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ENVR 403 Introduction to Environmental Chemistry. Philip C. Singer Department of Environmental Sciences and Engineering phil_singer@unc.edu 919-966-3865 April 2010. Topics. 1. Chemistry of Natural Waters 2. Purification of Water. 1. Chemistry of Natural Waters. Redox Equilibria
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ENVR 403Introduction to Environmental Chemistry Philip C. Singer Department of Environmental Sciences and Engineering phil_singer@unc.edu 919-966-3865 April 2010
Topics 1. Chemistry of Natural Waters 2. Purification of Water
1. Chemistry of Natural Waters Redox Equilibria -speciation -dissolved oxygen resources Acid-Base Equilibria -speciation -the carbonate system and alkalinity Solubility Equilibria -metal solubility -aqueous lead (Pb) solubility
pH and pE as Master Variables H2CO3 = HCO3- + H+ Ka1 pH = pKa1 + Log [HCO3-]/[H2CO3] NO3- + 10H+ + 8e- = NH4+ + 3H2O Kred pE = -1/8 (pKred) + 1/8 (Log [NO3-] [H+]10/[NH4+]) pEo = -1/n (Log Kred) pE = pEo + 1/n Log [Ox]/[Red]) Analogous to Nernst Equation EH = EoH + 2.3 RT/nF (Log [Ox]/[Red])
Electroneutrality Equation (Charge Balance) for Natural waters Major Cations: Na+, K+, Ca2+, Mg2+ Major Anions: Cl-, SO42- , NO3-, HCO3-, CO32- Charge Balance [Na+]+ [K +] + 2[Ca2+] + 2[Mg2+]+ [H +] = [Cl- ]+ 2[SO42-]+ [NO3-] + [HCO3-] + 2[CO32-]+ [OH-] Cb-Ca = Acid neutralizing Capacity = Alkalinity = [HCO3-] + 2[CO32-]+ [OH-] – [H +]
What is the pH of pure rainwater? Pure rainwater contains only dissolved CO2 in equilibrium with the atmosphere. H2CO3 = KH PCO2 ~ 10-5M What is the pH of a 10-5M solution of H2CO3? Cb-Ca = Alkalinity = 0 [H+] = [HCO3-] + 2[CO32-] + [OH-] = K1[H+]C/D + 2 K1K2C/D where D = [H+] 2 + K1 [H+] + K1K2 Answer: pH 5.65
If a body of water has a pH of 7.5 and an alkalinity of 2x10-3 equivalents/L, what is the equilibrium pH of the water if 10-3 eq./L of acid is added to the water? Use initial pH and given alkalinity to find total carbonate concentration C C = (Alkalinity + [H+] - [OH-]) /((K1[H+]/D) + 2K1K2/D) = 2.12x10-3 M Added acid consumes alkalinity New alkalinity = 2x10-3 – 10-3 = 10-3 eq/L Find new pH from alkalinity expression Alkalinity = [HCO3-] + 2[CO32-]+ [OH-] – [H +] 10-3 = ((K1[H+]/D) + 2K1K2/D) 2.12 x10-3 + [OH-] – [H +] Find pH = 6.35
Solubility of CaCO3(s) • Kso = [Ca2+][CO32-] • [Ca2+] = Kso/[CO32-] = Kso/α2[TOTCO3] • Solubility of Ca increases with decreasing pH
Lead in Drinking Water Pb(II) Pb(IV) OCl- Cl- Particulate CO32-, PO43- CO32-, PO43-, Cl- Pb2+ Pb 2+ Pb(IV)O2, Pb3(CO3)2OH2, PbCO3, Pb5(PO4)3OH Lead Pipe Pb(0) • Lead pipe (Pb0) develops scales of corrosion products. • The products formed (oxides, carbonates, and phosphates) depend on the water chemistry. • Changes in distribution system water chemistry can destabilize corrosion products. • Optimization of water chemistry can mitigate lead release to solution.
Solubility of Potential Lead-containing Phases Hydrocerussite (Pb3(CO3)2(OH)2) with 50 mg/L DIC Hydroxylpyromorphite (Pb5(PO4)3OH) with 1 mg/L P Plattnerite (PbO2) -2 -3 Litharge (PbO) -4 (M) -5 diss -6 Log[Pb] 15 ug/L Pb -7 -8 -9 4 5 6 7 8 9 10 11 pH
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Underlying Chemical Theory: Hydrolysis of Urea • NH2(CO)NH2+2H2O2NH4++CO3-2 • Urease is urea-hydrolyzing enzyme • Urease decomposes urea into ammonium and carbonate, causing the pH to increase • Increase in pH and addition of ammonium and carbonate leads to over-saturation with respect to several different minerals, including struvite (MgNH4PO4) and calcium hydroxyapatite Ca5(PO4)3(OH)2.
O P Mg Ca SEM Results for RDU Terminal A Composition Similar to Struvite, MgNH4PO4 (s)