Download Environmental Chemistry

Environmental Chemistry
Chapter 10:
The Chemistry of Natural Waters
Copyright © 2012 by DBS
Contents
•
•
•
Natural Waters
Oxidation and Reduction Chemistry in Natural Waters
Acid-Base Chemistry in Natural Waters
Natural Waters
The Blue Marble
0.001 %
water vapor
71 %
liquid water
The Blue Marble is a famous photograph of the Earth taken on 7 December 1972 by the crew
of the Apollo 17 spacecraft at a distance of about 29,000 km or about 18,000 miles. It is one of
the most widely distributed photographic images in existence. The image is one of the few to
show a fully lit Earth, as the astronauts had the Sun behind them when they took the image.
To the astronauts, Earth had the appearance of a child's glass marble (hence the name).
Natural Waters
Compartments
• Atmosphere
• Land
• Groundwater
• Rivers lakes
• Oceans
Water Cycle
Source: http://www.nasa.gov/vision/earth/environment/warm_wetworld.html
Where Does Potable
(fit for consumption) Drinking Water Come From?
Sources
Less than one
third of salt-free
water is liquid
Surface water: from lakes, rivers, reservoirs (< 0.01 % of total)
Ground water: pumped from wells drilled into underground aquifers (0.3 %)
Natural Waters
Uses of Water
World
• 70% irrigation
(agriculture)
• 20 % Industry
• 10% residences
World Resources 1998-99
The number of people living in countries facing severe or chronic water shortages is
projected to increase more than fourfold over the next 25 years. This will be from an
estimated 505 million people today to between 2.4 and 3.2 billion people by 2025.
Sources
< 1000 m3 per person per year
Engelman et al., 2000
Access to Water
Access to Water
Uneven distribution of water
Region
Total Renewable
Water Resources
(km3 yr-1)
Total Water
Withdrawals
(m3 yr-1)
Per Capita
(m3 person-1)
Average % of
Renewable
Resources
Average % Used
by Agriculture
Average %
Used by
Industry
World
43,249
3,414,000
650
-
71
20
Asia
11,321
1,516,247
1,028
29
79
10
Europe
6,590
367,449
503
9
25
48
518
303,977
754
423
80
5
4,850
512,440
1,720
14
27
58
Middle East/N.
Africa
N. America
Subject to contamination
Using water at a rate faster than it can be
supplied (>100 due to use of sea water)
Natural Waters
Role of Water in the Environment
• Water is an important constituent in our body and our survival
depends on natural waters
– transports substances into, within, and out of living organisms
– distributes soluble substances (e.g. pesticides, lead, mercury)
– reduces concentrations via dilution and dispersion
e.g. rainwater carries substances (e.g. acids) down to
earth’s surface, washes out (cleanses) the air but
pollutes waterways
Transport medium
volumes,
residence
times, Water Cycle
fluxes
Liquid Medium
Largest reservoir
– oceans
τ = 40,000 yr
Natural Waters
Representations of Water
Region of partial negative charge
Regions of partial positive charge
Polarity
Partial charges result
from bond polarization
A difference in the electronegativities
of the atoms in a bond creates a
polar bond
A polar covalent bond is a
covalent bond in which the
electrons are not equally shared,
but rather displaced toward the
more electronegative atom
Natural Waters
H-Bonding
Polarized bonds allow hydrogen
bonding to occur
•
•
•
A hydrogen bond is an
electrostatic attraction between an
atom bearing a partial positive
charge in one molecule and an
atom bearing a partial negative
charge in a neighboring molecule
The H atom must be bonded to an
O, N, or F atom
Hydrogen bonds typically are only
about one-tenth as strong as the
covalent bonds that connect atoms
together within molecules
H–bonds are intermolecular bonds
Covalent bonds are intramolecular bonds
Natural Waters
Unique Properties
•
•
•
•
•
•
•
•
Water shrinks on melting (ice floats
on water)
Unusually high melting point
Unusually high boiling point
Unusually high surface tension
Unusually high viscosity
Unusually high heat of vaporization
Unusually high specific heat
capacity
And more…
Natural Waters
Ice shrinks on melting as 15% H-bonds are lost
A certain mass of ice occupies more space than the same
mass of water
Natural Waters
Aqueous Chemistry
•
Organic content of water is dominated by redox chemistry (pE)
•
Inorganic content of water (dissolved ions) is controlled by acid-base and
solubility phenomena (pH-pK)
End
• Review
Redox Chemistry in Natural Waters
When the muddy sediments at the bottom of a river or lake are disturbed, two
things are noticed. First, there is a smell of rotten eggs, indicating the presence
of hydrogen sulfide. Secondly, the color of the sediment may change from a redbrown at the surface to a dark brown-black below. Both of these observations
are linked to a lack of oxygen.
The hydrogen sulfide will have come from the metabolism of sulfate by sulfatereducing bacteria, the dark coloration is due to the presence of metal sulfides in
the sediment which have been formed from this hydrogen sulfide
Redox Chemistry in Natural Waters
Most important oxidizing agent is dissolved O2 (atmospheric)
Acidic solution
O2 + 4H+ + 4e- → 2H2O
O2 is reduced from 0 to -2
state in H2O or OH−
Basic solution
O2 + 2H2O + 4e- → 4OH−
Concentration of O2 in water is low (10 ppm average), governed by Henry’s law:
O2(g) ⇌ O2(aq)
KH = [O2 (aq)]
PO2
At 25 °C, KH = 1.3 x10-3 mol L-1 atm-1
Dissolved O2 influences chemical
speciation of elements in natural
and polluted waters
Question
P9-1: Confirm by calculation the value of 8.7 mg L-1 for the solubility of oxygen in
water at 25 °C
At 25 °C, KH = 1.3 x10-3 mol L-1 atm-1
KH = [O2 (aq)] / PO2
[O2 (aq)] = KH x PO2
[O2 (aq)] = (1.3 x10-3 mol L-1 atm-1 ) x 0.21 atm = 2.7 x 10-4 mol L-1
[O2 (aq)]
= 2.7 x 10-4 mol L-1 x 32.00 g mol-1
= 8.7x 10-3 g L-1
= 8.7 mg L-1
= 8.7 ppm
Question
P9-2: Given the solubility cited above for O2 at O °C, calculate the value of KH at
this temperature
At O °C, [O2 (aq)] = 14.7 ppm
Since the equilibrium constant uses moles L-1 we must convert:
14.7 g O2 x 1 mol O2 x 1000 g H2O = 4.59 x 10-4 mols L-1
106 g H2O 32.0 g O2
1 L H2O
KH = [O2 (aq)] / PO2
KH = 4.59 x 10-4 mols L-1 / 0.21 = 2.2 x 10-3 mols L-1 atm-1
Depletion of DO
•
•
•
•
Temperature (inc)
Pressure (dec)
Salts (inc)
Organic matter (inc)
Dissolved O2 decreases with
increasing temperature
Redox Chemistry in Natural Waters
Oxygen Demand
•
The most common substance oxidized by DO in water is organic matter
(plant debris, dead animals etc.)
0 to -2
CH2O(aq) + O2(aq) → CO2(g) + H2O(aq)
0 to +4
•
•
•
Similarly DO is consumed by NH3 and NH4+ in the nitrification process
Water in streams and rivers are constantly replenished with oxygen
Stagnant water and deep lakes can have depleted oxygen
Redox Chemistry in Natural Waters
Oxygen Demand
Half reactions
Oxidation:
Reduction:
CH2O + H2O → CO2 + 4e- + 4H+
4H+ + O2 + 4e- → 2H2O
CH2O(aq) + O2(aq) → CO2(g) + H2O(aq)
In basic conditions?
O2 + 4H+ + 4e-  2 H2O
React with hydroxide
O2 + 4H+ + 4OH- + 4e-  2H2O + 4OHO2 + 4H2O + 4e-  2H2O + 4OHO2 + 2H2O + 4e-  4OH-
Same overall
Question
P9-4: Determine the balanced redox reaction for the oxidation of ammonia to nitrate
ion by O2 in alkaline solution (basic)
Does this reaction make the water more basic or less?
NH3 + O2  NO3- + H2O
Using standard redox balancing techniques:
NH3 + 2O2 + OH-  NO3- + 2H2O
The water becomes less basic since OH- is removed
Redox Chemistry in Natural Waters
Oxygen Demand
•
Humic acid is a fraction of the complex mixture known as humic substances
or Natural Organic Matter (NOM). Humic substances make up a large portion
of the dark matter in humus and are complex colloidal supramolecular
mixtures that have never been separated into pure components
•
•
•
Humic substances have been designated as either
humic acid, fulvic acid or humin.
Humic substances arise by the microbial degradation
of biomolecules (lipids, proteins, carbohydrates,
lignin) dispersed in the environment after the death of
living cells.
A modern structural description regards humic
material as a supramolecular structure of relatively
small bio-organic molecules
Tahquamenon Falls, Upper MI
Measures of amount of
organics/biological species in water
•
•
•
•
•
Biochemical Oxygen Demand (BOD)
Chemical Oxygen Demand (COD)
Total Organic Carbon (TOC)
Dissolved Organic Carbon (DOC)
(TOC)-(DOC) = Suspended carbon in water
Redox Chemistry in Natural Waters
Biological Oxygen Demand
•
•
The capacity of the organic and biological matter in a sample of natural water to
consume oxygen, a process usually catalyzed by bacteria, is called BOD
Procedure: measure O2 in the stream or lake. Take a sample and store at 25oC
for five days and remeasure O2 content. The difference is the BOD
– BOD5 corresponds to about 80% of the actual value. It is not practical to
measure the BOD for an infinite period of time
– Surface waters have a BOD of about 0.7 mg L-1 – significantly lower than
the solubility of O2 in water (8.7 mg L-1)
– Sewage has BOD of ~100 mg L-1
Redox Chemistry in Natural Waters
Chemical Oxygen Demand
O2 + 4H+ + 4e- → 2H2O
•
Dichromate ion, Cr2O72- dissolved in sulfuric acid is a powerful oxidizing agent. It
is used as an oxidant to determine COD
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7 H2O
•
Excess dichromate is added to achieve complete oxidation
Back titration with Fe2+ gives the desired endpoint value
# moles of O2 consumed = 6/4 x (#moles Cr2O7 consumed)
Note: Cr2O72- is a powerful oxidizing agent and can oxidize species that are
not usually oxidized by O2 - hence gives an upper limit
Question
P9-5: A 25 mL sample of river water was titrated with 0.0010 M Na2Cr2O7 and
required 8.7 mL to reach the endpoint. What is the COD (mg O2/L)?
No. moles Cr2O72- = 0.0010 mol L-1 x (8.7 x 10-3 L) = 8.7 x 10-6 mols
No. moles O2 = 1.5 moles Cr2O72- = 1.5 x (8.7 x 10-6 mols)
= 1.3 x 10-5 mols O2
1.3 x 10-5 mol x 32.00 g mol-1 = 4.2 x 10-4 g
0.42 mg / 0.025 L = 17 mg L-1
Redox Chemistry in Natural Waters
Decomposition of Organic Matter
•
In the absence of O2 bacteria present in water can induce chemical
transformations
2CH2O −bacteria  CH4 + CO2
disproportionation
1966-Michigan Swamp Gas Case
March 14-20, 1966: Southeastern Michigan. From about
3:50 a.m. on March 14 and for 2-1/2 hours thereafter,
Washtenaw County sheriffs and police in neighboring
jurisdictions reported disc-shaped objects moving at
fantastic speeds and making sharp turns, diving and
climbing, and hovering. At one point, four UFOs in
straight-line formation were observed.
Swamp Gas Mystery Solved
Bohannon, Science NOW 24 August 2005: 2
Redox Chemistry in Natural Waters
Aerobic vs. Anaerobic Conditions
Fe3+ in sediments gets converted into soluble Fe2+ under reductive
conditions
Fe3+ + eInsoluble Fe(III)
Fe2+
Soluble Fe(II)
epilimnion
Stratification:
In the same lake
aerobic and anaerobic
condition may exist
Hypolimnion – anoxic due to O2 consumption by
organic matter
Redox Chemistry in Natural Waters
The pE Scale
•
Oxidation and reduction are controlled by the concentrations of electrons which
are present:
pE = - log10[e-]
Low pE means electrons are available (reducing environment)
High pE means electrons are unavailable (oxidizing environment)
pE = E/0.0591
When a significant amount of O2 is dissolved, the reduction of O2 is the
dominant reaction determining e- availability:
¼ O2 + H+ + e- ⇌ ½ H2O
•
Under such circumstances, the pE of the water is related to its acidity and to the
partial pressure as follows:
pE = 20.75 + log([H+] PO2¼) OR pE = 20.75 – pH + ¼ log(PO2 )
Redox Chemistry in Natural Waters
The pE Scale
A convenient approach is to use Nernst Equation of electrochemistry
E = E0 – (RT/F) (log [products] / [reactants])
…for 1 electron redox process
E = E0 - 0.0591(log [products] / [reactants])
where E0 is the standard electrode potential for a one electron reduction
One can equate pE to the Electrode Potential E
pE = E/0.0591 or pE0= E0/0.0591
E/0.0591 = E0/0.0591- (log [products] / [reactants])
pE = pE0 - (log [products] / [reactants])
Redox Chemistry in Natural Waters
The pE Scale
¼ O2 + H+ + e- ⇌ ½ H2O
E0 = 1.23 V
pE0 = 1.23/0.0591
pE = pE0 - (log [products] / [reactants])
pE = 20.75 - log 1/ [reactants] = 20.75 + log ([reactants])
pE = 20.75 + log([H+] PO2¼) = = 20.75 + log([H+] + log (PO2¼)
pE = 20.75 – pH + ¼ log(PO2 )
pE = 20.75 – pH + ¼ log(PO2 )
For a neutral sample of water that is saturated with oxygen from air (PO2 = 0.21
atm) that is free from CO2 (pH = 7) the pE value corresponds to 13.9
…pE value decreases with decrease in O2 and increase in pH
Dominant redox equilibrium reaction determines pE of water
(O2 may not be dominant redox species!)
Question
What is the most oxidizing conditions possible in water?
PO2 cannot exceed 1, log(1) = 0
pE = 20.75 – pH
pE = 20.75 - pH
This can be drawn on a pE/pH
diagram as a boundary line,
When pE > 20.75 – pH water
will be oxidized
pE
A similar analysis gives boundary
below which water will be reduced
pE = - pH
pH
Redox Chemistry in Natural Waters
The pE Scale
Example
1/8NO3− + 5/4H+ + e- ⇌ 1/8NH4+ + 3/8H2O
E0 = +0.836 V
pE0 = E0/0.0591 = 0.836/0.0591 = +14.15
pE
=
pE0
– log
[NH4+]1/8
[NO3-]1/8[H+]5/4)
ax = x log a
log(1/b) = -log b
= 14.15 - 5/4pH -1/8log([NH4+]/[NO3-])
Note: Express the reactions as one electron reduction process
….. Follow the examples given on page 435
Question
9-7: Deduce the equilibrium ratio of concentrations of NH4+ to NO3- at a pH of 6.0
(a) for aerobic water having a pE = +11, and (b) for anaerobic water with pE = -3
pE = 14.15 – (5/4)pH – (1/8)log([NH4+] / [NO3-])
11 = 14.15 – (5/4) x 6 – (1/8)log([NH4+] / [NO3-])
log([NH4+] / [NO3-]) = -8(4.35) = -34.8
[NH4+] / [NO3-] = 1.6 x 10-35
pE = 14.15 – (5/4)pH – (1/8)log([NH4+] / [NO3-])
-3 = 14.15 – (5/4) x 6 – (1/8)log([NH4+] / [NO3-])
log([NH4+] / [NO3-]) = 8(9.65) = 77.2
[NH4+] / [NO3-] = 1.6 x 1077
Problem 9-7
pH = 6, pE = 11,
pH = 6, pE = -3,
Redox Chemistry in Natural Waters
The pE Scale
•
•
•
Ferrous iron (Fe2+) hydroxide is more soluble, than ferric (Fe3+) hydroxide.
Fe3+(aq) precipitates as the hydroxide under neutral or basic conditions
Fe2+ is readily oxidized to Fe3+ by dissolved oxygen
4Fe2+ + O2 + 2H2O
4[Fe3+ + 3OH4Fe2+ + O2 + 2H2O + 8OH-
4 Fe3+ + 4OH4Fe(OH)3(s)]
4Fe(OH)3(s)
Well water often deposits brown, rusty,
iron hydroxide when brought to the
surface and into contact with oxygen
Redox Chemistry in Natural Waters
The pE Scale
•
•
•
•
Oxidation in aquatic environment is controlled by MO’s
– biologically mediated
Low values of O2 found in reducing environments due to microbial catalyzed
fermentation of organic matter
In cases of low O2 solubility pE is determined by nitrate or sulfate
In the extreme case e- availability determined by methane and carbon dioxide
1/8CO2 + H+ + e- ⇌ 1/8CH4 + H2O
pE = 2.87 – pH + 1/8 log(PCO2/PCH4)
e.g. PCO2 = PCH4 and pH = 7, pE = -4.1
Highly oxygenated 0.81 V
(Oxic)
Chemistry (fate and transport) of
metals and organics changes
depending on E0
Redox Chemistry
Reactions and Equilibrium
Anaerobic -0.40 V
(Anoxic)
Redox Chemistry in Natural Waters
The pE Scale
Fe3+ + e- ⇌ Fe2+
•
•
For this reaction, pE0 = 13.2
pE = 13.2 + log([Fe3+] / [Fe2+])
NOT pH DEPENDENT!
e.g. Ratio of Fe3+ to Fe2+ when pE = -4.1 (reducing)
-4.1 = 13.2 + log([Fe3+] / [Fe2+])
log([Fe3+] / [Fe2+]) = -17.3
[Fe3+] / [Fe2+] = 5 x 10-18
•
Transition between dominance of one form over the other occurs at
[Fe3+] = [Fe2+], pE = 13.2 + log(1) = 13.2 + 0 = 13.2
Redox Chemistry in Natural Waters
pE – pH Stability Field Diagrams
Zone dominance of various oxidation states
pE independent
pH
independent
pE independent
•
Hydrated Fe2+/Fe3+ ions can only
exist under acidic conditions
•
At higher pH Fe3+ is present as
Fe(OH)3. Fe(OH)2 does not
precipitate until solution becomes
significantly basic
•
Changes in redox conditions govern
whether the iron will be in solution
or in the sediments
Redox Chemistry in Natural Waters
Sulfur Compounds
Highly reduced -2 state as found in H2S and DMS (CH3-S-CH3)
highly oxidized +6 state as encountered in SO42-
Similar to the oxidation in air (recall from acid rain chapter), H2S in water is
oxidized by O2 first to SO2 and then to H2SO4
H2S + 2O2 → H2SO4
Question
P9-10:
(a) Balance the reduction half-reaction that converts SO42- to H2S in acidic conditions
(b) Deduce the expression relating pE to pH, the concentration of sulfate ion, and the
partial pressure of hydrogen sulfide gas, given that for the half reaction, pE0 = +5.75
(c) Deduce the partial pressure of H2S when the sulfate ion concentration is 10-5 M and
the pH is 6.0 for water that is in equilibrium with atmospheric oxygen
(a) 10H++ SO42- + 8e-  H2S + 4H2O
(b) Reduce eqn. to 1 electron: 5/4H++ 1/8SO42- + e-  1/8H2S + 1/2H2O
pE
= pE0 – log(PH2S1/8 / [SO42-]1/8[H+]5/4)
= pE0 – -(5/4)(log[H+]) – (1/8)log(PH2S/[SO42-])
= pE0 – (5/4)pH – (1/8)log(PH2S/[SO42-])
= 5.75 – (5/4)pH – 1/8)log(PH2S/[SO42-])
Question
P9-10:
(c) Deduce the partial pressure of H2S when the sulfate ion concentration is 10-5 M and
the pH is 6.0 for water that is in equilibrium with atmospheric oxygen
By rearrangement:
-(1/8) log (PH2S/[SO42-]) = pE + pE0 + 5/4 pH
For any solution in equilibrium with atmospheric O2, we know that:
pE = 20.75 – pH + (1/4) log PO2
And because PO2 = 0.21 and pH = 6, here,
pE = 20.75 – 6 + (1/4) log (0.21) = +14.58
We substitute this value into the equation for pE of the H2S/SO42- system above to
obtain:
Question
P9-10:
(c) Deduce the partial pressure of H2S when the sulfate ion concentration is 10-5 M and
the pH is 6.0 for water that is in equilibrium with atmospheric oxygen
-(1/8) log (PH2S/[SO42-])
= 14.58 – pE0 + (5/4) x 6
=14.58 – 5.75 + 7.5 = 16.33
log (PH2S/[SO42-])
= -131
Since [SO42-] = 10-5, then
Since PH2S/[SO42-] = 10-131
PH2S = 10-131 x 10-5 = 10-136 atm
Thus the partial pressure of H2S under these conditions is absolutely negligible
Redox Chemistry in Natural Waters
Acid Mine Drainage
•
•
Mines (e.g., coal mines) that contain Ferrite (FeS2) often leach out iron in water
as a result of air oxidation
Under anaerobic condition FeS2 is insoluble in water, but when comes in contact
with air is oxidized from S22- (-1 oxidation state) to sulfate ion (+6):
2S22- + 7O2 + 2H2O → 4SO42- + 4H+
•
•
Since the sulfate of the Fe2+ is soluble in acidic water, the iron pyrite effectively
becomes solubilized
The reaction produces large amounts of acid, some of which is consumed in air
oxidation (catalyzed by MO’s) to form Fe3+
4Fe2+ + O2 + 4H+ → 4Fe3+ + 2H2O
4FeS2 + 15O2 + 2H2O → 4Fe3+ + 8SO42- + 4H+
i.e. 2Fe2(SO4)3 + 2H2SO4
Redox Chemistry in Natural Waters
Acid Mine Drainage
Overall reaction
4FeS2 + 15O2 + 2H2O → 4Fe3+ +8SO42- + 4H+
2Fe2(SO4)3 + 2H2SO4
In other words the insoluble FeS2 produces soluble iron(III) sulfate and sulfuric
acid
(Note: acidic conditions are required for the dissolution of Fe2(SO4)3)
•
•
If gets diluted with the natural water Fe2(SO4)3 gets deposited as Fe(OH)3
On the other hand, concentrated acid can also liberate toxic heavy metals from
the ores in the mines
Fe3+ can also serve as an oxidant for the oxidation of S22S22- + 14Fe3+ + 8H2O → 2SO42- +14Fe2+ + 16H+
The phenomenon of acid drainage is currently of importance in abandoned
mines
Redox Chemistry in Natural Waters
Nitrogen Compounds
Highly reduced -3 state as found in NH3 or NH4 and highly oxidized +5 state as
encountered in NO3-
The equilibrium between the highly reduced and oxidized state is given by
NH4+ + 3H2O ⇌ NO3- + 10H+ + 8e-
pE-pH Diagram
•
NH4+ + 3H2O ⇌ NO3- + 10H+ + 8e-
•
Oxidation is highly pH dependent,
disfavored in highly acidic environments
•
For the diagonal line when
[NH4+] = [NO3-]
pE
•
= 14.15 - 5/4pH -1/8log(1)
= 14.15 - 5/4pH
Nitrite ion (NO2-) dominates in relatively
small field
Sawyer et al., 1994
End
• Review
Chemical Aspects
pH
pH = - log [H+]
•
H+ usually surrounded by
water of hydration,
written H3O+
•
‘Master Variable’ –
controls parameters e.g.
speciation
•
Ranges 5.5 - 9
Acid-Base Chemistry in Natural Waters
The Carbonate System
•
The CO2-Carbonate System
CO2(g) + H2O(aq) ⇌ H2CO3(aq)
H2CO3(aq) ⇌ H+ + HCO3-
•
Calcium carbonate though insoluble,
leaches out small amounts of CO32-
CaCO3(s) ⇌ Ca2+ + CO32-
CO32- + H2O ⇌ HCO3- + OH-
Acid-Base Chemistry in Natural Waters
Water in Equilibrium with Solid Calcium Carbonate
•
•
Hypothetical body of water in equilibrium with excess CaCO3
Solubility product, Ksp = [Ca2+][CO32-] = 4.6 x 10-9 mols2 L-2
at 25 °C
From stoichiometry ,
Solubility of CaCO3, S = [Ca2+] = [CO32-] = 6.8 x 10-5 mol L-1
•
Assumes all other
reactions negligible
BUT! CO32- reacts as a base: CO32- + H2O ⇌ HCO3- + OH-
Kb = [HCO3-][OH-] = 2.1 x 10-4
[CO32-]
Since equilibrium lies to RHS with excess carbonate in natural waters (low OH-):
CaCO3(s) ⇌ Ca2+ + CO32CO32- + H2O ⇌ HCO3- + OHCaCO3(s) + H2O(aq) ⇌ Ca2+ + HCO3- + OH-
K = Kb x Ksp = 9.7 x 10-13
K = [Ca2+][HCO3-][OH-]
Acid-Base Chemistry in Natural Waters
Water in Equilibrium with Solid Calcium Carbonate
•
Hypothetical reaction:
CaCO3(s) + H2O(aq) ⇌ Ca2+ + HCO3- + OHIs the only process of relevance.
K = Kb x Ksp = 9.7 x 10-13 = [Ca2+][HCO3-][OH-]
From stoichiometry, S = [Ca2+] = [HCO3-] = [OH-]
S3 = 9.7 x 10-13
S = 9.9 x 10-5 mol L-1
>>>> 1 component model
Since the carbonate ion reacts with water molecules,
CaCO3(s) ⇌ Ca2+ + CO32- equilibrium shifts to RHS
Question
P9-12: Repeat the calculation of the solubility of calcium carbonate by the
approximate single equilibrium method using a realistic wintertime water
temperature of 5 °C; at that temperature, Ksp = 8.1 x 10-9 for CaCO3,
Ka = 2.8 x 10-11 for HCO3-, and Kw = 2.0 x 10-15. Does the solubility of calcium
carbonate increase or decrease with increasing temperature?
CaCO3(s) + H2O(aq) ⇌ Ca2+ + HCO3- + OHK = Kb x Ksp = Kw x Ksp = 2.0 x 10-15 x 8.1 x 10-9 = 5.8 x 10-13
Ka
2.8 x 10-11
From stoichiometry, S = [Ca2+] = [HCO3-] = [OH-]
S3 = 5.8 x 10-13
S = 8.3 x 10-5 mol L-1
i.e. Solubility of CaCO3 increases with temperature
Question
P9-13: Consider a body of water in equilibrium with calcium sulfate, CaSO4, for
which Ksp = 3.0 x 10-5 at 25 °C. Calculate the solubility of calcium sulfate in water
assuming that other reactions are negligible. From your answer, calculate the
fraction of the sulfate ion that reacts with water to form bisulfate ion:
SO42- + H2O ⇌ HSO4- + OHCaSO4(s) ⇌ Ca2+ + SO42-
Ka = 1.2 x 10-2
Ksp = [Ca2+][SO42-]
If no reaction of sulfate with water, [Ca2+] = [SO42-]
S = Solubility of CaSO4 ,
S2 = [SO42-]2 = 3.0 x 10-5
S = [SO42-] = 5.5 x 10-3 mol L-1
(0.75 g L-1)
Question
From your answer, calculate the fraction of the sulfate ion that reacts with water to
form bisulfate ion:
SO42- + H2O ⇌ HSO4- + OH-
Ka = 1.2 x 10-2
If reaction of sulfate with water occurs:
SO42- + H2O ⇌ HSO4- + OHKb = [HSO4-][OH-]
[SO42-]
Kb = Kw/Ka = 10-14 / 1.2 x 10-2 = 8.3 x 10-13
Since [HSO4-] = [OH-] and [SO42-] = 5.5 x 10-3 mol L-1
Kb = 8.3 x 10-13 = [HSO4-] 2 / 5.5 x 10-3 mol L-1
[HSO4-] = 6.8 x 10-8 mol L-1
6.8 x 10-8 mol L-1 / 5.5 x 10-3 mol L-1 x 100 = 0.0001%
Acid-Base Chemistry in Natural Waters
Water in Equilibrium with Both Calcium Carbonate and
Atmospheric CO2
•
System discussed here is unrealistic since it doesn’t include carbon dioxide and
carbonic acid
CO2(g) + H2O(aq) ⇌ H2CO3(aq)
H2CO3(aq) ⇌ H+ + HCO3CaCO3(s) ⇌ Ca2+ + CO32CO32- + H2O ⇌ HCO3- + OHH+ + OH- ⇌ H2O
KH
Ka
Ksp
Kb
1/Kw
CaCO3(s) + CO2(g) + H2O(aq) ⇌ Ca2+ + 2HCO3•
Giant titration of acid from atmospheric CO2 with base from carbonate ion in rocks
K = KspKbKHKa/Kw = 1.5 x 10-6 = [Ca2+][HCO3-]2
PCO2
Acid-Base Chemistry in Natural Waters
Water in Equilibrium with Both Calcium Carbonate and
Atmospheric CO2
•
CaCO3(s) + CO2(g) + H2O(aq) ⇌ Ca2+ + 2HCO3K = KspKbKHKa/Kw = 1.5 x 10-6 = [Ca2+][HCO3-]2
PCO2
If [Ca2+] = S,
[HCO3-] = 2S
1.5 x 10-6 = [Ca2+][HCO3-]2 = S (2S)2
PCO2
0.00037 atm
S = [CO2] = 5.2 x 10-4 mol L-1 (34 x amount calculated from Henry’s law)
S = [Ca2+] = 5.2 x 10-4 mol L-1 (this is 4x closed system)
[HCO3-] = 2S = 1.0 x 10-3 mol L-1
Acid-base reaction increase the solubility of both the gas and he solid – water that
contains CO2 more readily dissolves calcium carbonate
Acid-Base Chemistry in Natural Waters
Water in Equilibrium with Both Calcium Carbonate and
Atmospheric CO2
•
CO32-, H+ and OH- can be derived
Ksp = [Ca2+][CO32-]

[CO32-] = 8.8 x 10-6 mol L-1
Kb = [HCO3-][OH-]
[CO32-]

[OH-] = 1.8 x 10-6 mol L-1
Kw = [H+][OH-]

[H+] = 5.6 x 10-9 mol L-1
Conclude natural water at 25 °C with a pH determined by saturation with CO2 and
CaCO3 should be alkaline (pH = 8.3)
Actual value of calcereous waters is around 7-9
Acid-Base Chemistry in Natural Waters
Question
P9-15: In waters subject to acid rain the pH is not determined by the CO2-carbonate
system but rather by the strong acid from the precipitation. Assuming that equilibrium
with atmospheric carbon dioxide is in effect, calculate the concentration of HCO3- in
natural waters with pH = 6 and 4 at 25 °C, for which
KH = 3.4 x 10-2 and Ka for H2CO3 = 4.5 x 10-7
HCO3- + H2O ⇌ H2CO3 + OHKb = [H2CO3][OH-] = 2.2 x 10-8
[HCO3-]
[H2CO3] = 1.2 x 10-5
So
(from Henry’s law, KH = [H2CO3]/PCO2 = 3.4 x 10-2 / 0.00037
[HCO3-] = [H2CO3][OH-] / 2.2 x 10-8
[HCO3-] = 5.5 x 102[OH-]
Now at pH = 6, pOH = 8, [OH-] = 10-8
[HCO3-] = 5.5 x 102 x 10-8 = 5.5 x 10-6
Now at pH = 4, pOH = 10, [OH-] = 10-10
[HCO3-] = 5.5 x 102 x 10-10 = 5.5 x 10-8
End
• Review
Ions in Natural Waters and Drinking Waters
The Abundant Ions
•
•
Bicarbonate, HCO3Calcium, Ca2+
•
•
•
•
•
•
Magnesium ion, Mg2+
Sulfate, SO42Chloride, ClSodium, Na+
Fluoride, FPotassium, K+
most abundant ions in unpolluted waters
Found in 2:1 ratio
Ions in Natural Waters (mg L-1)
Na+
Ca2+
K+
Mg2+
Fe2+
Mn2+
0 - 100
Cl-
NO3-
0 - 25
Zn2+
Mn+
SO42-
0-1
PO43-
0 – 0.1
NO2-
HCO3-
Ions in Natural Waters and Drinking Waters
Ions in Natural Waters + Drinking Waters
Fluoride
•
•
•
•
•
•
[F-] < 0.01 ppm – 20 ppm
Natural source is fluorapatite, Ca5(PO4)3F
Fluorosilicic acid H2SiF6 is added to increase [F] to 1 ppm
Form of mass medication…reduces cavities?
Higher concentrations causes mottling of teeth
Maximum contaminant level (MCL) level is 4 ppm
http://en.epochtimes.com/news/5-5-16/28765.html
Question
What is 1 ppm F- in mols L-1?
1 ppm F- = 1 mg L-1 = 10-3 g
L
x
1 mol
= 5 x 10-5 mol L-1
19.00 g
Ions in Natural Waters + Drinking Waters
Alkalinity
•
•
•
Dissolution of limestone and other
minerals produces alkalinity
e.g.
CaCO3 ⇌ Ca2+ + CO32CO32- + H2O ⇌ HCO3- + OHWater supply with high total alkalinity
is resistant to pH change
Two samples with identical pH but
different alkalinity behave differently
on addition of acid
– Different capacity to neutralize acid
Mineral
Composition
Calcite
CaCO3
Magnesite
MgCO3
Dolomite
CaCO3.MgCO3
Brucite
Mg(OH)2
Ions in Natural Waters + Drinking Waters
Alkalinity
•
Measurement of the buffer capacity (resistance to pH change) also called acid
neutralizing capacity (ANC) / total alkalinity
e.g.
Carbonate neutralization reaction
CO32- + H+ ⇌ HCO3Bicarbonate neutralization reaction
HCO3- + H+ ⇌ H2O.CO2 ⇌ H2O + CO2
Lime (Ca(OH)2 and
soda ash (Na2CO3)
used to neutralize
industrial wastewater
Hydroxide neutralization reaction
H+ + OH- ⇌ H2O
•
•
Units are mg L-1 CaCO3 (regardless of species)
Measure by titration the quantity of acid or base needed to change the pH to 4.5
(methyl orange end-point)
CaCO3 + H2SO4 → H2CO3 + CaSO4
•
If pH < 4.5 there is no acid neutralizing capacity i.e. no need to measure
alkalinity
Question
What is the total alkalinity in mg L-1 and mmol L-1 CaCO3 for a sample
requiring 21.25 mL of 0.0100 M H2SO4?
0.02125 L x 0.0100 mol L-1 = 2.125 x 10-4 mol H2SO4
Mole ratio is 1:1
2.125 x 10-4 moles H2SO4 = 2.125 x 10-4 moles CaCO3
2.125 x 10-4 mol CaCO3 x 100.09 g / mol = 2.13 x 10-2 g = 21.3 mg
21.3 mg CaCO3 = 213 mg CaCO3 L-1
0.100 L
(= 3.28 mmol L-1)
Ions in Natural Waters + Drinking Waters
Hardness Index
Hard water contains high concentrations of dissolved
calcium and magnesium ions. Soft water contains
few of these dissolved ions.
Hardness = [Ca2+] + [Mg2+]
Counter ions of
alkalinity ions
Alkalinity is a
good indicator
of hardness and
vice-versa
A pipe with hard-water scale build up
Ions in Natural Waters + Drinking Waters
Hardness Index
•
•
Because calcium ions, Ca2+, are generally the largest contributors to hard water,
hardness is usually expressed in parts per million of calcium carbonate (CaCO3)
by mass
It specifies the mass of solid CaCO3 that could be formed from the Ca2+ in
solution, provided sufficient CO32- ions were also present:
Ca2+(aq) + CO32–(aq) → CaCO3(s)
•
A hardness of 10 ppm indicates that 10 mg of CaCO3 could be formed from the
Ca2+ ions present in 1 L of water
Ions in Natural Waters + Drinking Waters
Hardness Index
•
•
Deposition of white solid CaCO3 or MgCO3 when water is heated
– ‘furring-up blocks pipes and lowers efficiency of industrial
processes
Formation of scum (insoluble ppt) with soap and water
Ca2+(aq) + H3C-(CH2)10-COO-(aq)
•
[H3C-(CH2)10-COO-]2Ca2+(s)
– detergent action is blocked
Staining (due to transition metals)
A pipe with hard-water scale
build up
Types of Hardness
•
Solid deposit = carbonate hardness or temporary hardness
CaCO3(s) + H2CO3 ⇌ Ca2+ + HCO3-
(removed via boiling)
– Causes deposit in pipes and scales in boilers
– Temporary hard water has to be softened before it enters the boiler,
hot-water tank, or a cooling system
•
No solid = non-carbonate or permanent hardness
– Amount of metal ions that can not be removed by boiling
Total hardness = temporary hardness + permanent hardness
Ions in Natural Waters + Drinking Waters
Hardness Index
"According to the U.S. National Academy of Sciences (1977)
there have been more than 50 studies, in nine countries, that
have indicated an inverse relationship between water
hardness and mortality from cardiovascular disease”
Source: http://www.mgwater.com
Ions in Natural Waters + Drinking Waters
Hardness Index
Beneficial effects:
• Alkalinity lowers solubility of toxic metals
• Buffering action of CO32- lessens effects of acidic pollutants
• Non-toxic essential nutrients
• Reduces heart disease
• More pleasant to drink
Question
What is the value of the hardness index for a 500 mL sample of water
that contains 0.0040 g of calcium ion and 0.0012 g of magnesium ion?
0.0040 g Ca x 1 mol / 40.1 g = 0.0001 mol Ca
0.0012 g Mg x 1 mol / 24.3 g Mg = 0.00005 mol Mg
Since the volume is 0.5 L:
[Ca2+] = 0.0001 / 0.5 = 0.0002 mol L-1
[Mg2+] = 0.00005 / 0.5 = 0.0001 mol L-1
Total hardness = 0.0002 + 0.0001 = 0.0003 mol L-1
0.0003 mol L-1 x 100 g CaCO3 / mol = 30 mg CaCO3 L-1
Ions in Natural Waters + Drinking Waters
Aluminum
•
Solubility of Al in natural waters is small, [Al] ~10-6 M
Al(OH)3 ⇌ Al3+ + 3OHe.g.
Ksp = 10-33
Ksp = [Al3+][OH-]3 = 10-33
pH = 6, [Al3+] = 10-33/(10-8)3 = 10-9 M
pH = 5, [Al3+] = 10-6 M
pH = 4, [Al3+] = 10-3 M
•
•
Al is much more soluble in acidified rivers and lakes
Principle cation in waters of pH < 4.5 exceeding Ca2+ and Mg2+
– Precipitation of Al(OH)3 on fish gills prevents intake of O2
– Also contributes to forest decline
For every 1 pH
[Al3+] inc. 103
Question
What is the concentration of dissolved aluminum in water with a pH of 5.5?
Al(OH)3 ⇌ Al3+ + 3OH-
Ksp = 10-33
e.g. Ksp = [Al3+][OH-]3 = 10-33
pH + pOH = 14,  at pH = 5.5 pOH = 8.5
pOH = 8.5 = -log10 [OH-]
[OH-] = 3.2 x 10-9
pH = 5.5, [Al3+] = 10-33/(3.2 x 10-9)3 = 3 x 10-8 mol L-1
Mass Al = 8.2 x 10-7 g
Ions in Natural Waters + Drinking Waters
Perchlorate, ClO4•
Mostly released as ammonium
perchlorate – used as oxidizing
agents, rocket fuels, fire works,
airbags etc
•
Most of the release in the
southwestern part
•
No federal limit has been set,
although some states have set their
limits
Source: Logan, 2001
Ions in Natural Waters + Drinking Waters
Perchlorate, ClO4•
•
US drinking water ClO4- levels 5 – 20 ppb
No federal standard has been set – proposed 1 ppb standard considered too
low by NASA and DOD
•
High doses – affects hormone production by thyroid gland
– Maternal hormone levels vital to fetal brain development
•
Low doses – unknown
•
Difficult to remove from water systems