Water Quality Chem Solution Set 7

Transcription

1 Water Quality Chem Solution Set 7 1 Hydroxyapatite in equilibrium with pure H2O and atm. CO2 ph = 7.11 Ca E E CaCO3 (aq) 3.83E E CaH2PO E E CaHCO E E CaHPO4 (aq) 4.09E E CaOH+ 1.49E E CaPO4-1.41E E CO E E H E E H2CO3* (aq) 1.19E E H2PO4-1.97E E H3PO4 2.14E E HCO3-6.87E E HPO E E OH- 1.31E E PO E E 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 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 E E CaCO3 (aq) 1.34E E CaH2PO E E CaHCO E E CaHPO4 (aq) 8.24E E CaOH+ 5.17E E CaPO4-3.23E E CaSO4 (aq) 5.89E E CO E E H E E H2CO3* (aq) 1.19E E H2PO4-1.29E E H3PO4 1.23E E HCO3-7.80E E HPO E E HSO4-1.14E E Mg E E Mg2CO E E MgCO3 (aq) 4.66E E MgHCO E E MgHPO4 (aq) 7.92E E MgOH+ 6.86E E MgPO4-3.52E E MgSO4 (aq) 3.26E E OH- 1.48E E PO E E SO E E E-07 M 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 3 ph = 4 Ca E E CaH2PO E E CaHPO4 (aq) 1.15E E CaOH+ 4.95E E CaPO4-4.12E E CaSO4 (aq) 1.24E E H E E-04-4 H2PO4-2.83E E H3PO4 2.96E E HPO E E HSO4-3.76E E Mg E E MgHPO4 (aq) 5.67E E MgOH+ 3.37E E MgPO4-2.30E E MgSO4 (aq) 3.50E E OH- 1.31E E PO E E SO E E E-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) = 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 E E CaH2PO E E CaHPO4 (aq) 1.21E E CaOH+ 1.03E E CaPO4-4.86E E CaSO4 (aq) 1.61E E H E E H2PO4-1.66E E H3PO4 1.54E E HPO E E HSO4-2.73E E OH- 1.27E E PO E E SO E E

4 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 E E CaH2PO E E CaHPO4 (aq) 4.49E E CaOH+ 5.31E E CaPO4-1.08E E CaSO4 (aq) 5.64E E H E E H2PO4-1.12E E H3PO4 1.74E E HPO E E HSO4-1.74E E OH- 8.27E E PO E E SO E E 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 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 E E CaH2PO E E CaHPO4 (aq) 3.24E E CaOH+ 1.09E E CaPO4-1.58E E CaSO4 (aq) 1.95E E H E E H2PO4-2.74E E H3PO4 2.10E E HPO E E HSO4-2.04E E OH- 1.84E E PO E E (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 E etc H E PO E SO E 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 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 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 E E CaH2PO E E CaHPO4 (aq) 2.78E E CaOH+ 1.53E E CaPO4-2.01E E CaSO4 (aq) 2.15E E Fe(OH) E E Fe(OH)3 (aq) 2.36E E Fe(OH)4-1.28E E Fe(SO4)2-5.80E E Fe E E Fe2(OH) E E Fe3(OH) E E FeH2PO E E FeHPO E E FeOH E E FeSO E E H E E H2PO4-1.69E E H3PO4 8.78E E HPO E E HSO4-1.62E E OH- 2.72E E PO E E SO E E

7 1.90E-06 M 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 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.