CEE 370 Lecture #8 9/18/2019 Print version Updated: 18 September 2019 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 Cations: Na + , K + , Ca +2 , NH 4 + , H + , etc. Anions: NO 3 - , Cl - , SO 4 -2 , OH - , PO 4 -3 , HPO 4 -2 , H 2 PO 4 - , Ac - , HCO 3 - , CO 3 -2 , etc. Neutral species: NH 3 , HAc, H 3 PO 4 , H 2 CO 3 , etc. note that ionic salts (e.g., NaCl, KCl) completely dissociate in water an thus should not be listed. 2 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 1
CEE 370 Lecture #8 9/18/2019 Showing an example of 10 -3 NaCO 3 added to water Steps in Solving chemical equilibria (cont) 2. List all independent chemical equations that involve the species present, including: A. Chemical Equilibria [ H ][ HCO ] 6 . 3 K 3 10 1 [ H CO ] E.g., acid base equilibria 2 3 2 [ H ][ CO ] 10 . 3 K 3 10 2 [ HCO ] B. Mass Balance equations 3 Total amount of each element is conserved 2 3 Na 3 C carbonates [ H CO ] [ HCO ] [ CO ] 10 C sodium [ ] 10 and 2 3 3 3 C. Electroneutrality or charge balance All water solutions must be neutrally charged 2 [ H ] [ Na ] [ OH ] [ HCO ] 2 [ CO ] 3 3 3 CEE 370 L#8 David Reckhow 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 4 CEE 370 L#8 David Reckhow 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, [Na 2 CO 3 ] b) Sodium bicarbonate, [NaHCO 3 ] c) Sodium phosphate, [Na 3 PO 4 ] 5 CEE 370 L#8 David Reckhow Solution to 4.11 a) First the sodium carbonate will dissolve 2+ 2- Na CO (s) 2 Na + CO 2 3 3 Then the carbonate can become protonated + 2- - H + CO HCO 3 3 H + HCO + - H CO (aq) 3 2 3 H CO (aq) H O + CO (aq) 2 3 2 2 Finally, in an open system, the carbon dioxide can escape as a gas CO (aq) CO (g) 2 2 6 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 3
CEE 370 Lecture #8 9/18/2019 Solution to 4.11 b) First the sodium bicarbonate will dissolve NaHCO (s) Na + HCO + - 3 3 Then the bicarbonate can become protonated or deprotonated H + CO + 2- HCO - 3 3 + - H + HCO H CO (aq) 3 2 3 H CO (aq) H O + CO (aq) 2 3 2 2 Finally, in an open system, the carbon dioxide can escape as a gas CO (aq) CO (g) 2 2 7 CEE 370 L#8 David Reckhow Solution to 4.11 c) First the sodium phosphate will dissolve + 3- Na PO 3Na + PO 3 4 4 Then the phosphate can become protonated + 3- 2- H + PO HPO 4 4 + 2- - H + HPO H PO 4 2 4 + - H + H PO H PO (aq) 2 4 3 4 Finally, in an open system, there are no additional species to consider, because there are no gas-phase forms of phosphate 8 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 4
CEE 370 Lecture #8 9/18/2019 Carbon Forms: Definitions Inorganic Carbon (+IV oxidation state) CO 2 = carbon dioxide (dissolved and gas) H 2 CO 3 = carbonic acid (dissolved) HCO 3 - = bicarbonate (dissolved) -2 CO 3 = carbonate (dissolved) CaCO 3 = calcium carbonate (mineral) Organic Carbon (< +IV oxidation state) C 6 H 12 O 6 = glucose (a sugar) CH 3 COOH = acetic acid (a carboxylic acid) 9 CEE 370 L#8 David Reckhow The Carbonate System CO (aq) + H O H CO 2 2 2 3 • Major buffer ions • volatile: interaction with atmosphere • biologically active • Definitions: [CO (aq)] + [ H CO ] = [ H CO *] 2 2 3 2 3 * 2 [ H CO ] [ HCO ] [ CO ] C 2 3 3 3 T 10 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 5
CEE 370 Lecture #8 9/18/2019 Carbonate System (C T =10 -3 ) pK 2 pK 1 H + OH - 0 CO 3 -2 HCO 3 - -2 H 2 CO 3 C T Log H+ -4 Log H2CO3 Log C -6 Log HCO3- -8 Log CO3-2 -10 Log OH- -12 Compare with Figure 2-10 See next 4 -14 slides for 0 2 4 6 8 10 12 14 instructions on how this graph pH is made 11 CEE 370 L#8 David Reckhow Rapid Method for Log C vs. pH Graph 1. Plot diagonal [H + ] and [OH - ] lines 2. Draw a light horizontal line corresponding to log C T 3. Locate System Point i.e., pH = pK a , log C = log C T make a mark 0.3 units below system point 4. Draw 45º lines (slope = 1) below log C T 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 pK a s of polyprotic acids #2-#5 for other acid/base pairs 12 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 6
CEE 370 Lecture #8 9/18/2019 Diprotic acids: calculations Start with C T and K a equations 2 [ H ][ A ] [ H ][ HA ] K K 2 [ HA ] 1 [ H A ] 2 2 C T [ H A ] [ HA ] [ A ] K [ HA ] 2 [ A 2 ] 2 K [ H A ] [ H ] [ HA ] 1 2 [ H ] K K [ H A ] 1 2 2 2 [ H ] K [ H A ] K K [ H A ] C [ H A ] 1 2 1 2 2 T 2 2 [ H ] [ H ] K K K C [ H A ] 1 1 1 2 T 2 2 [ H ] [ H ] [ H A ] 1 2 K K K C 1 1 1 2 13 T CEE 370 L#8 David Reckhow 2 [ H ] [ H ] Diprotic acids: calculations (cont.) Use [H 2 A]/C T and K a equations to get other ’s [ H ][ HA ] [ H ][ A 2 ] K [ HA ] K [ A 2 ] K HA K 1 2 1 2 [ H A ] [ H ] [ H A ] [ HA ] [ H ] [ ] 2 2 For distribution diagrams 2 2 [ HA ] [ H A ] [ HA ] [ A ] [ HA ] [ A ] 2 C C [ H A ] C C [ HA ] T T 2 T T 1 K 1 K 1 2 K K K 1 [ H ] K [ H ] [ H ] 1 1 1 2 2 2 [ H ] [ H ] K [ H ] 1 [ H A ] 1 [ HA ] 1 [ A 2 ] 1 2 K K K C 1 C [ H ] K 2 1 C [ H ] [ H ] 1 1 2 1 2 T T 2 K T [ H ] [ H ] [ H ] K K K 1 1 2 2 0 1 2 Note: 0 + 1 + 2 = 1 14 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 7
CEE 370 Lecture #8 9/18/2019 Diprotic acids: calculations (cont.) 2 [ HA ] [ A ] [ H A ] 0 2 1 2 C C C T T T 1 1 1 [ H ] K 1 2 [ H ] [ H ] K K K 2 1 1 1 1 2 K [ H ] 2 1 K K K [ H ] [ H ] 1 2 2 If pH << pK 1 , or [H + ] >> K 1 K 1 K 2 /[H + ] 2 1 K 1 /[H + ] If pK 1 << pH << pK 2 , or K 1 >> [H + ] >> K 2 [H + ]/K 1 1 K 2 /[H + ] If pK 2 ,<< pH, or K 2 >> [H + ] 1 [H + ]/K 2 [H + ] 2 /K 1 K 2 15 CEE 370 L#8 David Reckhow Substances of low solubility Solubility product defines the limit of solubility For calcium sulfate we have: 2 2 4 . 6 K so [ Ca ][ SO ] 10 4 16 CEE 370 L#8 David Reckhow 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 ][ SO ] 10 4 2 4 . 6 Ca 10 2 2 . 3 [ Ca ] 10 M If you have a solution of 10 -2 M NaSO4 which is entirely dissolved, and to this you add an excess of CaSO 4 crystals, how much of the calcium sulfate will dissolve at equilibrium. Present your answer in moles per liter 17 CEE 370 L#8 David Reckhow Organic Chemistry Definitions & Intro Properties and Nomenclature Alkanes Alkenes Alkynes Alicyclics Aromatics Functional Groups 18 CEE 370 L#8 David Reckhow Lecture #8 Dave Reckhow 9
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