DISSOLVED CO2 AND ACIDBASE EQUILIBRIA INVOLVING CARBONATE

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DISSOLVED CO2 AND ACID-BASE EQUILIBRIA INVOLVING
CARBONATE
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• CO2 is the most prevalent source of acidity in
  natural waters (unperturbed by man) and causes
  minerals to dissolve in weathering processes,
  e.g.
• CaCO3(calcite) CO2(g) H2O(l) ? Ca2 2HCO3-
• NaAlSi3O8(albite) CO2(g) 11/2H2O(l) ? Na
  HCO3- 2H4SiO40 1/2Al2Si2O5(OH)4(kaolinite)
• Sources of CO2 volcanic emissions, respiration,
  fossil fuel combustion, respiration,
  decomposition of organic matter, precipitation of
  carbonates.
• Sinks of CO2photosynthesis, mineral dissolution.

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CLOSED SYSTEM

• Treat carbonic acid as a non volatile acid.
• Species involved CO2(aq), H2CO30, HCO3-, CO32-,
  H, OH-.
• Four equilibrium relations
• Hydration of CO2(aq) CO2(aq) H2O(l) ? H2CO30
• First and second dissociation reactions of
  H2CO30
• H2CO30 ? H HCO3-
• HCO3- ? H CO32-
• Ionization of water H2O(l) ? H OH-
• Electroneutrality H OH- HCO3-
  2CO32-
• An appropriate mass balance.

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A SIMPLIFICATION

• A common simplification is to define a species
  H2CO3 such that
• H2CO3 ? CO2(aq) H2CO30
• H2CO3 ? HCO3- H K1
• H2CO30 ? HCO3- H K10
• H2CO30 ? CO2(aq) H2O(l) K ? 650

Thus, H2CO30 is actually a much stronger acid
than H2CO3 (by a factor of 650 at 25?C and 1
bar!).
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Figure 4.1a-d from Stumm and Morgan 25C, 1
atm. I 10-3 M CT 10-3 M pcKA,1 6.3 pcKA,2
10.25
Diagrams relevant to a closed system involving
CO2 in a freshwater system.
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Diagrams relevant to CO2 equilibria in closed
system in seawater
Figure 4.1e-f from Stumm Morgan 10C, 1 atm. CT
2.3 x 10-3 M pcKA,1 6.1 pcKA,2 9.3 BT 4.1
x 10-4 M pK(boric acid) 8.8
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Figure 4.2 from Stumm Morgan pH of
carbonate-bearing solutions in a closed system
as a function of CT at 25C and 1 atm.
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OPEN SYSTEM

• Same species as closed system.
• Same charge-balance expression.
• Same ionization constants.
• No mass balance, instead we have a Henrys law
  expression CO2(g) H2O(l) ? H2CO3
• H2CO3 KHpCO2 10-1.5pCO2

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• Charge balance H HCO3- 2CO32- OH-
• At a given temperature, we can define the entire
  system by specifying pCO2. For a system composed
  of pure water and CO2 with pCO2 10-3.5 atm, the
  pH is given by point P in Figure 4.3, because
  H ? HCO3-. This point corresponds to a pH
  5.65. Thus, pure rainwater in equilibrium with
  CO2 as the only acid (i.e., natural rainwater
  unaffected by pollution) will have an acidic pH!

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Figure 4.3 from Stumm Morgan pH of
carbonate-bearing solutions in an open
system. 25C and 1 atm. pCO2 10-3.5 atm. pKH
1.5 pKA,1 6.3 pKA,2 10.25
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• For a system with excess strong acid or base, the
  charge balance becomes
• CB H HCO3- 2CO32- OH- CA
• where CB concentration of strong base added
  concentration of counter cation, and CA
  concentration of strong acid added
  concentration of counter anion. Rearranging and
  substituting we get
• CB - CA CT(?1 2?2) OH- - H ANCf
  0 Alk

In this case we define the system by specifying
pCO2 and Alk.
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• If other bases are present, they too must be
  added to the charge-balance expression.
• Groundwaters or soil waters tend to have higher
  pCO2 than surface waters, primarily because of
  respiration and bacterial decomposition of
  organic matter.
• Problem Estimate the pH of the following
  solutions at pCO2 10-3.5 atm i) 10-3 M KOH
  ii) 5 x 10-4 M Na2CO3 iii) 10-3 M NaHCO3 iv) 5
  x 10-4 M MgO.
• Solution In each of the above cases, Alk
  10-3 eq/L, so all have the same pH at the given
  pCO2. From Figure 4.3 we can read off pH 8.3.

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ANOTHER EXAMPLE

• One liter of a solution of 2 x 10-3 M NaHCO3
  solution is brought into contact with 10 mL of
  gaseous N2. How much CO2 will be in the gas phase
  after equilibration at 25C (pKH 1.5 pK1
  6.3 pK2)?
• Start with Henrys law constant and ideal gas law

R 0.820 atm L K-1 mol-1
And now a mass balance Note that Vgas/Vsoln
0.01.
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• If Vgas is very small relative to Vsoln, then it
  can be assumed that ?0 does not change much.
  Recall that

So we need to calculate pH, so we can get ?0 and
then get CO2(g) from the mass-balance
expression.
To calculate the pH we start with charge- and
mass-balances for the system Charge balance
Na H HCO3- 2CO32- OH- Mass
balance Na CT H2CO3 HCO3-
CO32- Proton condition H2CO3 H
CO32- OH- and from speciation diagram we
see that H2CO3 ? CO32- OH-
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Speciation diagram for the carbonate system with
CT 2 x 10-3 M
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• H2CO3 CT?0 CO32- CT?2 OH- KW/H
• so the proton condition becomes
• CT?0 ? CT?2 KW/H
• And the solution to this equation leads to pH
  8.23, and ?0 1.15 x 10-2. Note that if we had
  assumed the proton condition to reduce to
  H2CO3 ? CO32- and then pH 8.275. Now

CO2(aq) 2.3 x 10-5 M
CO2(g) (2.3 x 10-5)/0.75 3.07 x 10-5
moles/L
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Figure 4.4 from Stumm Morgan. Schematic
titration of a carbonate solution with a strong
acid.
The endpoints of titration of carbonate solutions
to H2CO3 and CO32- occur at pH 4.5 and 10.3,
respectively. These pH limits also represent the
approximate limits beyond which life cannot
normally proceed as usual.
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ALKALINITY IN CARBONATE SYSTEMS

• ACID-NEUTRALIZING CAPACITY (ANC)
• Caustic alkalinity (f 2)
• OH-Alk OH- - HCO3- - 2H2CO3 - H
• p-Alkalinity (f 1)
• p-Alk OH- CO32- - H2CO3 - H
• Alkalinity (f 0)
• Alk HCO3- 2CO32- OH- - H

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ACIDITY IN CARBONATE SYSTEMS

• BASE-NEUTRALIZING CAPACITY (BNC)
• Mineral acidity (f 0)
• H-Acy H - HCO3- - 2CO32- -OH-
• CO2-Acidity (f 1)
• CO2-Acy H2CO3 H - CO32- - OH-
• Acidity (f 2)
• Acy 2H2CO3 HCO3- H - OH-

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RELATIONSHIPS AMONG ACIDITY AND ALKALINITY

• Alk H-Acy 0
• Acy OH-Alk 0
• p-Alk CO2-Acy 0
• Alk CO2-Acy CT
• Alk Acy 2CT
• Alk - p-Alk CT
• CO2-Acy - H-Acy CT

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OTHER CONTRIBUTIONS TO ALKALINITY

• Other bases that could be titrated as alkalinity
  silicate, borate, ammonia, organic bases (e.g.,
  acetate), sulfides and phosphates.
• Usually concentration of these is small compared
  to carbonate.
• Borate most important in seawater (10-3.4 M) and
  silicate in freshwater (10-4 - 10-3 M).
• In most waters borate and silicate have only
  minor effects on alkalinity owing to high pKA
  values
• pK(B(OH)3) 8.9 pK(H4SiO4) 9.5

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• Alk CT(?1 2?2) OH- - H
• BT?B(OH)4- SiT?H3SiO4-
• ?B(OH)4- B(OH)4-/BT ?H3SiO4- H3SiO4-/SiT
• Thus, at pH lt 9, ?B(OH)4- and ?H3SiO4- ltlt 1, so
  silica and borate are rarely significant
  contributors to alkalinity.

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GRAN TITRATION METHODFigure 4.13 from Stumm
Morgan
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ALKALINITY TITRATIONGRAN METHOD

• Difficulties that can be encountered in
  alkalinity titrations
• Difficulty seeing endpoint of weak acid.
• Impurities.
• Changes after sampling (best to do titrations in
  the field).
• To get around these problems, we do a Gran
  titration, or linearize the titration curve. In
  essence, you perform the titration just as
  before, but you make measurements well past the
  endpoint (g 2, f 0).

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BASIS OF GRAN METHOD

• Let v0 original volume of sample, v volume of
  strong acid added at each point in titration, CA
  normality (eq/L) of strong acid, v2 volume of
  strong acid needed to get to endpoint g 2, v1
  volume of strong acid needed to get to endpoint g
  1.
• Alk HCO3- 2CO32- OH- - H
• p-Alk OH- CO32- - H2CO3 - H
• v0Alk v2CA
• v0p-Alk v1CA

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THE GRAN FUNCTIONS

• There are four Gran functions to calculate. In
  the region well past the endpoint g 2 we have
• F1 (v0 v)10-pH ? (v - v2)CA
• Note that F1 0 at v v2
• In the region between g 1 and g 2 we have
• F2 (v2 - v)10-pH (v - v1)K1
• (we need to have v2 to get v1)
• F3 (v - v1)10pH (v2 - v)/K1
• In the region g lt 1 we have
• F4 (v2 - 2v1 v)10pH (v1 - v)/K2

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GRAN TITRATION METHODFigure 4.13 from Stumm
Morgan
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DETAILED STEPS TO GET Alk FROM GRAN TITRATION

• 1) Titrate a known volume of sample well past
  endpoint g 2 using standardized strong acid.
• 2) Calculate F1 for data near and beyond g 2.
  Plot F1 vs. v.
• 3) Fit a straight line to the linear part of
  curve and extrapolate it to v-axis, read v2 from
  intersection of line with v-axis.
• 4) Calculate Alk according to Alk v2CA/v0.
• 5) Calculate F2 and plot it vs. v using v2
  obtained in step 3.
• 6) Fit straight line to F2 and extrapolate to
  v-axis to get v1.
• 7) Calculate p-Alk as p-Alk v1CA/v0.
• 8) Plot F3 and F4 to check values of v1 and v2
  obtained.

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GRAN TITRATION OF A SAMPLE OF MINE DRAINAGE
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CONSERVATIVE PROPERTIES

• Conservative properties are those that are
  independent of pressure and temperature, e.g.,
  ANC, BNC and CT (if expressed as moles/Kg).
• Examples of properties that are not conservative
  pH and the concentration of any individual
  species, e.g., H2CO3.
• These properties are also independent of certain
  changes in chemical composition. For example,
  addition of H2CO3 or change of pCO2 will
  decrease the pH, and increase Acy and CT, but
  will not affect Alk because H2CO3 is the
  reference by which Alk is defined.

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• Alk HCO3- 2CO32- OH- - H
• Acy 2H2CO3 HCO3- H - OH-
• Acidity is unaffected by changes in concentration
  of CO32-, so addition of salts such as Na2CO3(s)
  or CaCO3(s) will not affect acidity!

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CAPACITY DIAGRAMS

• Capacity diagrams - plots with conservative
  properties, e.g., CT, Alk or Acy, as
  coordinates and contoured with pH or activities
  of carbonate species.
• Constructed from the equation
• Alk CT(?1 2?2) OH- - H
• Addition or removal of acids or bases can be
  represented by vectors.
• For any given pH, Alk is a linear function of
  CT.
• Vertical lines on these diagrams yield
  acidimetric or alkalimetric titration curves.

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Fig. 4.5 from Stumm Morgan Alkalinity vs. CT
capacity diagram
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Fig. 4.6 from Stumm Morgan Acidity vs. CT
capacity diagram.
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EXAMPLE - MIXING OF TWO WATERS

• Two waters (A pH 6.1, Alk 1.0 meq/L B pH
  9, Alk 2 meq/L) are mixed in equal
  proportions. What is the pH of the mixture if no
  CO2 is lost?
• Reading from graph CT(A) 2.8 mmoles/L, CT(B)
  1.9 mmoles/L.
• Mixing in equal proportions yields
• CT (2.8 1.9)/2 2.35 mmoles/L
• Alk (2 1)/2 1.5 meq/L
• From graph we read pH 6.6

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EXACT SOLUTION TO PROBLEM

• Alk CT(?1 2?2) OH- - H
• CT (Alk - OH- H)/(?1 2?2)
• For A, ?1 0.387 ?2 3.07 x 10-5 so
• CT ? 10-3/(0.387) 2.6 mM
• For B, ?1 0.939 ?2 0.0592 so
• CT ? (2 x 10-3/(0.939 2 x 0.0592)) 1.9 mM
• For the mixture CT (2.6 1.9)/2 2.25 mM
• and Alk 1.5 meq and
• ?1 2?2 ? ?1 ? Alk/CT 1.5/2.25 0.667
• and the above occurs at pH 6.6, where ?1
  0.666 and ?2 1.67 x 10-4.

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EXAMPLE - INCREASE IN pH BY ADDITION OF BASE OR
REMOVAL OF CO2

• The pH of a surface water with Alk0 1 meq/L
  and pH0 6.5 is to be raised to pH 8.3.
  Calculate the compositional changes required if
  this is accomplished by
• 1) Addition of NaOH - From the graph we read CT ?
  1.7 mM. Vertical line up to pH 8.3 gives us 0.7
  meq/L NaOH.
• 2) Addition of Na2CO3 - Need to draw line with
  slope 2 from initial point (pH 6.5 and Alk
  1 meq/L to intersection with pH 8.3 contour.
  About 1.4 meq/L or 0.7 mM Na2CO3 required.

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• 2) (continued) This problem could have been
  solved more easily using acidity diagram. A water
  with pH 6.5 and CT 1.7 mM has Acy 2.35
  meq/L. Because addition of Na2CO3 cannot change
  acidity, the length a horizontal line segment
  over to pH 8.3 yields 0.7 mM.
• 3) By removal of CO2 - Use alkalinity diagram
  because alkalinity is independent of CO2. The
  length of a horizontal line segment at Alk 1
  meq/L from pH 6.5 to pH 8.3 corresponds to
  0.7mM CO2.

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PHOTOSYNTHESIS AND RESPIRATION

• Alkalinity cannot change due to changes in PCO2
  alone. However, associated uptake or excretion of
  ions during these processes change charge balance
  and hence alkalinity, e.g., NO3-, NH4, HPO42-.
• What is the pH change resulting from aerobic
  decomposition of organic matter (6 ?g OC) in 1 mL
  interstitial lake water (10C)? Alk0 1.2
  meq/L, pH0 6.90 and I 3 x 10-3 M.
• Assume reaction C106H263O110N16P 106O2 14H
  ? 106 CO2 16NH4 HPO42- 106H2O
• Redfield composition C106H263O110N16P

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• Estimate change in two-step process
• 1) CO2 increase at constant Alk
• 2) Alk change at constant CT
• 6 mg/L organic carbon ? 0.5 x 10-3 M CO2
• pcK1 6.43 pcK2 10.39 pcKw 14.53
• Alk CT(?1 2?2) OH- - H
• at pH0 6.9, ?1 0.747 ?2 ltlt 1
• Alk CT(0.747) 10-7.63 - 10-6.9
• CT 1.61 x 10-3 M
• However, the CO2 added by decomposition of OC
  adds 0.5 x 10-3 M to CT, so
• CT 1.61 x 10-3 0.5 x 10-3 2.11 x 10-3 M

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• ?Alk 5 x 10-4(14/106) 6.6 x 10-5
• Alkf 1.2 x 10-3 6.6 x 10-5 1.266 x 10-3
  eq/L
• Alk ? CT?1
• ?1 0.600

H 2.5 x 10-7 pH 6.61