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CEE 370 Environmental Engineering Principles CEE 370 Lecture #8 9/18/2019 Updated: 18 September 2019 Print version CEE 370 Environmental Engineering Principles Lecture #8 Environmental Chemistry VI: Acids- bases III, Organic Nomenclature Reading: Mihelcic & Zimmerman, Chapter 3 Davis & Masten, Chapter 2 Mihelcic, Chapt 3 David Reckhow CEE 370 L#8 1 Steps in Solving chemical equilibria 1. List all chemical species or elemental groupings that are likely to exist + + +2 + + Cations: Na , K , Ca , NH4 , H , etc. - - -2 - -3 -2 - Anions: NO3 , Cl , SO4 , OH , PO4 , HPO4 , H2PO4 , - - -2 Ac , HCO3 , CO3 , etc. Neutral species: NH3, HAc, H3PO4, H2CO3, etc. note that ionic salts (e.g., NaCl, KCl) completely dissociate in water an thus should not be listed. David Reckhow CEE 370 L#8 2 Lecture #8 Dave Reckhow 1 CEE 370 Lecture #8 9/18/2019 -3 Showing an example of 10 NaCO3 added to water Steps in Solving chemical equilibria (cont) 2. List all independent chemical equations that involve the species present, including: [H ][HCO ] A. Chemical Equilibria K 3 106.3 1 [H CO ] E.g., acid base equilibria 2 3 [H ][CO 2 ] K 3 1010.3 2 [HCO ] B. Mass Balance equations 3 Total amount of each element is conserved 2 3 3 Ccarbonates [H 2CO3 ][HCO3 ][CO3 ] 10 and Csodium [Na ] 10 C. Electroneutrality or charge balance All water solutions must be neutrally charged 2 [H ][Na ] [OH ][HCO3 ] 2[CO3 ] David Reckhow CEE 370 L#8 3 Steps in Solving chemical equilibria (cont) 3. Solve the equations You should have as many independent equations as chemical species Often it is easiest to solve for H+ and then use that concentration to calculate all other species David Reckhow CEE 370 L#8 4 Lecture #8 Dave Reckhow 2 CEE 370 Lecture #8 9/18/2019 Example 4.11 Determine the species present if the following compounds are dissolved into water, in both open and closed systems: a) Sodium carbonate, [Na2CO3] b) Sodium bicarbonate, [NaHCO3] c) Sodium phosphate, [Na3PO4] David Reckhow CEE 370 L#8 5 Solution to 4.11 a) First the sodium carbonate will dissolve 2+ 2- Na23 CO (s) 2 Na + CO3 Then the carbonate can become protonated + 2- - H + CO3 HCO3 + - H + HCO3 HCO23(aq) HCO23(aq) H 2O + CO 2(aq) Finally, in an open system, the carbon dioxide can escape as a gas CO22(aq) CO (g) David Reckhow CEE 370 L#8 6 Lecture #8 Dave Reckhow 3 CEE 370 Lecture #8 9/18/2019 Solution to 4.11 b) First the sodium bicarbonate will dissolve + - NaHCO3 (s) Na + HCO3 Then the bicarbonate can become protonated or deprotonated + 2- - H + CO3 HCO3 + - H + HCO3 HCO23(aq) HCO23(aq) H 2O + CO 2(aq) Finally, in an open system, the carbon dioxide can escape as a gas CO22(aq) CO (g) David Reckhow CEE 370 L#8 7 Solution to 4.11 c) First the sodium phosphate will dissolve + 3- Na34 PO 3Na + PO4 Then the phosphate can become protonated + 3- 2- H + PO4 HPO4 + 2- - H + HPO4 HPO24 + - H + HPO24 HPO34(aq) Finally, in an open system, there are no additional species to consider, because there are no gas-phase forms of phosphate David Reckhow CEE 370 L#8 8 Lecture #8 Dave Reckhow 4 CEE 370 Lecture #8 9/18/2019 Carbon Forms: Definitions Inorganic Carbon (+IV oxidation state) CO2 = carbon dioxide (dissolved and gas) H2CO3 = carbonic acid (dissolved) - HCO3 = bicarbonate (dissolved) -2 CO3 = carbonate (dissolved) CaCO3 = calcium carbonate (mineral) Organic Carbon (< +IV oxidation state) C6H12O6 = glucose (a sugar) CH3COOH = acetic acid (a carboxylic acid) David Reckhow CEE 370 L#8 9 The Carbonate System CO2223(aq) + H O HCO • Major buffer ions • volatile: interaction with atmosphere • biologically active • Definitions: [CO22323(aq)] + [ HCO] = [ HCO*] * 2 [H 2CO3 ][HCO3 ][CO3 ] CT David Reckhow CEE 370 L#8 10 Lecture #8 Dave Reckhow 5 CEE 370 Lecture #8 9/18/2019 -3 Carbonate System (CT=10 ) pK pK H+ 1 2 OH- 0 - -2 -2 H2CO3 HCO3 CO3 CT -4 Log H+ Log H2CO3 -6 Log HCO3- -8 Log C Log Log CO3-2 -10 Log OH- -12 Compare with Figure 2-10 -14 See next 4 slides for 02468101214instructions on pH how this graph is made David Reckhow CEE 370 L#8 11 Rapid Method for Log C vs. pH Graph 1. Plot diagonal [H+] and [OH-] lines 2. Draw a light horizontal line corresponding to log CT 3. Locate System Point i.e., pH = pKa, log C = log CT make a mark 0.3 units below system point 4. Draw 45º lines (slope = 1) below log CT line, and aimed at system point 5. Approximate curved sections of species lines 1 pH unit around system point 6. Repeat steps as necessary for more complex graphs #3-#5 for additional pKas of polyprotic acids #2-#5 for other acid/base pairs David Reckhow CEE 370 L#8 12 Lecture #8 Dave Reckhow 6 CEE 370 Lecture #8 9/18/2019 Diprotic acids: calculations Start with C and K equations T a 2 [H ][HA ] [H ][A ] K2 K1 [HA ] [H 2 A] 2 CT [H 2 A][HA ][A ] K [HA ] [A2 ] 2 K [H A] [HA ] 1 2 [H ] [H ] K K [H A] 1 2 2 [H ]2 K [H A] K K [H A] C [H A] 1 2 1 2 2 T 2 [H ] [H ]2 K K K C [H A]1 1 1 2 T 2 2 [H ] [H ] [H A] 1 2 C 1 K1 K1K2 David Reckhow T [H ] [HCEE ]2 370 L#8 13 Diprotic acids: calculations (cont.) Use [H2A]/CT and Ka equations to get other ’s [H ][HA ] K [HA ] K [A2 ] [H ][A2 ] K 1 2 K 1 2 [H 2 A] [H ] [H 2 A] [H ] [HA ] [HA ] For distribution diagrams 2 2 [HA ] [H 2 A] [HA ] [A ] [HA ] [A ] CT CT [H 2 A] CT CT [HA ] 1 K1 1 K2 K K K 1 1 2 [H ] K2 1 2 [H ] 1 [H ] [H ] [H ] K1 [H ] [H A] 1 [HA ] 1 [A2 ] 1 2 K K K 2 1 1 2 C [H ] K2 [H ] [H ] CT 1 2 T K 1 CT 1 [H ] [H ] 1 [H ] K1K2 K2 0 1 2 David Reckhow CEE 370 L#8 14 Note: 0 + 1 + 2 = 1 Lecture #8 Dave Reckhow 7 CEE 370 Lecture #8 9/18/2019 Diprotic acids: calculations (cont.) 2 [H 2 A] [HA ] [A ] 0 1 2 CT CT CT 1 1 1 [H ] K2 2 K1 K1K2 [H ] [H ] 1 1 2 K1 [H ] 1 [H ] [H ] K1K2 K2 + If pH << pK1, or [H ] >> K1 + K K /[H+]2 1 K1/[H ] 1 2 + If pK1 << pH << pK2, or K1>> [H ] >> K2 [H+]/K + 1 1 K2/[H ] + If pK2,<< pH, or K2 >> [H ] + 2 [H+]/K 1 [H ] /K1K2 2 David Reckhow CEE 370 L#8 15 Substances of low solubility Solubility product defines the limit of solubility For calcium sulfate we have: 2 2 4.6 K so [Ca ][SO4 ] 10 David Reckhow CEE 370 L#8 16 Lecture #8 Dave Reckhow 8 CEE 370 Lecture #8 9/18/2019 Solubility Solubility product constants The solubility product constant for the dissolution of CaSO4 is about 10-4.6. If you add 100 g of CaSO4 (GFW=136) to 1 liter of water, what will the calcium concentration be? 2 2 4.6 K so [Ca ][SO4 ] 10 Ca 2 104.6 [Ca 2 ] 102.3 M If you have a solution of 10-2 M NaSO4 which is entirely dissolved, and to this you add an excess of CaSO4 crystals, how much of the calcium sulfate will dissolve at equilibrium. Present your answer in moles per liter David Reckhow CEE 370 L#8 17 Organic Chemistry Definitions & Intro Properties and Nomenclature Alkanes Alkenes Alkynes Alicyclics Aromatics Functional Groups David Reckhow CEE 370 L#8 18 Lecture #8 Dave Reckhow 9 CEE 370 Lecture #8 9/18/2019 Nomenclature: Intro Carbon can form a nearly limitless diversity of compounds. One reason for this is carbon's ability to bind covalently with itself in long chains: C C C C C C C C In the above structure, each carbon atom (C) is surrounded by four single bonds. This is a consequence of carbon's tendency to form four covalent bonds each. These extra bonds not used to join the carbon chain may be linked to hydrogen atoms or other structures. The particular structure shown above is an aliphatic chain. The carbons are linked in a linear fashion, without forming rings or cycles. David Reckhow CEE 370 L#8 19 Nomenclature: Alkanes 1. Unbranched Alkanes An homologous series of simple aliphatic organic compounds is then the following: H H H H H H H H H H H H H H H H C H H C C H H C C C H H C C C C H H C C C C C H H H H H H H H H H H H H H H H Methane Ethane Propane Butane Pentane The series continues: hexane (6C), heptane (7C), octane (8C), nonane (9C), decane (10C), etc.. All alkanes have the general empirical formula, CnH2n+2. David Reckhow CEE 370 L#8 20 Lecture #8 Dave Reckhow 10 CEE 370 Lecture #8 9/18/2019 Nomenclature: Branched Alkanes 2. Branched Alkanes & IUPAC Nomenclature The smallest branched aliphatic is called Isobutane because it is an isomer of butane (often referred to n-butane to distinguish it from isobutane).
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