JYU Applied Geochemistry & Lab Ch.6 Solubility Diagram Part 1

 Solubility: the property of a solid, liquid, or gaseous substance called solute to dissolve in another solid, liquid, or gaseous substance called solvent to form a solution. This chapter specifically deals with solid solute in water to form aqueous solutions. The solubility also often means the extent of dissoslution.  Types of dissolution  Congruent dissolution: the composition of the dissolved solid matches that of solutes produced  Incongruent dissolution: doesn’t match 1. Definitions

 For a dissolution reaction  ML = M + + L -  K sp = [M + ][L - ]  If there is neither M + nor L - preexisting in the solution  Let solubility=m,  Then, m=(K sp ) 0.5  If there is L - in [L - ]’preexisting in the solution  [L - ]=[L - ]’+[M + ]  Then, K sp = [M + ][L - ] = [M + ] 2 + [M + ] [L - ]’  [M + ]=m, solve for m 2. Calculation of the Solubility

 Precipitation vs. dissolution  Undersaturation  dissolution  Oversaturation (supersaturation)  precipitation  Saturation  “equilibrium”  Expression of the saturation  Saturation ratio (SR) = IAP/Ksp  SR<1: US  SR>1: OS  SR=1: EQ  Saturation index (SI) = log(SR)  SI<0: US  SI>0: OS  SI=0: EQ  Affinity (A) = -RT*SI  A>0: US  A<0: OS  A=0: EQ 3. Saturation

 Represent the boundaries of saturation of a solid in pH- metal conc. space  Procedures  Setup a system of interest  e.g. Al-O-H system  List possible species (including hydroxy complexes)  e.g. Al 3+, AlOH 2+, Al(OH) 2 +, Al(OH) 3 0, Al(OH) 4 -, Al(OH) 5 2-  Construct reactions between the solid phase and the each of the above species  Al(OH) 3 (gibbsite) +3H + = Al H 2 O(R1)  Al(OH) 3 (gibbsite) +2H + = AlOH H 2 O(R2)  Al(OH) 3 (gibbsite) +1H + = Al(OH) H 2 O(R3)  Al(OH) 3 (gibbsite) = Al(OH) 3 0 (R4)  Al(OH) 3 (gibbsite) +H 2 O = Al(OH) H + (R5)  Al(OH) 3 (gibbsite) + 2H 2 O = Al(OH) H + (R6) 4. Solubility Diagram

 Calculate K values and express them in terms of pH and Al-conc for the above reactions  (R1)  G r o = kcal/mole  K = (by law of mass action),  log[Al 3+ ]=-3pH  (R2)  G r o = kcal/mole  K = ,  log[AlOH 2+ ]=-2pH  (R3)  G r o = 1.39 kcal/mole  K = ,  log[Al(OH) 2 + ]=-pH  (R4)  G r o = 9.16 kcal/mole  K = ,  log[Al(OH) 3 0 ]=-6.72  (R5)  G r o = 20.5 kcal/mole  K = ,  log[Al(OH) 4 - ]=pH  (R6)  G r o = kcal/mole  K = ,  log[, Al(OH) 5 2- ]=2pH -26.0