Chemical Equilibrium

A condition in which the system is at its minimum attainable chemical energy level and hence has no tendency to undergo chemical change.

Analogous Physical Equilibria:

1

2

1 1 2 2gh gh 2 1

1 2

hK

h

The Equilibrium Constant, Keq

For any reaction with stoichiometry:

aA + bB cC + dD

where (i) is a normalized (non-dimensional) concentration of chemical i, and the values are all measured in a system at equilibrium

C D

A B

c d

eq a bK

The Equilibrium Constant, Keq

C D

A B

c d

eq a bK

1

2

h1

h2

Analogous to defining an equilibrium constant relating liquids 1 and 2 as the ratio of the liquid column heights (h2/h1) for a system that has reached static equilibrium.

Note that one need not know the densities of the fluids (or understand how they relate to to the heights) if the “K” value is given. Also, h2/h1 K if the system is not at equilibrium.

aA + bB cC + dD

The Equilibrium Constant, Keq

• If the actual ratio of normalized concentrations (Q) equals Keq, the reaction is at equilibrium

• If Q does not equal Keq, the reaction will proceed in the direction that causes Q to approach Keq

• Evaluation/use of Keq requires knowledge of:

– the chemical reaction (stoichiometry)– conventions for quantifying and normalizing the

concentrations of the reactants and products

The Equilibrium Constant, Keq

• Concentration Conventions – For dilute dissolved species (e.g., Na+, Cl in water),

concentrations are expressed in mol/L, and the normalizing concentration is 1.0 mol/L

– For the main constituents of a condensed phase (e.g., water, or a pure solid), concentrations are expressed as the mole fraction of that phase that the species represents (moles of i/total moles in phase). This fraction is typically so close to 1.0 that we can approximate the normalized concentration as exactly 1.0.

– For gases, concentrations are expressed as the pressure that the species exerts (the partial pressure), in atmospheres, and the normalizing concentration is 1.0 atm.

The Equilibrium Constant for Water Dissociation, Kw

H2O H+ + OH

pH = log10(H+)

+ +

14.0

2

H OH H OH10

H O 1.0eq wK K

Some Important Equilibria in Water Treatment Involving Solids

• CaCO3(s) Ca2+ + CO32 Hardness

• Mg(OH)2(s) Mg2+ + 2 OH Hardness

• PbCO3(s) Pb2+ + CO32 Lead & Copper Rule (LCR)

• Cu(OH)2(s) Cu2+ + 2 OH LCR

• Cu2(OH)2CO3(s) 2Cu2+ + 2 OH + CO32 LCR

• Fe(OH)3(s) Fe3+ + 3 OH Coagulation

• Al(OH)3(s) Al3+ + 3 OH Coagulation

• SiO2(s) + 2 H2O H4SiO4 Membrane fouling, Arsenic treatment

Equilibrium Constants Involving Solids

XY2Z(s) X + 2 Y + Z

• “Solubility product” (Ksp or Kso) is Keq for dissolution of the solid into its constituents

• The solid (and sometimes H2O) are the only chemicals on the left side

2 2

2

X Y Z X Y Z or

1.0XY Zeq sp soK K Ks

Some Solids are Very Soluble or Very Insoluble

• Some solids (e.g., NaCl, CaCl2, Na2SO4) are so soluble under normal water treatment conditions that we never consider the possibility that the solid will be present in an equilibrium solution.

• Other solids (e.g., Fe(OH)3, MnO2) are so insoluble under normal water treatment conditions that we assume 100% of the limiting species (e.g., Fe, Mn) precipitates.

• Some solids (CaCO3) are “slightly soluble” under normal water treatment conditions, so neither assumption applies. Others (PbCO3, AlPO4) are very insoluble, but the trace amount that dissolves is still of concern.

Issues that Arise Regarding Slightly Soluble Solids in Water Treatment

• Is Qso greater than, less than, or equal to Kso initially?

• How much does the concentration of some constituent have to change to cause Qso to equal or exceed Kso (i.e., for precipitation to begin)?

• How much of some chemical must be added to cause Qso to equal or exceed Kso?

• Will a solid precipitate if a specified amount of some chemical is added? If so, how much?

• If a solution with known composition comes into contact with a solid, how much solid will dissolve?

Example: Dissolution of Gypsum (CaSO4(s))

• Kso for gypsum is 104.6

• 1.0 g of gypsum is dispersed in 1.0 L of water containing no Ca2+ or SO4

2. How much solid (if any) remains after the system reaches equilibrium?

• Repeat the analysis if the solution initially contains 200 mg/L SO4

2.

Dissolution of CaSO4(s) if (Ca2+)init = (SO4

2)init = 0

2 2 2 2

4 44.6 2 24

4

Ca SO Ca SO10 Ca SO

1.0CaSOspK s

24.6 210 Ca

0.52 4.6 2.3 3 24Ca 10 10 5.0x10 SO

2 24Ca SO

2 2.3 mol g gCa 10 40 0.20

L mol L

2 2.34

mol g gSO 10 96 0.48

L mol L

CaSO4(s) dissolved = 0.68 g/L

CaSO4(s) remaining undissolved = 0.32 g/L (32%)

Dissolution of CaSO4(s) if (Ca2+)init = (SO4

2)init = 0

Dissolution of CaSO4(s) with (SO4

init

2 34

init

mg 1 mol molSO 200 2.08x10

L 96,000 mg L

4.6 2 2 34final final

10 Ca SO 2.08x10spK x x

34.08x10x

2 3

final

mol mg mgCa 4.08x10 40,000 163

L mol Lx

2 3 34 final

mol mol mg mgSO 2.08x10 6.16x10 96,000 591

L L mol Lx

CaSO4(s) dissolved = 163 mg/L + (591 200)mg/L = 554 mg/L

CaSO4(s) remaining undissolved = 446 mg/L (44.6%)

Dissolution of CaSO4(s) with (SO4

init

The Carbonate Chemical Group in Water

• Dissolved Carbonate-containing species

– Carbonic Acid: H2CO3. Can form by combination of a carbon dioxide molecule and water:

CO2 + H2O H2CO3

– Bicarbonate Ion: HCO3. Can form by “dissociation”

of carbonic acid:

H2CO3 HCO3 + H+

2 3

2 2

H CO

CO H OeqK g

+3

12 3

HCO H

H COeq aK K

The Carbonate Chemical Group

• Dissolved Carbonate-containing species

– Carbonate Ion: CO32. Can form by dissociation of

bicarbonate ion:

HCO3 CO3

2 + H+

• Other Commonly Defined Quantities

– Total Dissolved Carbonate: TOTCO3

TOTCO3 = (H2CO3) + (HCO3+ (CO3

2) + [others?]

– Alkalinity: ALK

ALK = (HCO3+ 2(CO3

2) + (OH(H+)

2 +3

2

3

CO H

HCOeq aK K

Units for Expressing Alkalinity

• Equivalents: One equivalent (equiv) of alkalinity refers to one mole of "H+ binding capacity."

– one mole of HCO3 has the capacity to combine with

one mole of H+, so one mole of HCO3 is one

equivalent of alkalinity

– one mole of CO32 has the capacity to combine with two

moles of H+, so one mole of CO32 is two equivalents of

alkalinity

Units for Expressing Alkalinity

• mg/L as CaCO3: The number of mg/L of CaCO3 that would have to be dissolved into pure water for that water to have the same alkalinity as the water of interest

– one mole of CaCO3 can combine with two moles of H+, so one mole of CaCO3 is two equivalents of alkalinity

– the MW of CaCO3 is 100, so 50 g of CaCO3 is one equivalent. An alkalinity of “75 mg/L as CaCO3” is therefore 1.5 meq/L

Equilibrium Among Dissolved Carbonate Species

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

2 3 4 5 6 7 8 9 10 11 12

pH

Co

nc

'n /

TO

TC

O3

CO32

/TOTCO3HCO3/TOTCO3

H2CO3/TOT CO3

Other Types of Precipitation & Dissolution Problems

• In preceding example (precipitation of CaSO4(s)), TOTCa & TOTSO4 varied due to solid precipitation, but under all conditions, TOTCa=(Ca2+) and TOTSO4=(SO4

2)

• In other commonly encountered systems, the concentration of one or both precipitating ions depends not only on the amount of solid formed, but also other variables (especially pH).

Example: Precipitation & Dissolution of a Phosphate Solid

• Drinking water at pH 8.0 contains 3.22x106 mol/L TOTPO4 (0.1 mg/L TOTPO4-P); Corrosion of galvanized pipe releases Zn2+

• TOTPO4 distributed among four HxPO4x3 species:

H3PO4 H2PO4 + H+ Ka1=102.2

H2PO4 HPO4

2 + H+ Ka2=107.2

HPO42

PO43 + H+ Ka3=1012.4

• Question: What is the maximum concentration of Zn2+ that can be dissolved in the solution, considering possible precipitation of Zn3(PO4)2(s) (Ks0=1036.7)

2 4 2.2

13 4

H PO H10

H POaK

234 19.6

2 3

2 4

PO H10

H POa aK K

24 7.2

2

2 4

HPO H10

H POaK

34 12.4

3 24

PO H10

HPOaK

321.8 33 4 4

2.2 34

H PO 10 PO H

10 PO

2 12.4 34 4

4.4 34

HPO 10 PO H

10 PO

219.6 32 4 4

3.6 34

H PO 10 PO H

10 PO

334 21.8

1 2 33 4

PO H10

H POa a aK K K

2 34 3 4 2 4 4 4

2.2 3 3.6 3 4.4 3 34 4 4 4

4 34

PO H PO H PO HPO PO

10 PO 10 PO 10 PO PO

2.91x10 PO

TOT

1/3 1/336.7

2 602 23 10

4

10Zn 2.56x10

PO 1.11x10

sK

6

3 104 4

3.22x10PO 1.11x10

2.91x10

Example: Water Softening

• Drinking water at pH 7.0 contains 3x103 mol/L Ca2+

(120 mg/L), and 1.0x103 Alk; Ks0 for CaCO3(s) is 108.3

– Is the solution supersaturated, undersaturated, or at equilibrium with respect to CaCO3(s)?

– What is the stoichiometric dose of lime (Ca(OH)2) required to convert H2CO3 and HCO3

to CO32?

– If TOTCO3 is essentially all converted to CO32, will CaCO3(s)

precipitate? If so, how much, and what will the final concentration of Ca2+ be?

23 3HCO 2 CO OH HALK

Determine TOTCO3 and Distribution of Carbonate Species

3 23

3

HCO1.0x10 HCO 2 H

H H

a wK K

10.3 14.033 7.0

3 7.0 7.0

HCO 10 101.0x10 HCO 2 10

10 10

331.0x10 1.001 HCO

3

33

1.0x10HCO 1.0x10

1.001

Determine TOTCO3 and Distribution of Carbonate Species

3.0 10.33 2 6.3

3 7.0

HCO 10 10CO 10

10H

aK

3.0 7.0

3 3.7 42 3 6.3

1

HCO H 10 10H CO 10 2x10

10aK

23 2 3 3 3CO H CO HCO COTOT

4 3 7 32.0x10 1.0x10 5x10 1.2x10

2+ 2 3 6.3 8.83 0Ca CO 3x10 10 10 sK

Thus, CaCO3(s) is undersaturated in the initial solution

Estimate Stoichiometric Dose of Lime

23 3 2HCO OH CO H O 2

2 3 3 2H CO 2OH CO 2H O

2 3 323 3

Moles of OH required to2 H CO HCO OH H

cause CO COTOT

4 3 3.2 7.0 7.0 10.8

3

2 2.0x10 1.0x10 10 10 10 10

2.03x10

OH will be used to (1) convert HCO3 and H2CO3 to CO3

2, and (2) decrease free H+ and increase free OH . Assume required pH is ~10.8.

Estimate Stoichiometric Dose of Lime

2 23 3 3 3

3

3

Moles of lime required to Moles of OH required to1cause CO CO 2 cause CO CO

12.03x10

2

1.02x10

TOT TOT

After Dosing with Lime, the Solution is Supersaturated

3 23 3CO 1.2x10 COTOT

2 2 2 3 3 3Ca Ca Ca 3.0x10 1.0x10 4.0x10init lime

2 2 3 3 6 5.323 0Ca CO 4.0x10 1.2x10 4.8x10 10 sK

How Much Solid Will Precipitate?

Precipitation removes 99.8% of TOTCO3 but lowers TOTCa by only 7% compared to initial value (2.8x103 vs 3.0x103).

2 2 2 20 3 3Ca CO Ca COs equil equil init init

K x x

If x is the amount (mol/L) of CaCO3(s) that precipitates, the concentrations of TOTCa and TOTCO3 in solution will each decrease by x. If (CO3

2)  TOTCO3 after precipitation (i.e., if the pH is >10.8), then x is also the decrease in (CO3

2). In such a case, we can write:

8.3 3 310 4.0x10 1.2x10x x

31.198x10x 6 3

3,CO 2x10 Ca 2.8x10final finalTOT TOT

Determine How Much Soda Ash to AddTo remove more Ca2+, we must add more TOTCO3. A frequently cited, but somewhat outdated rule of thumb is to add Na2CO3 to make TOTCa=TOTCO3. If that is done:

2 3 3,init

3 3 3

Na CO Ca CO

4.0x10 1.2x10 2.8x10

doseTOT TOT

22 2 2 8.33Ca CO Ca 10

2 8.3 4.15 5Ca 10 10 7.1x10 2.8 mg/L

Then, when the solution equilibrates with the solid:

This Ca2+ concentration is very low. Since Na2CO3 is expensive, the actual dose is usually less than the dose based on the rule of thumb.