Water Quality Chem Solution Set 7 1 Hydroxyapatite in equilibrium with pure H2O and atm. CO2 ph = 7.11 Ca+2 6.10E-05 5.71E-05-4.243 CaCO3 (aq) 3.83E-09 3.83E-09-8.417 CaH2PO4+ 2.52E-08 2.48E-08-7.606 CaHCO3+ 5.00E-08 4.92E-08-7.308 CaHPO4 (aq) 4.09E-07 4.09E-07-6.388 CaOH+ 1.49E-10 1.46E-10-9.835 CaPO4-1.41E-08 1.39E-08-7.857 CO3-2 4.31E-08 4.04E-08-7.393 H+1 7.97E-08 7.84E-08-7.106 H2CO3* (aq) 1.19E-05 1.19E-05-4.924 H2PO4-1.97E-05 1.94E-05-4.713 H3PO4 2.14E-10 2.14E-10-9.67 HCO3-6.87E-05 6.76E-05-4.17 HPO4-2 1.67E-05 1.57E-05-4.805 OH- 1.31E-07 1.28E-07-6.891 PO4-3 9.77E-11 8.43E-11-10.074 Significant species in boldface 3.64E-05 M 3.46 mg/l For comparison, a highly eutrophic lake would have total P on the order of 0.1 mg/l, so this is a pollution concern.
2 Ca-H 30 mgcaco3/l = 3.0E-04 M Mg-H 20 mgcaco3/l = 2.0E-04 M SO4-S 67.2 mg/l = 2.1E-03 M Alk 8.00E-04 eq/l ph = 8.13 Ca+2 2.38E-04 1.75E-04-3.758 CaCO3 (aq) 1.34E-06 1.34E-06-5.874 CaH2PO4+ 5.06E-11 4.68E-11-10.33 CaHCO3+ 1.74E-06 1.61E-06-5.794 CaHPO4 (aq) 8.24E-09 8.25E-09-8.083 CaOH+ 5.17E-09 4.78E-09-8.32 CaPO4-3.23E-09 2.99E-09-8.524 CaSO4 (aq) 5.89E-05 5.90E-05-4.229 CO3-2 6.28E-06 4.61E-06-5.336 H+1 7.93E-09 7.34E-09-8.134 H2CO3* (aq) 1.19E-05 1.19E-05-4.924 H2PO4-1.29E-08 1.20E-08-7.922 H3PO4 1.23E-14 1.23E-14-13.908 HCO3-7.80E-04 7.22E-04-3.142 HPO4-2 1.41E-07 1.03E-07-6.986 HSO4-1.14E-09 1.06E-09-8.976 Mg+2 1.66E-04 1.22E-04-3.915 Mg2CO3+2 3.62E-10 2.66E-10-9.576 MgCO3 (aq) 4.66E-07 4.67E-07-6.331 MgHCO3+ 9.73E-07 9.01E-07-6.045 MgHPO4 (aq) 7.92E-09 7.93E-09-8.101 MgOH+ 6.86E-08 6.35E-08-7.197 MgPO4-3.52E-11 3.26E-11-10.487 MgSO4 (aq) 3.26E-05 3.26E-05-4.486 OH- 1.48E-06 1.37E-06-5.863 PO4-3 1.19E-11 5.94E-12-11.227 SO4-2 2.01E-03 1.47E-03-2.832 1.73E-07 M 0.016 mg/l Total P now over 100x lower due to higher ph This total P is still fairly high for a lake but might be acceptable for discharge to a suitable water body.
3 ph = 4 Ca+2 5.49E-02 1.89E-02-1.724 CaH2PO4+ 1.20E-02 9.15E-03-2.038 CaHPO4 (aq) 1.15E-04 1.18E-04-3.926 CaOH+ 4.95E-11 3.79E-11-10.421 CaPO4-4.12E-09 3.15E-09-8.501 CaSO4 (aq) 1.24E-03 1.27E-03-2.895 H+1 1.31E-04 1.00E-04-4 H2PO4-2.83E-02 2.17E-02-1.664 H3PO4 2.96E-04 3.05E-04-3.516 HPO4-2 4.00E-05 1.37E-05-4.862 HSO4-3.76E-06 2.88E-06-5.541 Mg+2 1.96E-04 6.74E-05-4.172 MgHPO4 (aq) 5.67E-07 5.84E-07-6.234 MgOH+ 3.37E-12 2.58E-12-11.589 MgPO4-2.30E-13 1.76E-13-12.755 MgSO4 (aq) 3.50E-06 3.61E-06-5.442 OH- 1.31E-10 1.01E-10-9.997 PO4-3 6.40E-13 5.80E-14-13.237 SO4-2 8.57E-04 2.95E-04-3.531 4.07E-02 M 3865 mg/l YOW!!! No wonder the rock phosphate crumbles so well. This is not only a major pollutant souce but also a waste of P 4 Mix HAp and H2SO4 Finite Hap = (31 g/l)/(502 g/mol) = 0.060 M (In problem statement Hap conc is rounded to 0.1 M but the difference does not really affect your answer) H2SO4 = 0.5 M Initial run: NO SOLIDS allowed to ppt: ph = 0.99 Ca+2 1.34E-01 4.05E-02-1.393 CaH2PO4+ 1.50E-02 1.11E-02-1.953 CaHPO4 (aq) 1.21E-07 1.40E-07-6.855 CaOH+ 1.03E-13 7.63E-14-13.118 CaPO4-4.86E-15 3.61E-15-14.443 CaSO4 (aq) 1.61E-01 1.86E-01-0.731 H+1 1.39E-01 1.03E-01-0.987 H2PO4-1.66E-02 1.23E-02-1.91 H3PO4 1.54E-01 1.78E-01-0.749 HPO4-2 2.50E-08 7.56E-09-8.122 HSO4-2.73E-01 2.02E-01-0.694 OH- 1.27E-13 9.45E-14-13.025 PO4-3 4.56E-19 3.09E-20-19.51 SO4-2 6.64E-02 2.01E-02-1.698
Hap now UNDERsat'd by 20 orders of magnitude (SI = -20) Gypsum and anhydrite are oversat'd, but anhydrite is the "anhydrous" (water-free) form of CaSO4 so it cannot form in aqueous soln. Therefore, allow only gypsum to ppt: ph = 0.8 Ca+2 1.10E-02 3.22E-03-2.492 CaH2PO4+ 8.07E-04 5.93E-04-3.227 CaHPO4 (aq) 4.49E-09 4.79E-09-8.32 CaOH+ 5.31E-15 3.90E-15-14.409 CaPO4-1.08E-16 7.95E-17-16.1 CaSO4 (aq) 5.64E-03 6.01E-03-2.221 H+1 2.18E-01 1.60E-01-0.795 H2PO4-1.12E-02 8.23E-03-2.084 H3PO4 1.74E-01 1.86E-01-0.732 HPO4-2 1.12E-08 3.26E-09-8.487 HSO4-1.74E-01 1.28E-01-0.894 OH- 8.27E-14 6.08E-14-13.216 PO4-3 1.37E-19 8.57E-21-20.067 SO4-2 2.80E-02 8.16E-03-2.089 HAp now undersatd' by 27 orders of magnitude (SI = -27) 100% of PO4 is soluble; 94% of it is phosphoric acid. Gypsum pptn lowers the soluble Ca2+ by 94% and SO4 by about 59% (which is helpful to waste disposal); 0.29 mol/l of gypsum precipitate out as a finite solid = 50 g/l RATIO: Gypsum/HAp Mass Ratio = (50 g/l)/(31 g/l) = 1.6 tonnes gypsum/tonnehap Note that although gypsum takes out almost all Ca2+, it leaves about 40% of the SO4, mostly as free sulfuric acid, hence the extreme ph. CONCLUSION: The dissolution of phosphate bearing minerals is complete and most of the Ca and SO4 are converted to solid gypsum for disposal as "phosphogypsum" (gypsum contaminated with residual phosphates and acid.
5 Looking now just at leachate itself (no solids in contact with it) Therefore, omit gypsum since total Ca2+ and SO4 given. ph = 5.2 Ca+2 5.23E-03 2.98E-03-2.526 CaH2PO4+ 1.83E-05 1.59E-05-4.799 CaHPO4 (aq) 3.24E-06 3.26E-06-5.487 CaOH+ 1.09E-10 9.48E-11-10.023 CaPO4-1.58E-09 1.37E-09-8.862 CaSO4 (aq) 1.95E-03 1.96E-03-2.708 H+1 7.26E-06 6.31E-06-5.2 H2PO4-2.74E-04 2.38E-04-3.623 H3PO4 2.10E-07 2.11E-07-6.675 HPO4-2 4.20E-06 2.39E-06-5.621 HSO4-2.04E-06 1.77E-06-5.751 OH- 1.84E-09 1.60E-09-8.797 PO4-3 5.67E-13 1.60E-13-12.796 (Output shows leachate slightly understa'd w/r/t gypsum which makes sense if the leachate washes through before it can completely equilibrate) The acidity is the sum of the species protonated at the equivalence point: [H+] + 3[H3PO4] + 2[H2PO4-] + [HPO4-2] + [HSO4-] + 2[CaH2PO4+] + [CaHPO4] (aq) Sum = 6.02E-04 eq/l = to Acidity (= -Alk) This should be equal to the TotH+ in the system. You can check this directly in MINTEQ by going to the Equilibrated Mass Distribution output page, and looking at the total concentration of the COMPONENT (not speecies) H+. On that page here you'd see: Component Total Diss % Diss Ca+2 7.20E-03 100 etc H+1 6.02E-04 100 PO4-3 3.00E-04 100 SO4-2 7.00E-03 100 Ah ha! Find the exact same value! So no need to do tedious adding of spp. Just find TotH+ here. This is the amount of excess strong acid present in solution compared to a solution with no net acidity or alkalinity. Thus is you added 6.04e-04 eq/l of base to this leachate the ph would rise to 7.0 in the absence of CO2 and to 5.6 if in equlibrium with atmospheric CO2.
6 Pickle liquor is 1.3 lb/gal Fe-total = 2.8 mol/l Fe-total = 1.0 mol/l Fe-active SO4 is 3/2(Fe) but unclear which Fe to use. Does not really affect results, though. I'll use 3/2(Fe-total) SO4 = 4.2 M Residual acid = 1% w/w = 10 g/kg ~ 10 g/l = 0.10 M H2SO4 = 0.20 M H+-total To get a mix of leachate and pickle liquor with Fe3+T = PO4T =0.3 mm will need to mix in pickle liquor at a ratio of 3e-04 to 1. (dilute P.L with leachate at a factor of 3e-04) FeT = 0.3 mm SO4 = 1.3 mm + pre 7.0 mm = 8.3 mm H+ 0.06 mm + pre 0.6 mm = 0.66 mm NOTE: You could easily forget that there was some (not insignificant) acidity from previous step;since this may have been unclear I will accept a nswers for either H+ = 0.66 mm or H+ = 0.06 mmm For H+ = 0.66 mm Find quite a few oversat'd Fe solids including strengite (FePO4) From problem statement you could figure out to allow Strengite to ppt. Note that although 2 Fe-hydroxide solids are more oversat'd than strengite, they are both highly crystalline forms that are unlikely to precipitate compared to strengite, The mostl likely Fe(OH )3 solid to form is ferrihydrite (amorphous stuff) which is much less oversat'd than strengite. From the Distribution of Components output page we see that Fe and PO4 are both 99.4 precipitated so it looks like it works well. Solution speciation after pptn is: ph = 3.37 Ca+2 5.05E-03 2.82E-03-2.549 CaH2PO4+ 1.07E-07 9.26E-08-7.033 CaHPO4 (aq) 2.78E-10 2.80E-10-9.553 CaOH+ 1.53E-12 1.32E-12-11.878 CaPO4-2.01E-15 1.74E-15-14.76 CaSO4 (aq) 2.15E-03 2.16E-03-2.665 Fe(OH)2+ 2.09E-07 1.81E-07-6.743 Fe(OH)3 (aq) 2.36E-13 2.37E-13-12.625 Fe(OH)4-1.28E-17 1.10E-17-16.957 Fe(SO4)2-5.80E-08 5.02E-08-7.3 Fe+3 6.89E-08 1.87E-08-7.729 Fe2(OH)2+4 2.47E-11 2.42E-12-11.616 Fe3(OH)4+5 3.75E-15 9.93E-17-16.003 FeH2PO4+2 9.25E-10 5.17E-10-9.286 FeHPO4+ 3.80E-08 3.29E-08-7.483 FeOH+2 7.43E-07 4.16E-07-6.381 FeSO4+ 8.10E-07 7.01E-07-6.154 H+1 4.95E-04 4.28E-04-3.368 H2PO4-1.69E-06 1.47E-06-5.834 H3PO4 8.78E-08 8.82E-08-7.054 HPO4-2 3.88E-10 2.17E-10-9.664 HSO4-1.62E-04 1.40E-04-3.854 OH- 2.72E-11 2.35E-11-10.629 PO4-3 7.89E-19 2.13E-19-18.671 SO4-2 5.98E-03 3.35E-03-2.475
1.90E-06 M 0.181 mg/l This is less than 1% of original leachate p. Although it is high for a P level in a lake, it may be good enough for discharge to a diluting water body. For H+ = 0.06 mm ph = 4.4 (a bit higher, of course) 7.20E-07 M 0.068 mg/l Even better than above because the ph is (erroneously) higher. Suggests that adding base would improve P removal. This is likely true, although with too much base, you might get Fe precipitating as hydroxides instead of FePO4. Practical solution would be to increase dose of Fe (pickle liquor is cheap), and do some lab tests to find the optimal ph range for the desired P removal.