CEE 370 Environmental Engineering Principles Lecture #8 - PDF document

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