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JYU Applied Geochemistry & Lab Ch.6 Solubility Diagram Part 1
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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
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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
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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
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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 3+ + 3H 2 O(R1) Al(OH) 3 (gibbsite) +2H + = AlOH 2+ + 2H 2 O(R2) Al(OH) 3 (gibbsite) +1H + = Al(OH) 2 + + H 2 O(R3) Al(OH) 3 (gibbsite) = Al(OH) 3 0 (R4) Al(OH) 3 (gibbsite) +H 2 O = Al(OH) 4 - + H + (R5) Al(OH) 3 (gibbsite) + 2H 2 O = Al(OH) 5 2- + 2H + (R6) 4. Solubility Diagram
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Calculate K values and express them in terms of pH and Al-conc for the above reactions (R1) G r o = -11.29 kcal/mole K = 10 8.28 (by law of mass action), log[Al 3+ ]=-3pH +8.28 (R2) G r o = -4.44 kcal/mole K = 10 3.26, log[AlOH 2+ ]=-2pH +3.26 (R3) G r o = 1.39 kcal/mole K = 10 -1.02, log[Al(OH) 2 + ]=-pH -1.02 (R4) G r o = 9.16 kcal/mole K = 10 -6.72, log[Al(OH) 3 0 ]=-6.72 (R5) G r o = 20.5 kcal/mole K = 10 -15.1, log[Al(OH) 4 - ]=pH -15.1 (R6) G r o = 35.41 kcal/mole K = 10 -26.0, log[, Al(OH) 5 2- ]=2pH -26.0
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