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1 22

The Equilibrium Constant, K eq 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

The Equilibrium Constant, K eq 11 22 h1h1 h2h2 Analogous to defining an equilibrium constant relating liquids 1 and 2 as the ratio of the liquid column heights (h 2 /h 1 ) 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, h 2 /h 1  K if the system is not at equilibrium. aA + bB  cC + dD

The Equilibrium Constant, K eq If the actual ratio of normalized concentrations (Q) equals K eq, the reaction is at equilibrium If Q does not equal K eq, the reaction will proceed in the direction that causes Q to approach K eq Evaluation/use of K eq requires knowledge of: –the chemical reaction (stoichiometry) –conventions for quantifying and normalizing the concentrations of the reactants and products

The Equilibrium Constant, K eq 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, K w H 2 O  H + + OH  pH =  log 10 (H + )

Some Important Equilibria in Water Treatment Involving Solids CaCO 3 (s)  Ca 2+ + CO 3 2  Hardness Mg(OH) 2 (s)  Mg OH  Hardness PbCO 3 (s)  Pb 2+ + CO 3 2  Lead & Copper Rule (LCR) Cu(OH) 2 (s)  Cu OH  LCR Cu 2 (OH) 2 CO 3 (s)  2Cu OH  + CO 3 2  LCR Fe(OH) 3 (s)  Fe OH  Coagulation Al(OH) 3 (s)  Al OH  Coagulation SiO 2 (s) + 2 H 2 O  H 4 SiO 4 Membrane fouling, Arsenic treatment

Equilibrium Constants Involving Solids XY 2 Z(s)  X + 2 Y + Z “Solubility product” (K sp or K so ) is K eq for dissolution of the solid into its constituents The solid (and sometimes H 2 O) are the only chemicals on the left side

Some Solids are Very Soluble or Very Insoluble Some solids (e.g., NaCl, CaCl 2, Na 2 SO 4 ) 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, MnO 2 ) are so insoluble under normal water treatment conditions that we assume 100% of the limiting species (e.g., Fe, Mn) precipitates. Some solids (CaCO 3 ) are “slightly soluble” under normal water treatment conditions, so neither assumption applies. Others (PbCO 3, AlPO 4 ) are very insoluble, but the trace amount that dissolves is still of concern.

Issues that Arise Regarding Slightly Soluble Solids in Water Treatment Is Q so greater than, less than, or equal to K so initially? How much does the concentration of some constituent have to change to cause Q so to equal or exceed K so (i.e., for precipitation to begin)? How much of some chemical must be added to cause Q so to equal or exceed K so ? 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 (CaSO 4 (s)) K so for gypsum is 10  g of gypsum is dispersed in 1.0 L of water containing no Ca 2+ or SO 4 2 . How much solid (if any) remains after the system reaches equilibrium? Repeat the analysis if the solution initially contains 200 mg/L SO 4 2 .

Dissolution of CaSO 4 (s) if (Ca 2+ ) init = (SO 4 2  ) init = 0

CaSO 4 (s) dissolved = 0.68 g/L CaSO4(s) remaining undissolved = 0.32 g/L (32%) Dissolution of CaSO 4 (s) if (Ca 2+ ) init = (SO 4 2  ) init = 0

Dissolution of CaSO 4 (s) with (SO 4   init 

CaSO 4 (s) dissolved = 163 mg/L + (591  200)mg/L = 554 mg/L CaSO4(s) remaining undissolved = 446 mg/L (44.6%) Dissolution of CaSO 4 (s) with (SO 4   init 

The Carbonate Chemical Group in Water Dissolved Carbonate-containing species –Carbonic Acid: H 2 CO 3. Can form by combination of a carbon dioxide molecule and water: CO 2 + H 2 O  H 2 CO 3 –Bicarbonate Ion: HCO 3 . Can form by “dissociation” of carbonic acid: H 2 CO 3  HCO 3  + H +

The Carbonate Chemical Group Dissolved Carbonate-containing species –Carbonate Ion: CO 3 2 . Can form by dissociation of bicarbonate ion: HCO 3   CO 3 2  + H + Other Commonly Defined Quantities –Total Dissolved Carbonate: TOTCO 3 TOTCO 3 = (H 2 CO 3 ) + (HCO 3   + (CO 3 2  ) + [others?] –Alkalinity: ALK ALK = (HCO 3   + 2(CO 3 2  ) + (OH   (H + )

Units for Expressing Alkalinity Equivalents: One equivalent (equiv) of alkalinity refers to one mole of "H + binding capacity." –one mole of HCO 3  has the capacity to combine with one mole of H +, so one mole of HCO 3  is one equivalent of alkalinity –one mole of CO 3 2  has the capacity to combine with two moles of H +, so one mole of CO 3 2  is two equivalents of alkalinity

Units for Expressing Alkalinity mg/L as CaCO 3 : The number of mg/L of CaCO 3 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 CaCO 3 can combine with two moles of H +, so one mole of CaCO 3 is two equivalents of alkalinity –the MW of CaCO 3 is 100, so 50 g of CaCO 3 is one equivalent. An alkalinity of “75 mg/L as CaCO 3 ” is therefore 1.5 meq/L

Equilibrium Among Dissolved Carbonate Species

Other Types of Precipitation & Dissolution Problems In preceding example (precipitation of CaSO 4 (s)), TOTCa & TOTSO 4 varied due to solid precipitation, but under all conditions, TOTCa=(Ca 2+ ) and TOTSO 4 =(SO 4 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 3.22x10  6 mol/L TOTPO 4 (0.1 mg/L TOTPO 4 -P); Corrosion of galvanized pipe releases Zn 2+ TOTPO 4 distributed among four H x PO 4 x  3 species:  H 3 PO 4  H 2 PO 4  + H + K a1 =10  2.2  H 2 PO 4   HPO 4 2  + H + K a2 =10  7.2  HPO 4 2   PO 4 3  + H + K a3 =10  12.4 Question: What is the maximum concentration of Zn 2+ that can be dissolved in the solution, considering possible precipitation of Zn 3 (PO 4 ) 2 (s) (K s0 =10  36.7 )

Example: Water Softening Drinking water at pH 7.0 contains 3x10  3 mol/L Ca 2+ (120 mg/L), and 1.0x10  3 Alk; K s0 for CaCO 3 (s) is 10  8.3 –Is the solution supersaturated, undersaturated, or at equilibrium with respect to CaCO 3 (s)? –What is the stoichiometric dose of lime (Ca(OH) 2 ) required to convert H 2 CO 3 and HCO 3  to CO 3 2  ? –If TOTCO 3 is essentially all converted to CO 3 2 , will CaCO 3 (s) precipitate? If so, how much, and what will the final concentration of Ca 2+ be?

Determine TOTCO 3 and Distribution of Carbonate Species

Thus, CaCO 3 (s) is undersaturated in the initial solution

Estimate Stoichiometric Dose of Lime OH  will be used to (1) convert HCO 3  and H 2 CO 3 to CO 3 2 , and (2) decrease free H + and increase free OH . Assume required pH is ~10.8.

Estimate Stoichiometric Dose of Lime

After Dosing with Lime, the Solution is Supersaturated

How Much Solid Will Precipitate? Precipitation removes 99.8% of TOTCO 3 but lowers TOTCa by only 7% compared to initial value (2.8x10  3 vs 3.0x10  3 ). If x is the amount (mol/L) of CaCO 3 (s) that precipitates, the concentrations of TOTCa and TOTCO 3 in solution will each decrease by x. If (CO 3 2  )  TOTCO 3 after precipitation (i.e., if the pH is >  10.8), then x is also the decrease in (CO 3 2  ). In such a case, we can write:

Determine How Much Soda Ash to Add To remove more Ca 2+, we must add more TOTCO 3. A frequently cited, but somewhat outdated rule of thumb is to add Na 2 CO 3 to make TOTCa=TOTCO 3. If that is done: Then, when the solution equilibrates with the solid: This Ca 2+ concentration is very low. Since Na 2 CO 3 is expensive, the actual dose is usually less than the dose based on the rule of thumb.