Introduction, Concepts in Aqueous Chemistry. Why is Environmental Chemistry a required course for environmental engineers/scientists?

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1 Introduction, Concepts in Aqueous Chemistry Introduction: Why is Environmental Chemistry a required course for environmental engineers/scientists? Environmental Engineers 1) Predict, model & measure changes in the natural/ manmade environment as a result of human activities. 2) Design and operate treatment systems to minimize pollution problems. Chemistry is central to both of these goals - for example Acid rain - SO 2 emissions from a coal fired power plant SO 2 + H 2 O H 2 SO 3 H 2 SO 4 (atmospheric chemistry) Acid rain into certain lakes - ph decreases - fish die, other lakes ph constant - no effect on biota Controls - wet scrubbing of SO 2 with alkaline solutions - SO 2 + OH - SO 4 = + H + 1

2 Water Chemistry What is water? 2

3 Differences of Water Chemistry to Chemistry classes taught in the Chemistry Department: Environmental engineer - a dilute aqueous solution/suspension of inorganic/organic compounds that constitutes various types of aquatic systems Chemists & chemical engineers - work with concentrated solutions - > 1 M (> 1 mole dissolved /L) 3

4 Nature - sea water M (most concentrated) usually M as low as M 4

5 Chemistry Basics Element - distinct chemical substance of which all chemical compounds are composed ie - gold, silver, O 2, not H 2 O Atom - smallest particle of an element - made up of nucleus, protons and neutrons Atomic number - number of positively charged particles in an atom's nuclei (used in the periodic table - inside front cover of book.) Molecule - smallest particle of a pure substance that has the composition of that substance and is capable of independent existence - H 2 O, S 8 (crystalline sulfur) Mole - Contains an Avagadro's number of parts (6.024x10 23 ) Atomic weight - mass/mole - periodic table - inside front cover of book Ex x O atoms = 16 g O = 1 mole 6.02 x C atoms = 12 g C = 1 mole 6.02 x CO 2 molecules = 44 g CO 2 = 1 mole (2 O + 1 C) Average life of a man = 2.2 x 10 9 seconds - so counting 1 atom or molecule a second - takes 3 x lives to count Avagadro's number. 5

6 Methods of expressing concentrations Mass Concentration mg / L = mass of Subs tan ce( mg) Volume ofsolution ( L) ppm m = mass of substance (mg) if density = 1 (true in most environmental systems except mass of solution (kg) seawater ( g/ml)) then mg/l = ppm Unit 1 ppm (1 mg/l) 1 ppb (1 µg/l) Length 1 inch/16 miles 1 inch/16,000 miles Time 1 minute/2 years 1 second/32 years Money 1 cent/$10,000 1 cent/$10,000,000 Volume 1 drop vermouth/80 fifths gin 1 drop vermouth/500 barrels gin Weight 1 oz/32 tons potato chips 1 pinch salt/10 tons potato chips Area 1 sq ft/23 acres 1 sq ft/36 sq miles Action 1 bogey/56,000 rounds 1 bogey/56,000,000 rounds Quality 1 bad apple/2,000 barrels 1 bad apple/2,000,000 barrels Rate 1 dented fender/10 car lifetimes 1 dented fender/10,000 car lifetimes Molarity (M) - # moles of solute / # L of solution (720 mg NH 4 + /L) (1 mmole / 18 mg NH 4 + ) = 40 mm = 0.04 M 6

7 Composite Parameters, Concentrations Reported as X Report the concentration of a single element that is common to all species/molecules. E.g., nitrogen speies found in aquatic environmental samples include molecular nitrogen (N 2 ), nitrate (NO - 3 ), nitrite ion (NO - 2 ), ammonium ion (NH + 4 ) and nitrogen incorporated in organic unspecified compounds. Ex. Solution of NH 4 Cl - dissolves in water to form NH 4 + and Cl -. Calculate N concentrations in a 100 ml solution containing 72 mg NH 4 +. (traditional and analytically may not know if it was NH 3 or NH 4 + which have different weights) Equivalents & Normal Concentrations Equivalent weights (EW) - defined so that equal numbers of equivalents of 2 substances react exactly with each other reaction dependent. Normality = N = (mass of substance / L) / equivalent wt. [eq/l or meq/l] 7

8 1. Ion charge: EW = molecular weight / ion charge Ex. Mg 2+ + S 2- MgS if 48 mg Mg 2+ /L 2. Acid-base rxns: EW = MW/n where n = # of protons (H + ) or hydroxyl ions (OH - ) that react Ex. H 3 PO 4 H + + H 2 PO 4-3. Oxidation - reduction rxns: EW = MW/# of e - transferred per mole (1 mole of electrons = 1 eq of electrons) Ex. NO H e - NH H 2 O How can you tell if it s a redox reaction? Look at oxidation states. How do you determine oxidation states (or number electrons missing or in excess)? If it exists as a single ion - look at change in charge Ex. Fe 2+, Fe 3+,Fe(II) Fe(III) O 2 has no charge - oxidation state of each O atom is zero. 8

9 Rules to consider If O present in a molecule other than O 2 its -2 If H present in a molecule other than H 2 its +1 (exception H 2 O 2 - O -1) remember H 2 O is zero N has an oxidation number of -3 when bonded only to H or C, as is in many organic compounds S has an oxidation state of -2 when bonded only to H or C, as is in many organic compounds. All first column metals +1 - second column metals +2 ClO - 4 perchlorate Cl +7 oxidized ClO - 3 chlorate Cl +5 ClO - 2 chlorite Cl +3 ClO - hypochlorite Cl +1 Cl 2 chlorine Cl 0 Cl - chloride Cl -1 reduced "LEO the lion goes GER" Loss of electrons oxidation (reducing agent) Gain of electrons reduction (oxidizing agent) Is this a redox reaction? AlCl 3 + NaOH NaAlO 2 + NaCl CH O 2 CO H 2 O 9

10 Balancing oxidation reduction reactions 1. Identify half reactions 2. Balance non H & O atoms 3. Balance O with H 2 O 4. Balance H with H + 5. Balance charge with e- 6. Multiply half reactions to contain the same number of e- 7. Add 2 balanced half reactions 8. Eliminate H + if in caustic solution (ph>7; alkaline soln) Example: Oxidize ethyl alcohol to acetic acid using dichromate in a caustic soln. Half rxns CH 3 CH 2 OH CH 3 COOH Cr 2 O 7 2- Cr 3+ 10

11 Balancing non-redox reactions Most fairly obvious AlCl 3 + NaOH NaAlO 2 + NaCl (Al +3 on both sides) inspect: notice no H on right hand side - add H 2 O AlCl 3 + a NaOH b NaAlO 2 + c NaCl + d H 2 O take 1 mole AlCl 3 (arbitrary and unimportant) "trial and error" - Cl c=3, Al b=1, Na a=4, H d=2, O - checks "mass balance" take 1 mole AlCl 3 (arbitrary and unimportant) Al: b=1 Na: a=b+c a=4 H: a=2d d=2 O: a=2b+d b=1 Cl: c=3 Checking: AlCl NaOH 1 NaAlO NaCl + 2 H 2 O Balanced!!! 11