Pb 10 3 1 310 3 1 Pb 2 1 PbOH 1 PbOH 2 1 PbCl 1 7 Cl 10 2 2 Cl 1 PbCl 1 8 Note

Pb 10 3 1 310 3 1 pb 2 1 pboh 1 pboh 2 1 pbcl 1 7 cl

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Pb (10 -3 )(1) + (3*10 -3 )(1) = {Pb 2+ }(1) + {PbOH + }(1) + {Pb(OH) 0 2 }(1) + { PbCl + }(1) 7 Cl (10 -2 )(2) = {Cl - }(1) + {PbCl + }(1) 8 Note that the type d-species PbCl + had to be included twice – that is in the MB equations of both Cd and Cl Step-4 : Write the charge balance (CB) . 4
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The equation to be written here is a balance in concentration of charges and not a balance of the concentration of individual chemical species. Aquatic systems are electrically neutral. Therefore, the 2 sides of the equations below should be equal. {H + }(1) + {Pb 2+ }(2) + {PbOH + }(1) + {PbCl + }(1) = {OH - }(1) + {CH 3 COO - }(1) + {Cl - }(1) Eq.9 . The coefficient (2) following Pb 2+ reflects the fact that each mole of Pb 2+ accounts for 2 positive charges and should be used that way. 5
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Step-5 : MINTEQ Problem set up for MINTEQ a. pH to be calculated b. Ionic strength to be calculated c. Select concentrations units (e.g. mol/L) d. Temperature at default value 25 degrees C e. Components i. Check the “show organic components” box ii. Enter the following PbCl 2 = Pb 2+ + 2Cl - 0.01 0.01 2*(0.01)=0.02 M Pb(Ac) 2 = Pb 2+ + 2Ac - 0.003 0.003 2*(0.003)=0.006 M Pb+2…………………………… (0.01 + 0.003) = 0.013 M Cl-1……………………………... (2*0.010) = 0.020 M Acetate -1...……………………. (2*0.003) = 0.006 M f. View/Edit List H+1…………………….……….0.00 Pb+2…………………………… 0.013 Cl-1…………………………….. 0.020 Acetate -1...…………………….0.006 g. Run MINTEQ and MINTEQ OUTPUT (see next page) 6
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  • Fall '08
  • CHADIK
  • Chemistry, #, Electric charge, 0.02 m, 0.020 M, 0.013 m

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