Lecture #13 Acids & Bases: Polyprotics Benjamin, Chapter 4 - PowerPoint PPT Presentation

Print version Updated: 11 February 2020 Lecture #13 Acids & Bases: Polyprotics Benjamin, Chapter 4 (Stumm & Morgan, Chapt.3 ) David Reckhow CEE 680 #13 1

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 David Reckhow CEE 680 #13 2

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 − = 2 [ A ] 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 David Reckhow CEE 680 #13 3 T + 2 [ H ] [ H ]

Diprotic acids: calculations (cont.)  Use [H 2 A]/C T and K a equations to get other α ’s + − + − − 2 [ H ][ HA ] − 2 [ H ][ A ] K [ HA ] K [ A ] = = K + = + = HA K 1 2 1 2 − − [ H A ] [ HA ] [ H ] [ H A ] [ 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 ] + + [ H ] [ H ] K     1 1 1 2 2 + + 2 K + [ H ] [ H ] [ H ] 1 − − [ H A ] 1 [ HA ] 1 2 [ A ] 1 = = 2 = + K + + K K + C C + + K + + 1 [ H ] 2 1 C + + [ H ] [ H ] 1 1 2 1 2 T T + 2 T K + [ H ] [ H ] K K K [ H ] 1 1 2 2 α 1 α 2 α 0 Note: α 0 + α 1 + α 2 = 1 David Reckhow CEE 680 #13 4

Diprotic acids: calculations (cont.) − − 2 [ HA ] [ A ] [ H A ] α ≡ α ≡ α 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 ] + K K K 2 1 [ 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 David Reckhow CEE 680 #13 5

Carbonate System (C T =10 -3 ) OH - 0 H + CO 3 -2 -2 H 2 CO 3 HCO 3 - Log H+ -4 Log H2CO3 Log C -6 Log HCO3- -8 Log CO3-2 -10 Log OH- -12 -14 0 2 4 6 8 10 12 14 pH David Reckhow CEE 680 #13 6

 To next lecture DAR David Reckhow CEE 680 #13 7

0 -1 OH - H + -2 -3 -4 -5 -6 Log C -7 -8 -9 -10 -11 -12 -13 -14 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH David Reckhow CEE 680 #13 8