|
CEE 680 |
16 November 1999 |
SECOND EXAM
Closed book, two pages of notes allowed.
Answer all questions. Please state any additional assumptions you made, and show all work.
Problem #1: (40%) Carbonate System.
Two raw drinking waters are mixed in equal proportions as they enter the headworks of a water treatment plant. The two are characterized as follows:
|
Water |
Alkalinity (mg/L as CaCO3) |
pH |
|
A |
70 |
7.31 |
|
B |
220 |
9.12 |
What will the pH of the blended water be?
Solution to Problem #1
This is a closed system problem. Therefore the total carbonate concentrations (CT's) must be determined and treated as conservative. Likewise the alkalinities are conservative, and then the final pH can be determined from the blended CT and alkalinity.
for either water:
first, I would determine the alpha's at the two pH's. Recall the general equations for a diprotic acid are:
This results in the following values (assuming pKs of 6.3 and 10.3:
|
pH |
alpha-1 |
alpha-2 |
|
7.31 |
0.9101 |
0.00093 |
|
9.12 |
0.9367 |
0.0619 |
so, for water A:
CT = {(70/50,000) - (10-6. 69) + (10-7..31)}/(0.9101 + 2*0.00093) = 0.001535 M
and for water B:
CT = {(220/50,000) - (10-4.88) + (10-9.12)}/(0. 9367 + 2*0.0619) = 0.004161 M
now we need to calculate the blended alkalinities and total carbonates:
Alk = (70 + 220)/2/50,000 = 0.00290 equ/L
CT = (0.001535 + 0.004161)/2 = 0.002848 M
Now calculate the pH from the earlier equation for total carbonates, and making the following simplifying assumption:
which becomes:
Now use the quadratic equation:
which is simplified to:
and (Alk – CT) is 0.00290–0.002848 = 0.000052
and (Alk – 2CT) is 0.00290–2(0.002848) = –0.002796
or
pH = 8.66
Problem #2: (60%) Precipitation/Dissolution
Aluminum Solubility
|
Equilibria |
Log K |
|
AlOH3(s) (amorphous) + 3H+ = Al+3 + 3 H2O |
10.8 |
|
Al+3 + H2O = AlOH+2 + H+ |
-4.97 |
|
Al+3 + 2H2O = Al(OH)2+ + 2H+ |
-9.3 |
|
Al+3 + 3H2O = Al(OH)3(aq) + 3H+ |
-15.0 |
|
Al+3 + 4H2O = Al(OH)4- + 4H+ |
-23.0 |
|
AlPO4•2H2O(s) (variscite) = Al+3 + PO4-3 + 2H2O |
-21 |
|
H3PO4 = H+ + H2PO4- |
-2.15 |
|
H2PO4- = H+ + HPO4-2 |
-7.2 |
|
HPO4-2 = H+ + PO4-3 |
-12.3 |
Solution to Problem #2
The above equilibria can be re-arranged to get a series of equations expressing the log concentration of each species as a function of pH.
First do this for the species in equilibrium with amorphous aluminum hydroxide.
|
Components |
Log K |
|||
|
Species |
Al(OH)3 (S) |
H+ |
PO4 2 |
|
|
Al 2 |
1 |
3 |
0 |
10.8 |
|
Al(OH) |
1 |
2 |
0 |
5.83 |
|
Al(OH)2 |
1 |
1 |
0 |
1.5 |
|
Al(OH)3 |
1 |
0 |
0 |
-4.2 |
|
Al(OH)4 |
1 |
-1 |
0 |
-12.2 |
|
H+ |
0 |
1 |
0 |
|
|
OH- |
0 |
-1 |
-14 |
Which gives rise to the following lines:
|
Log[ |
Al 2 |
] = |
+ 10.8 |
-3 |
pH |
|
Log[ |
Al(OH) |
] = |
+ 5.83 |
-2 |
pH |
|
Log[ |
Al(OH)2 |
] = |
+ 1.5 |
-1 |
pH |
|
Log[ |
Al(OH)3 |
] = |
- 4.2 |
0 |
pH |
|
Log[ |
Al(OH)4 |
] = |
- 12.2 |
+1 |
pH |
|
Log[ |
H+ |
] = |
0 |
-1 |
pH |
|
Log[ |
OH- |
] = |
- 14 |
+1 |
pH |
Next do this for the species in equilibrium with aluminum phosphate.
|
Components |
Log K |
|||
|
Species |
AlPO4 |
H+ |
PO4 2 |
|
|
Al |
1 |
0 |
-1 |
-21 |
|
Al(OH) |
1 |
-1 |
-1 |
-25.7 |
|
Al(OH)2 |
1 |
-2 |
-1 |
-30.3 |
|
Al(OH)3 |
1 |
-3 |
-1 |
-36 |
|
Al(OH)4 |
1 |
-4 |
-1 |
-44 |
|
H+ |
0 |
1 |
0 |
|
|
OH- |
0 |
-1 |
-14 |
Which gives rise to the following lines:
|
Log[ |
Al |
] = |
-21 |
+0 |
pH |
-1 |
Log PO4 |
|
Log[ |
Al(OH) |
] = |
-25.7 |
+1 |
pH |
-1 |
Log PO4 |
|
Log[ |
Al(OH)2 |
] = |
-30.3 |
+2 |
pH |
-1 |
Log PO4 |
|
Log[ |
Al(OH)3 |
] = |
-36 |
+3 |
pH |
-1 |
Log PO4 |
|
Log[ |
Al(OH)4 |
] = |
-44 |
+4 |
pH |
-1 |
Log PO4 |
|
Log[ |
H+ |
] = |
0 |
-1 |
pH |
0 |
Log PO4 |
|
Log[ |
OH- |
] = |
-14 |
+1 |
pH |
0 |
Log PO4 |
