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DISSOLVED CO2 AND ACID BASE EQUILIBRIA INVOLVING CARBONATE (PowerPoint)

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DISSOLVED CO2 AND ACID BASE EQUILIBRIA INVOLVING CARBONATE (PowerPoint) Powered By Docstoc
					DISSOLVED CO2 AND ACIDBASE 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+ + 2HCO3NaAlSi3O8(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 CO2:photosynthesis, mineral dissolution.

2

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-] + 2[CO32-] • An appropriate mass balance.
3

A SIMPLIFICATION
A common simplification is to define a species H2CO3* such that: [H2CO3*]  [CO2(aq)] + [H2CO30]

H2CO3*  HCO3- + H+ H2CO30  HCO3- + H+ H2CO30  CO2(aq) + H2O(l)
K1  K1
0

K1 K10 K  650
0

1 K



K1 K

0



K1

650

Thus, H2CO30 is actually a much stronger acid than H2CO3* (by a factor of 650 at 25C and 1 bar!).
4

Diagrams relevant to a closed system involving CO2 in a freshwater system.

Figure 4.1a-d from Stumm and Morgan 25°C, 1 atm. I = 10-3 M CT = 10-3 M

pcKA,1 = 6.3 pcKA,2 = 10.25

5

Diagrams relevant to CO2 equilibria in closed system in seawater

Figure 4.1e-f from Stumm & Morgan

10°C, 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

6

Figure 4.2 from Stumm & Morgan: pH of carbonate-bearing solutions in a closed system as a function of CT at 25°C and 1 atm.

7

OPEN SYSTEM
• Same species as closed system. • Same charge-balance expression. • Same ionization constants.

• No mass balance, instead we have a Henry’s law expression: CO2(g) + H2O(l)  H2CO3*

[H2CO3*] = KHpCO2 = 10-1.5pCO2
CT  1

0

K H p CO 2

8

[ HCO

 3

]

1 0

K H p CO 2 

K A ,1 [H ]


K H p CO 2

[ CO

2 3

]

2 0

K H p CO 2 

K A ,1 K A , 2 [H ]
 2

K H p CO 2

Charge balance: [H+] = [HCO3-] + 2[CO32-] + [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! 9

Figure 4.3 from Stumm & Morgan: pH of carbonatebearing solutions in an open system. 25°C and 1 atm. pCO2 = 10-3.5 atm. pKH = 1.5
pKA,1 = 6.3

pKA,2 = 10.25

10

For a system with excess strong acid or base, the charge balance becomes: CB + [H+] = [HCO3-] + 2[CO32-] + [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+] = [ANC]f = 0 = [Alk]
[ alk ]  K H p CO 2

0

( 1  2 2 )  [ OH



]  [H ]



In this case we define the system by specifying p CO2 and
[Alk].
11

• 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.

12

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 25°C (pKH = 1.5; pK1 = 6.3; pK2)? Start with Henry’s law constant and ideal gas law:
KH  [ CO 2 ( aq )] p CO 2
[ CO 2 ( g )] 
[ CO 2 ( aq )] [ CO 2 ( g )]
[ CO 2 ( aq )]

n V



p CO 2 RT

R = 0.820 atm L

K-1

mol-1

 K H RT  0 . 75
3

And now a mass balance: Note that Vgas/Vsol’n = 0.01.
0

0
 2 x10

 0 . 01[ CO 2 ( g )]  2 x10
3

[ CO 2 ( aq )]

13

If Vgas is very small relative to Vsol’n, then it can be assumed that 0 does not change much. Recall that:
0 
1 1 K1 [H ]




K 1K 2 [H ]
 2

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-] + 2[CO32-] + [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 C T = 2 x 10-3 M
-2 H 2C O 3 -4
A p p ro xim a te so lu tio n
*

HCO3

-

CO3

2-

-6

lo g [i]
-8 -1 0 OH -1 2 0 2 4 6 8 10 12 14
15
-

H

+

pH

[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
[ CO 2 ( aq )]

0

 2 x10

3

[ CO 2 ( aq )] 1 . 15 x10
2

 2 x10

3

[CO2(aq)] = 2.3 x 10-5 M

[CO2(g)] = (2.3 x 10-5)/0.75 = 3.07 x 10-5 moles/L
16

Figure 4.4 from Stumm & Morgan. Schematic titration of a carbonate solution with a strong acid.

The endpoints of titration of carbonate solutions to H 2CO3* 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. 17

ALKALINITY IN CARBONATE SYSTEMS
ACID-NEUTRALIZING CAPACITY (ANC) Caustic alkalinity (f = 2) [OH-Alk] = [OH-] - [HCO3-] - 2[H2CO3*] - [H+]

p-Alkalinity (f = 1) [p-Alk] = [OH-] + [CO32-] - [H2CO3*] - [H+] Alkalinity (f = 0) [Alk] = [HCO3-] + 2[CO32-] + [OH-] - [H+]
18

ACIDITY IN CARBONATE SYSTEMS
BASE-NEUTRALIZING CAPACITY (BNC) Mineral acidity (f = 0) [H-Acy] = [H+] - [HCO3-] - 2[CO32-] -[OH-]

CO2-Acidity (f = 1) [CO2-Acy] = [H2CO3*] + [H+] - [CO32-] - [OH-] Acidity (f = 2) [Acy] = 2[H2CO3*] + [HCO3-] + [H+] - [OH-]
19

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
21

[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 < 9, B(OH)4- and H3SiO4- << 1, so silica and borate are rarely significant contributors to alkalinity.

22

GRAN TITRATION METHOD Figure 4.13 from Stumm & Morgan

23

ALKALINITY TITRATION GRAN 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).

24

BASIS OF GRAN METHOD
Let v0 = original volume of sample, v = volume of strong acid added at each point in titration, C A = 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-] + 2[CO32-] + [OH-] - [H+] [p-Alk] = [OH-] + [CO32-] - [H2CO3*] - [H+] v0[Alk] = v2CA v0[p-Alk] = v1CA
25

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)10”pH” = (v2 - v)/K1 • In the region g < 1 we have: F4 = (v2 - 2v1 + v)10”pH” = (v1 - v)/K2
26

GRAN TITRATION METHOD Figure 4.13 from Stumm & Morgan

27

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.
28

GRAN TITRATION OF A SAMPLE OF MINE DRAINAGE
0 .0 0 3 5 0 .0 0 3 0 0 .0 0 2 5 0 .0 0 2 0

F1
0 .0 0 1 5 0 .0 0 1 0 0 .0 0 0 5 0 .0 0 0 0 0 5 10 15 20 25 30

v2

m L titra n t a d d e d

29

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.
30

[Alk] = [HCO3-] + 2[CO32-] + [OH-] - [H+] [Acy] = 2[H2CO3*] + [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!

31

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. 32

Fig. 4.5 from Stumm & Morgan: Alkalinity vs. C T capacity diagram
33

Fig. 4.6 from Stumm & Morgan: Acidity vs. CT capacity diagram.

34

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
35

B

M ixtu re

A

36

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. 37

EXAMPLE - INCREASE IN pH BY ADDITION OF BASE OR REMOVAL OF CO2
The pH of a surface water with [Alk]0 = 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.
38

Ad d in g Na O H

Re m o vin g C O 2

39

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.

40

Ad d itio n o f Na 2 C O 3

41

PHOTOSYNTHESIS AND RESPIRATION
• Alkalinity cannot change due to changes in P CO2 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 (10°C)? [Alk]0 = 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 42

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 << 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

43

[Alk] = 5 x 10-4(14/106) = 6.6 x 10-5 [Alk]f = 1.2 x 10-3 + 6.6 x 10-5 = 1.266 x 10-3 eq/L [Alk]  CT1 1= 0.600
1 
1 [H ] K1


1

K2 [H ]




1 [H ] 10

 6 . 43 

1

10

 10 . 39 

[H ]

1 . 667 

[H ] 10
 6 . 43

1

[H+] = 2.5 x 10-7

pH = 6.61
44


				
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