1998, p. 140) for some values. a = the effective
diameter of the ionic species in , see Table 10.4 in Faure (1998,
p. 141) for some values. There are some controversies over the
values of a for some chemical species. Drever (1997, p. 28) uses
more recent references than Faure (1998, Table 10.4, p. 141) and
lists the following values for a:
Mg2+ = 5.5 Na+ = 4.0 K+, Cl- = 3.5 Ca2+, Sr2+, Ba2+, SO42+ = 5.0
HCO3-, CO32- = 5.4 NH4+ = 2.5 Fe2+, Mn2+, Li+ = 6.0 H+, Al3+, Fe3+
= 9.0
The second equation, the Davies equation, is effective in
estimating activity coefficients in solutions with I < 0.5 m.
There are different forms of the Davies equation. The following
form from Drever (1997, p. 28) appears to be more accurate than
equation 10.47 from Faure (1998, p. 141): __ __ log i = [(-Azi2 I )
/ (1 + BaI )] + bI where: b usually equals 0.3. Some other values
for b from Drever (1997, p. 28) are: Ca2+ = 0.165 Mg2+ = 0.20 Na+ =
0.075
-
K+, Cl- = 0.015 SO42+ = -0.04 HCO3-, CO32- = 0.0 The other
variables are the same as in the Debye-Hckel equation.
Unfortunately, the Debye-Hckel and Davies equations do not work
with seawater and brines. As discussed on page 141 in Faure (1998)
and his references, there are methods for estimating activity
coefficients in seawater and brines. However, they involve complex
mathematics and the equations require values for the different
chemical species, which are often unknown. Some examples: Calculate
the ionic strength of an aqueous solution with 0.04 M NaCl and 0.01
M MgSO4. This solution would have the following concentrations: Na+
= 0.04 moles/liter Mg2+ = 0.01 moles/liter Cl- = 0.04 moles/liter
SO42- = 0.01 moles/liter Assume M = m. I = 1/2 mi (z2) I = 1/2
[0.04 (1)2 + 0.04 (-1)2 + 0.01 (2)2 + 0.01 (-2)2] I = 1/2 [0.04 +
0.04 + 0.04 + 0.04] = 0.08 Whats the of Mg2+ for the solution at
25oC using the Davies equation? Using the equation from Drever
(1997): __ __ log i = [(-Az2 I ) / (1 + BaI )] + bI
-
I = 0.08 A at 25oC from Table 10.3 (Faure, 1998, p. 140): 0.5085
z for Mg2+ = +2 b for Mg2+ from Drever (1997, p. 28) = 0.20 B at
25oC from Table 10.3 (Faure, 1998, p. 140): 0.3281 a for Mg2+ from
Table 10.4 (Faure, 1998, p. 141): 8; Drever (1997, p. 28) says 5.5
Using Drever's value for a:
log i = -0.36
i = 0.43 Since a = m, the activity of Mg2+ is 43% of the
concentration. Using a = 8 from Faure (1998, p. 141), i = 0.48.
References: Drever, J.I., 1997, The Geochemistry of Natural Waters,
3 ed., Prentice Hall, Upper Saddle River, NJ 07458.
rd
Krauskopf, K.B. and D.K. Bird, 1995, Introduction to
Geochemistry, 3rd ed., McGraw-Hill, Boston. Langmuir, D., 1997,
Aqueous Environmental Geochemistry, Prentice Hall, Upper Saddle
River, NJ.
530Lec5Sketches of H3O+ and hydrogen-bonded H5O2+ and H7O3+:
530Lec6Brief Comments on Field and Laboratory Blanks and
Standards
530Lec9530Lec12