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 Elemental classification  Lewis acid/base  Pearson’s hard/soft metals  Ionic and covalent index  Ionic potentials  Electron negativity  Earth Scientist's.

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Presentation on theme: " Elemental classification  Lewis acid/base  Pearson’s hard/soft metals  Ionic and covalent index  Ionic potentials  Electron negativity  Earth Scientist's."— Presentation transcript:

1  Elemental classification  Lewis acid/base  Pearson’s hard/soft metals  Ionic and covalent index  Ionic potentials  Electron negativity  Earth Scientist's Periodic Table of the Elements and Their Ions  Know common minerals and aqueous species in soils - Pb, Cd, Cu and Zn  Familiar with equilibrium reactions  Use log activity-pH graphs Trace metals review

2 Adriano (2001) /metal HSAB-the 13 th most cited paper in JACS low En, large size, high polarization, easily oxidized

3 Lindsay, 1979 CaCO 3 + 2 H +  Ca 2+ + CO 2 + H 2 O log(Ca 2+ ) = log k° - 2 pH – log (CO 2 )

4 PbO + 2H + Pb 2+ + H 2 O K° = (Pb 2+ )/(H + ) 2 log K° = log (Pb 2+ ) - 2 log (H + ) log K° = log (Pb 2+ ) + 2 pH log (Pb 2+ ) = log K° - 2 pH PbSO 4 Pb 2+ + SO 4 2- K° = (Pb 2+ )*(SO 4 2- ) log K° = log (Pb 2+ ) + log (SO 4 2- ) log (Pb 2+ ) = log K° - log (SO 4 2- )

5 PbCO 3 + 2 H +  Pb 2+ + CO 2 + H 2 O log(Pb 2+ ) = log k° - 2 pH – log (CO 2 ) Pb 5 (PO 4 ) 3 Cl+ 6 H +  5Pb 2+ + 3 H 2 PO 4 - + Cl - log (Pb 2+ )= 1/5 [log K° - 6 pH –3 log (H 2 PO 4 - ) - log (Cl - )]

6 Use log activity-pH graphs Pb 5 (PO 4 ) 3 Cl + 6 H +  5Pb 2+ + 3 H 2 PO 4 - + Cl - log (Pb 2+ ) = 1/5 [log K° - 6 pH – 3 log (H 2 PO 4 - ) - log (Cl - )] Log (Cl-) = -3 M Ca 5 (PO 4 ) 3 OH + 8 H +  5Ca 2+ + 3 H 2 PO 4 - + H 2 O Hydroxyapatite log (H 2 PO 4 - ) = 1/3 [log K° - 8 pH – 5 log (Ca 2+ ) ] Soil-Ca = 10 -2.5 MLog (Ca 2+ ) = -2.5 M CaCO 3 + 2 H +  Ca 2+ + CO 2 + H 2 O log(Ca 2+ ) = log k° - 2 pH – log (CO 2 )

7 Use log activity-pH graphs Pb 2+ H 2 PO 4 -  FePO 4 Ca 5 (PO 4 ) 3 OH Soil-Ca CaCO 3 Soil-Fe

8 Pb 5 (PO 4 ) 3 Cl+ 6 H +  5Pb 2+ + 3 H 2 PO 4 - + Cl - log (Pb 2+ )= 1/5 [log K° - 6 pH –3 log (H 2 PO 4 - ) - log (Cl - )]

9 Adriano (2001)

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11 Pb 2+ + 4OH - Pb(OH) 4 2- K  = (Pb(OH) 4 2- )/(Pb 2+ )*4(OH - ) log K  = log (Pb(OH) 4 2- ) - 2 log (Pb 2+ ) + 4 pH log (Pb(OH) 4 2- ) = log K  - 2 log (Pb 2+ ) + 4 pH

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