1. Mixing in water Solutions dominated by water (1 L=55.51 moles H 2 O) a A =k H X A where K H is Henry’s Law coefficient – where is this valid? Low concentration of A Mol fraction AH2OH2OA Activity 0.0 1.0 0.01.0 Ideal mixing aAaA aH 2 O Raoult’s Law – higher concentration ranges (higher X A ):  A =  A 0 +RTln  A X A where  A is Rauolt’s law activity coefficient

2. Activity Activity, a, is the term which relates Gibbs Free Energy to chemical potential:  i -G 0 i = RT ln a i Why is there now a correction term you might ask… –Has to do with how things mix together –Relates an ideal solution to a non-ideal solution

3. Activity II For solids or liquid solutions: a i =X i  i For gases: a i =P i  i = f i For aqueous solutions: a i =m i  i X i =mole fraction of component i P i = partial pressure of component i m i = molal concentration of component i

4. Activity Coefficients Where do they come from?? We think of ‘ideal’ as the standard state, but for dissolved ions, that is actually an infinitely dilute solution Gases, minerals, and bulk liquids (H 2 O) are usually pretty close to 1 in waters Dissolved molecules/ ions are have activity coefficients that change with concentration (ions are curved lines relating concentration and activity coefficients, molecules usually more linear relation)

5. Application to ions in solution Ions in solutions are obviously nonideal mixtures! Use activities (a i ) to apply thermodynamics and law of mass action a i =  i m i The activity coefficient,  i, is found via some empirical foundations

6. Dissolved species  i First must define the ionic strength (I) of the solution the ion is in: Where m i is the molar concentration of species i and z i is the charge of species I

7. Activity Coefficients Debye-Huckel approximation (valid for I: Where A and B are constants (depending on T, see table 10.3 in your book), and a is a measure of the effective diameter of the ion (table 10.4)

8. Different ways to calculate  i Limiting law Debye-Huckel Davies TJ, SIT models Pitzer, HKW models

10. Neutral species Setchnow equation: Log  n =k s I For activity coefficient (see table 4-2 for selected coefficients)

11. Law of Mass Action Getting ‘out’ of the standard state: Accounting for free energy of ions ≠ 1:  =  0 + RT ln P Bear in mind the difference between the standard state G 0 and  0 vs. the molar property G and  (not at standard state  25 C, 1 bar, a mole) G P – G 0 = RT(ln P – ln P 0 )G P – G 0 = RT ln P

12. Equilibrium Constant For a reaction of ideal gases, P becomes: for aA + bB  cC + dD Restate the equation as:  G R –  G 0 R = RT ln Q AT equilibrium,  G R =0, therefore:  G 0 R = -RT ln K eq where K eq is the equilibrium constant

13. Assessing equilibrium If  G R –  G 0 R = RT ln Q, and at equilibrium  G 0 R = 0, then: Q=K Q  reaction quotient, aka Ion Activity Product (IAP) is the product of all products over product of all reactants at any condition K  aka K eq, same calculation, but AT equilibrium

14. Solubility Product Constant For mineral dissolution, the K is K sp, the solubility product constant Use it for a quick look at how soluble a mineral is, often presented as pK (table 10.1)  G 0 R = RT ln K sp Higher values  more soluble CaCO 3(calcite)  Ca 2+ + CO 3 2- Fe 3 (PO 4 ) 2 *8H 2 O  3 Fe 2+ + 2 PO 4 3- + 8 H 2 O

15. Ion Activity Product For reaction aA + bB  cC + dD: For simple mineral dissolution, this is only the product of the products  activity of a solid phase is equal to one CaCO 3  Ca 2+ + CO 3 2- IAP = [Ca 2+ ][CO 3 2- ]

16. Saturation Index When  G R =0, then ln Q/K eq =0, therefore Q=K eq. For minerals dissolving in water: Saturation Index (SI) = log Q/K or IAP/K eq When SI=0, mineral is at equilibrium, when SI<0 (i.e. negative), mineral is undersaturated

17. Calculating K eq  G 0 R = -RT ln K eq Look up G 0 i for each component in data tables (such as Appendix F3-F5 in your book) Examples: CaCO 3(calcite) + 2 H +  Ca 2+ + H 2 CO 3(aq) CaCO 3(aragonite) + 2 H +  Ca 2+ + H 2 CO 3(aq) H 2 CO 3(aq)  H 2 O + CO 2(aq) NaAlSiO 4(nepheline) + SiO 2(quartz)  NaAlSi 3 O 8(albite)