Summary of a few general rules At the intersection of sequential - PDF document

CEE 680 Lecture #30 3/23/2020 Print version Updated: 23 March 2020 Lecture #30 Coordination Chemistry: case studies (Stumm & Morgan, Chapt.6: pg.305 ‐ 319) Benjamin; Chapter 8.1 ‐ 8.6 David Reckhow CEE 680 #30 1 Summary of a few general rules  At the intersection of sequential alphas,  i and  i+1 : 𝑀𝑝𝑕 𝑀 � 𝑞𝐿 ���  At the peak of an intermediate alpha,  i , where 𝑗 � 0, 𝑝𝑠 𝑢ℎ𝑓 𝑑𝑝𝑝𝑠𝑒𝑗𝑜𝑏𝑢𝑗𝑝𝑜 𝑜𝑣𝑛𝑐𝑓𝑠 1.0 0.8  -max value 0.6 𝑀𝑝𝑕 𝑀 � � � 𝑞𝐿 � � 𝑞𝐿 ��� 0.4 0.2 0.0 -1 0 1 2 𝑀𝑝𝑕𝐿 � � 𝑀𝑝𝑕𝐿 ��� LogK x -LogK x+1  This peak is also usually near intersection of the previous and following alphas (i.e.,  i ‐ 1 and  i+1 ), and its maximum height is estimated from: David Reckhow CEE 680 #30 2 1

CEE 680 Lecture #30 3/23/2020 Cadmium Complexes  Bisulfide Ligand Species log K Log Beta Log  1 = 10.17 CdL Log K 1 = 10.17 Log  2 = 16.53 CdL 2 Log K 2 = 6.36 Log  3 = 18.71 CdL 3 Log K 3 = 2.18 Log  4 = 20.90 CdL 4 Log K 4 = 2.19  Cyanide Ligand Species log K Log Beta Log  1 = 5.32 CdL Log K 1 = 5.32 Log  2 = 10.37 CdL 2 Log K 2 = 5.05 CdL 3 Log  3 = 14.83 Log K 3 = 4.46 Log  4 = 18.29 CdL 4 Log K 4 = 3.46 David Reckhow CEE 680 #30 3 Cd-HS system 1.2 1.1 1.0 0.9 0.8 0.7 Alpha 0.6 0.5 0.4 0.3 0.2 0.1 0.0 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 Log [L] David Reckhow CEE 680 #30 4 2

CEE 680 Lecture #30 3/23/2020 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 5 Cd-HS system 1.2 1.1         1.0 0.9 0.8 0.7 Alpha 0.6 0.5 0.4 0.3   0.2 0.1 0.0 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 Log [L] 6 3

CEE 680 Lecture #30 3/23/2020 1.2 Cd + CN 1.1     1.0 0.9 0.8 0.7   Alpha 0.6   0.5   0.4 0.3 0.2 0.1 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 7 Cadmium Bisulfide Species log K Log Beta Log  1 = 10.17 CdL Log K 1 = 10.17 Log  2 = 16.53 CdL 2 Log K 2 = 6.36 Log  3 = 18.71 CdL 3 Log K 3 = 2.18 Log  4 = 20.90 CdL 4 Log K 4 = 2.19  Specific Problem  5 x 10 ‐ 4 M Cd total  10 ‐ 3 M HS total David Reckhow CEE 680 #30 8 4

CEE 680 Lecture #30 3/23/2020 Cd-HS system 1.2 1.1         1.0 0.9 0.8 0.7 Alpha 0.6 0.5 0.4 0.3   0.2 0.1 0.0 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 Log [L] 9 5x10 -4 M Cd Total + 10 -3 M HS Total 1.2 6.0 1.1 5.5         1.0 5.0 0.9 4.5 n-bar (equ) 0.8 4.0 0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 n-bar (mass balance) 0.4 2.0   0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 Log [L] David Reckhow CEE 680 #30 10 5

CEE 680 Lecture #30 3/23/2020 5x10 -4 M Cd Total + 2.5x10 -4 M Cu Total + 10 -3 M HS Total 1x10 -3 M Cd Total + 5x10 -4 M HS Total 1.2 6.0 1.1 5.5         1.0 5.0 0.9 4.5 0.8 4.0 n-bar (equ) 0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0   0.3 1.5 0.2 1.0 n-bar (mass balance) 0.1 0.5 0.0 0.0 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 Log [L] David Reckhow CEE 680 #30 11 Cadmium Cyanide Species log K Log Beta Log  1 = 5.32 CdL Log K 1 = 5.32 Log  2 = 10.37 CdL 2 Log K 2 = 5.05 Log  3 = 14.83 CdL 3 Log K 3 = 4.46 Log  4 = 18.29 CdL 4 Log K 4 = 3.46  Specific Problem  10 ‐ 5 M Cd total  2.5 x 10 ‐ 5 M CN total David Reckhow CEE 680 #30 12 6

CEE 680 Lecture #30 3/23/2020 1.2 Cd + CN 1.1     1.0 0.9 0.8 0.7   Alpha 0.6   0.5   0.4 0.3 0.2 0.1 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 13 10 -5 M Cd Total + 2.5x10 -5 M CN Total 1.2 6.0 1.1 5.5   1.0 5.0 0.9 4.5   0.8 4.0   n-bar (equ) 0.7 3.5 Alpha n-bar 0.6 3.0 n-bar (mass balance) 0.5 2.5 0.4 2.0   0.3   1.5 0.2 1.0 0.1 0.5 0.0 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 14 7

CEE 680 Lecture #30 3/23/2020 10 -5 M Cd Total + 2.5x10 -5 M CN Total 1.2 6.0 1.1 5.5   1.0 5.0 0.9 4.5   0.8 4.0   n-bar (equ) 0.7 3.5 Alpha n-bar 0.6 3.0 n-bar (mass balance) 0.5 2.5 0.4 2.0     0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 Log [L] David Reckhow CEE 680 #30 15 Zinc Cyanide Species Betas Log  1 = 5.7 ZnL ZnL 2 Log  2 = 11.1 Log  3 = 16.1 ZnL 3 Log  4 = 19.6 ZnL 4  Specific Problem  10 ‐ 4 M Zn total  5 x 10 ‐ 4 M CN total David Reckhow CEE 680 #30 16 8

CEE 680 Lecture #30 3/23/2020 1.2 Zn + CN 1.1     1.0  as 0.9 0.8   0.7 Alpha 0.6 0.5 0.4     0.3 0.2 0.1 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 17 10 -4 M Zn Total + 5x10 -4 M CN Total 1.2 6.0 1.1 5.5   n-bar (mass balance) 1.0 5.0 0.9 4.5   0.8 4.0   n-bar (equ) 0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0     0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -8 -7 -6 -5 -4 -3 -2 -1 Log [L] David Reckhow CEE 680 #30 18 9

CEE 680 Lecture #30 3/23/2020 Complexation 10 -4 M Zn Total + 5x10 -4 M CN Total 1.2 6.0 1.1 5.5   n-bar (mass balance) 1.0 5.0 0.9 4.5   0.8 4.0   n-bar (equ) 0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0   0.3   1.5 0.2 1.0 0.1 0.5 0.0 0.0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 19 Log [L] Complexation 10 -2 M Hg Total + 3x10 -2 M Cl Total Hg ‐ Cl example 1.2 1.1     1.0   0.9 0.8 0.7 Alpha 0.6 0.5     0.4 0.3 0.2 0.1 0.0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 Log [L] 20 10

CEE 680 Lecture #30 3/23/2020 Complexation 10 -2 M Ag Total + 10 -2 M Br Total Ag ‐ Br example 1.2 1.1   1.0     0.9 0.8   0.7 Alpha 0.6 0.5 0.4   0.3 0.2 0.1 0.0 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 Log [L] 21 Aluminum Fluoride system  Significance  Aluminum in natural waters  Aluminum in coagulation  Thermodynamic Values: Smith & Martel (Benjamin)  Log K 1 = 6.16 (7.01)  Log K 4 = 2.71 (2.70)  Log K 2 = 5.05 (5.74)  Log K 5 = 1.46 (1.08)  Log K 3 = 3.91 (4.27)  Log K 6 = 0 ( ‐ 0.3)  Now calculate alpha’s   [ M ]  1        2    n 1 [ L ] [ L ] [ L ]  0 1 2 n C M [ ML ]     0  n n [ L ] n n C David Reckhow CEE 680 #30 22 M 11

CEE 680 Lecture #30 3/23/2020 Aluminum Fluoride Aluminum Fluoride: alphas alone 1.2 6.0 1.1 5.5   1.0 5.0   0.9 4.5   0.8 4.0         0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0 0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 David Reckhow CEE 680 #30 Log [L] 23 Aluminum Fluoride Aluminum Fluoride – alpha diagram 1.2 6.0 1.1 5.5   1.0 5.0 n-bar (equ)   0.9 4.5   0.8 4.0         0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0 0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 6.16 5.05 3.91 2.71 1.46 0 The pK’s David Reckhow CEE 680 #30 Log [L] 24 12

CEE 680 Lecture #30 3/23/2020 1.2 6.0 1.1 5.5   1.0 5.0 n-bar (equ)   0.9 4.5   0.8 4.0         0.7 3.5 Alpha n-bar 0.6 3.0 0.5 2.5 0.4 2.0 0.3 1.5 0.2 1.0 0.1 0.5 0.0 0.0 -8 -6 -4 -2 0 6.16 5.05 3.91 2.71 1.46 0 The pK’s Log [L] David Reckhow CEE 680 #30 25 Natural Fluoride Figure 1.4, pg. 9 in Benjamin, 2015 10 -5.5 10 -3.8 Figure 15.1, pg.873 in Stumm & Morgan, 1996 David Reckhow CEE 680 #30 26 13

CEE 680 Lecture #30 3/23/2020 Fluoride addition  Balance between Dental Caries and Fluorosis http://fluoride-math-tutorial.blogspot.com/  Recommended dose Fig. 15.3 from Water  0.7 to 1.2 mg/L, Based on temperature Quality & Treatment, 1999 (5 th edition) David Reckhow CEE 680 #30 27 Al ‐ F Problems & Discussion  Typical WT Situation  Alum dose = 33 mg/L  Total Fluoride = 1.9 mg/L  High Fluoride pulse & high alum dose  Alum dose = 660 mg/L  Total Fluoride = 190 mg/L  Impacts of OH complexation? David Reckhow CEE 680 #30 28 14