1. Bjerrum plot showing the activities of inorganic carbon species as a function of pH for a value of total inorganic carbon of 10 -3 mol L -1. In most natural waters, bicarbonate is the dominant carbonate species!

2. SPECIATION IN OPEN CO 2 -H 2 O SYSTEMS - I In an open system, the system is in contact with its surroundings and components such as CO 2 can migrate in and out of the system. Therefore, the total carbonate concentration will not be constant. Let us consider a natural water open to the atmosphere, for which p CO 2 = 10 -3.5 atm. We can calculate the concentration of H 2 CO 3 * directly from K CO 2 : Note that M H 2 CO 3 * is independent of pH!

3. SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - II The concentration of HCO 3 - as a function of pH is next calculated from K 1 : but we have already calculated M H 2 CO 3 * : so

4. SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - III The concentration of CO 3 2- as a function of pH is next calculated from K 2 : but we have already calculated M HCO 3 - so: and

5. SPECIATION IN OPEN CO 2 - H 2 O SYSTEMS - IV The total concentration of carbonate C T is obtained by summing:

6. Plot of log concentrations of inorganic carbon species H + and OH -, for open-system conditions with a fixed p CO 2 = 10 -3.5 atm.

7. Plot of log concentrations of inorganic carbon species H + and OH -, for open-system conditions with a fixed p CO 2 = 10 -2.0 atm.

8. Methods of solving equations that are ‘linked’ Sequential (stepwise) or simultaneous methods Sequential – assume rxns reach equilibrium in sequence: 0.1 moles H 3 PO 4 in water: –H 3 PO 4 = H + + H 2 PO 4 2- pK=2.1 –[H 3 PO 4 ]=0.1-x, [H + ]=[HPO 4 2- ]=x –Apply mass action: K=10 -2.1 =[H + ][HPO 4 2- ] / [H 3 PO 4 ] –Substitute x  x 2 / (0.1 – x) = 0.0079  x 2 +0.0079x-0.00079 = 0, solve via quadratic equation –x=0.024  pH would be 1.61 Next solve for H 2 PO 4 2- =H + + HPO 4 - …

9. Calcite Solubility CaCO 3 -> Ca 2+ + CO 3 2- log K=8.48 We consider minerals to dissolve so that 1 Ca 2+ dissolves with 1 CO 3 2- If dissolving into dilute water (effectively no Ca 2+ or CO 3 2- present): x 2 =10 -8.48, x= a Ca2+ = a CO32- If controlled by atmospheric CO 2, substitute CO 3 2- for expression What happens in real natural waters??

10. Charge Balance Principle of electroneutrality  For any solution, the total charge of positively charged ions will equal the total charge of negatively charged ions. –Net charge for any solution must = 0 Charge Balance Error (CBE) –Tells you how far off the analyses are (greater than 5% is not good, greater than 10% is terrible…) Models adjust concentration of an anion or cation to make the charges balance before each iteration!

11. Using K eq to define equilibrium concentrations  G 0 R = -RT ln K eq K eq sets the amount of ions present relative to one another for any equilibrium condition AT Equilibrium

12. Speciation Any element exists in a solution, solid, or gas as 1 to n ions, molecules, or solids Example: Ca 2+ can exist in solution as: Ca ++ CaCl + CaNO 3 + Ca(H 3 SiO 4 ) 2 CaF + CaOH + Ca(O-phth) CaH 2 SiO 4 CaPO 4 - CaB(OH) 4 + CaH 3 SiO 4 + CaSO 4 CaCH 3 COO + CaHCO 3 + CaHPO 4 0 CaCO 3 0 Plus more species  gases and minerals!!

13. How do we know about all those species?? Based on complexation  how any ion interacts with another ion to form a molecule, or complex (many of these are still in solution) Yet we do not measure how much CaNO 3 +, CaF +, or CaPO 4 - there is in a particular water sample We measure Ca 2+  But is that Ca 2+ really how the Ca exists in a water??

14. Aqueous Complexes Why do we care?? 1.Complexation of an ion also occuring in a mineral increases solubility 2.Some elements occur as complexes more commonly than as free ions 3.Adsorption of elements greatly determined by the complex it resides in 4.Toxicity/ bioavailability of elements depends on the complexation

15. Defining Complexes Use equilibrium expressions:  G 0 R = -RT ln K eq cC + lHL  CL + lH+ Where B is just like K eq !

16. Mass Action & Mass Balance mCa 2+ =mCa 2+ +MCaCl + + mCaCl 2 0 + CaCL 3 - + CaHCO 3 + + CaCO 3 0 + CaF + + CaSO 4 0 + CaHSO 4 + + CaOH + +… Final equation to solve the problem sees the mass action for each complex substituted into the mass balance equation

17. Mineral dissolution/precipitation To determine whether or not a water is saturated with an aluminosilicate such as K-feldspar, we could write a dissolution reaction such as: KAlSi 3 O 8 + 4H + + 4H 2 O  K + + Al 3+ + 3H 4 SiO 4 0 We could then determine the equilibrium constant: from Gibbs free energies of formation. The IAP could then be determined from a water analysis, and the saturation index calculated.

18. INCONGRUENT DISSOLUTION Aluminosilicate minerals usually dissolve incongruently, e.g., 2KAlSi 3 O 8 + 2H + + 9H 2 O  Al 2 Si 2 O 5 (OH) 4 + 2K + + 4H 4 SiO 4 0 As a result of these factors, relations among solutions and aluminosilicate minerals are often depicted graphically on a type of mineral stability diagram called an activity diagram.

19. ACTIVITY DIAGRAMS: THE K 2 O-Al 2 O 3 -SiO 2 -H 2 O SYSTEM We will now calculate an activity diagram for the following phases: gibbsite {Al(OH) 3 }, kaolinite {Al 2 Si 2 O 5 (OH) 4 }, pyrophyllite {Al 2 Si 4 O 10 (OH) 2 }, muscovite {KAl 3 Si 3 O 10 (OH) 2 }, and K-feldspar {KAlSi 3 O 8 }. The axes will be a K + /a H + vs. a H 4 SiO 4 0. The diagram is divided up into fields where only one of the above phases is stable, separated by straight line boundaries.

20. Activity diagram showing the stability relationships among some minerals in the system K 2 O-Al 2 O 3 -SiO 2 -H 2 O at 25°C. The dashed lines represent saturation with respect to quartz and amorphous silica.

21. Seeing this, what are the reactions these lines represent?