1. Acid-Base Geochemistry Arrhenius’ definition: –Acid  any compound that releases a H + when dissolved in water –Base  any compound that releases an OH - when dissolved in water Bronstead-Lowry’s definition: –Acid  donates a proton –Base  receive/accept a proton Lewis’ definition: –Acid  electron pair donor acceptor –Base  electron pair donor

2. Conjugate Acid-Base pairs Generalized acid-base reaction: HA + B  A + HB A is the conjugate base of HA, and HB is the conjugate acid of B. More simply, HA  A - + H + HA is the conjugate acid, A - is the conjugate base H 2 CO 3  HCO 3 - + H +

3. Hydrolysis M z + H 2 O  M(OH) z-1 + H + Reaction of a cation, which generates a H + from water is a hydrolysis reaction Described by the equilibrium constant Ka Hydrolysis also describes an organic reaction in which the molecule is cleaved by reaction with water…

4. AMPHOTERIC SUBSTANCE Now consider the acid-base reaction: NH 3 + H 2 O  NH 4 + + OH - In this case, water acts as an acid, with OH - its conjugate base. Substances that can act as either acids or bases are called amphoteric. Bicarbonate (HCO 3 - ) is also an amphoteric substance: Acid: HCO 3 - + H 2 O  H 3 O + + CO 3 2- Base: HCO 3 - + H 3 O +  H 2 O + H 2 CO 3 0

5. Strong Acids/ Bases Strong Acids more readily release H+ into water, they more fully dissociate –H 2 SO 4  2 H + + SO 4 2- Strong Bases more readily release OH- into water, they more fully dissociate –NaOH  Na + + OH - Strength DOES NOT EQUAL Concentration!

6. Acid-base Dissociation For any acid, describe it’s reaction in water: –H x A + H 2 O  x H + + A - + H 2 O –Describe this as an equilibrium expression, K (often denotes K A or K B for acids or bases…) Strength of an acid or base is then related to the dissociation constant  Big K, strong acid/base! pK = -log K  as before, lower pK=stronger acid/base!

7. LOTS of reactions are acid-base rxns in the environment!! HUGE effect on solubility due to this, most other processes Geochemical Relevance?

8. Dissociation of H 2 O H 2 O  H + + OH - K eq = [H + ][OH - ] log K eq = -14 = log K w pH = - log [H+] pOH = - log [OH-] pK = pOH + pH = 14 If pH =3, pOH = 11  [H + ]=10 -3, [OH - ]=10 -11 Definition of pH

9. pH Commonly represented as a range between 0 and 14, and most natural waters are between pH 4 and 9 Remember that pH = - log [H + ] –Can pH be negative? –Of course!  pH -3  [H + ]=10 3 = 1000 molal? –But what’s   ?? Turns out to be quite small  0.002 or so…

10. pK x ? Why were there more than one pK for those acids and bases?? H 3 PO 4  H + + H 2 PO 4 - pK 1 H 2 PO 4 -  H + + HPO 4 2- pK 2 HPO 4 1-  H+ + PO 4 3- pK 3

11. 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 - …

12. BUFFERING When the pH is held ‘steady’ because of the presence of a conjugate acid/base pair, the system is said to be buffered In the environment, we must think about more than just one conjugate acid/base pairings in solution Many different acid/base pairs in solution, minerals, gases, can act as buffers…

13. Henderson-Hasselbach Equation: When acid or base added to buffered system with a pH near pK (remember that when pH=pK HA and A- are equal), the pH will not change much When the pH is further from the pK, additions of acid or base will change the pH a lot

14. Buffering example Let’s convince ourselves of what buffering can do… Take a base-generating reaction: – Albite + 2 H 2 O = 4 OH- + Na + + Al 3+ + 3 SiO 2(aq) –What happens to the pH of a solution containing 100 mM HCO3- which starts at pH 5?? –pK 1 for H 2 CO 3 = 6.35

15. Think of albite dissolution as titrating OH - into solution – dissolve 0.05 mol albite = 0.2 mol OH - 0.2 mol OH-  pOH = 0.7, pH = 13.3 ?? What about the buffer?? –Write the pH changes via the Henderson-Hasselbach equation 0.1 mol H 2 CO 3(aq), as the pH increases, some of this starts turning into HCO3 - After 12.5 mmoles albite react (50 mmoles OH-): –pH=6.35+log (HCO3 - /H 2 CO 3 ) = 6.35+log(50/50) After 20 mmoles albite react (80 mmoles OH - ): –pH=6.35+log(80/20) = 6.35 + 0.6 = 6.95

16. Bjerrum Plots 2 D plots of species activity (y axis) and pH (x axis) Useful to look at how conjugate acid-base pairs for many different species behave as pH changes At pH=pK the activity of the conjugate acid and base are equal

17. Bjerrum plot showing the activities of reduced sulfur species as a function of pH for a value of total reduced sulfur of 10 -3 mol L -1.

18. 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!