CEE 680
Spring 2016
Water Chemistry
Homework Set #9
Work in groups of 3-4 for this homework
Aluminum Solubility Continued
Below are the relevant chemical equilibria for amorphous aluminum hydroxide and
aluminum phosphate. These will be used in the problems below.
Equilibria
Log K
3H+
Al+3
AlOH3(s) (amorphous) +
=
+ 3 H 2O
Al+3 + H2O = AlOH+2 + H+
Al+3 + 2H2O = Al(OH)2+ + 2H+
Al+3 + 3H2O = Al(OH)3(aq) + 3H+
Al+3 + 4H2O = Al(OH) - + 4H+
10.8
-4.97
-9.3
-15.0
-23.0
AlPO4•2H2O(s) (variscite) = Al+3 + PO4-3 + 2H2O
H PO = H+ + H PO -
-21
-2.15
H2PO4- = H+ + HPO4-2
HPO -2 = H+ + PO -3
-7.2
4
3
4
4
2
4
-12.3
4
9.1 Solubility and Mixed Solids (4 points)
Prepare a solubility diagram for Aluminum in water. Assume that the hydroxide phase
that forms is amorphous. Also assume that 0.5 mM total phosphate is present in the
system. Present the diagram in the usual form (log C vs pH). Outline the zones of
precipitation and mark the identity of the precipitates.
Solution
Determine lines for all soluble Al species for amorphous aluminum hydroxide as with
previous problems of this type. Do the same for equations based on Aluminum
phosphate. This requires that one assume a total phosphate concentration of 0.5 mM, and
consider the α3, value. Line segments are determined between the pKa’s of the phosphate
system, and smooth curved portions are drawn between them.
1
Summary for Al(OH)3-based lines on Log C vs pH diagram
Species
+3
Al
AlOH+2
Al(OH)2+
Al(OH)3o
Al(OH)4-
Intercept Slope
10.8
5.8
1.5
-4.2
-12.2
-3
-2
-1
0
+1
Summary for AlPO4–based lines on Log C vs pH diagram
Species
Al+3
AlOH+2
Al(OH)2+
Al(OH)3(aq)
Al(OH) 4
pH < 2.15
Intercept
3.95
-1.02
-5.35
-11.05
-19.05
Slope
-3
-2
-1
0
+1
pH =2.15-7.2
Intercept Slope
1.8
-2
-3.17
-1
-7.50
0
-13.20
+1
-21.2
+2
pH = 7.2-12.3
Intercept Slope
-5.4
-1
-10.37
0
-14.70
+1
-20.4
+2
-28.4
+3
pH > 12.3
Intercept Slope
-17.7
0
-22.7
+1
-27.0
+2
-32.7
+3
-40.7
+4
For more on how the AlPO4-based lines are determined see the ferrous carbonate example done in class (lecture #42). The methods
are analogous.
2
Many of you did this using a spreadsheet without determining the individual line
segements, which is OK.
0.0001 M total phosphate with amorphous Al(OH)3
0
H+
-1
AlTotal
+3
-2
AlTotal
Al
-3
Al(OH)4
-4
-
AlOH+2
Al(OH)3
-5
Log C
-6
Al(OH)2+
-7
Al(OH)2
+
-8
-9
Al(OH)3
-10
+2
+3 AlOH
-11
Al
-
-12
OH
Al(OH)4
-13
-
-14
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
pH
From the above graph it can be seen that the total Al concentrations based on the two
solid phases intersect at about pH 8.1. At this same pH all lines representing the same
specific Al species should also intesect. Draw a vertical line from the intersecion of the
total Al lines. Decide which solid phase controls on each side of that line (i.e., which one
is least soluble), and erase those portions of each line that do not give the minimum
solubility (i.e., that represent the non-controlling solid).
3
0.0005 M total phosphate with amorphous Al(OH)3
0
H+
-1
-2
-3
Al(OH)4
-
-4
Al(OH)3
+2
AlOH
-5
Log C
-6
Al(OH)2+
-7
Al(OH)2
-8
+
-9
-10
Al(OH)3
AlOH+2
-11
-
-12
OH
Al+3
Al(OH)4
-13
+3
Al
-14
0
1
2
3
4
5
6
7
8
9
10
pH
Finally, mark off and label the precipitation regions accordingly.
4
11
12
13
14
9.2. Predominance Diagram (4 points)
Prepare a predominance diagram based on the Aluminum Hydroxide / Aluminum
Phosphate equilibria in the table above. Use a soluble aluminum concentration (AlT of 1
mM). Again, you should outline the zones of precipitation and mark the identity of the
precipitates. Outside of these zones, you should indicated the principal soluble species.
Solution
5
The predominance diagram must have pH on the x-axis and log PT (log total phosphate)
on the y-axis.
Type A lines
Line
A1
A2
A3
A4
Species
Al+3/AlOH+2
AlOH+2/Al(OH)2+2
Al(OH)2+/Al(OH)3o
Al(OH)3o/Al(OH)4-
pH
4.97
4.33
5.7
8.0
Notes
Later found to be under Al(OH)3 precipitate
Doesn’t appear as dihydroxide never predominates
Later found to be under Al(OH)3 precipitate
Later found to be under Al(OH)3 precipitate
Type B lines for hydroxide precipitate
B1: Determine the equations defining the lower and upper pH boundary on aluminum
hydroxide precipitation
Equilibrium between Al(OH)3 and Al+3 is:
pH =4.65
assumption is OK as this is below the A1 line. Now we turn to the other side of the
hydroxide precipitation zone.
B4: Equilibrium between Al(OH)3 and Al(OH)4- is:
pH = 9.2
This is also OK as it is at a pH above that of the A4 line where the tetra hydroxide
predominates.
Type B lines for the phosphate precipitate
pH for L <2.15
2.15-7.2
7.2-12.3
>12.3
pH for M
>4.97
4.97-5.7
5.7-8.0
>8.0
Dominant Al+3
Al+3
Al(OH)2+ Al(OH)3o Al(OH)3o Al(OH)4- Al(OH)4Species
H3PO4 H2PO4
H2PO4H2PO4HPO4-2
HPO4-2
PO4-3
<2.15
2.15-4.97
4.97-5.7
5.7-7.2
7.2-8.0
8.0-12.3
>12.3
pH range
Equ #
B5a
B5b
B7b
B8b
B8c
B9c
B9d
Next, determine boundary between AlPO4 and Al+3. This requires separate equations
below and above pH 2.15.
First the B5a line:
6
Log PT = 3.65 - 3pH
Next the B5b line:
Log PT = 1.5 - 2pH
And finally, we need to know the boundary between AlPO4 and Al(OH)4-, all of which
falls between the second and third pKa where HPO4-2 is the dominant phosphate
species:
So the B9c line is:
Log PT = -28.7 + 3pH
Type C lines:
Next determine boundary between Al(OH)3 and AlPO4. This must be evaluated from pH
4.65 to 7.2, and then from 7.2 to 9.2
The C1b line is:
Log PT = -12.3 + pH
The C1c line is:
Log PT = -19.5 + 2pH
This gives us the following predominance diagram.
7
Total Al = 1mM; amorphous Al(OH)3
0
#B9c
-1
#B5a
-2
AlPO4 (s)
-3
#C1c
-4
#B5b
-5
Log C
-6
#C1b
-7
-8
+3
Al
-9
Al(OH)4
Al(OH)3 (s)
-
-10
-11
-12
#B1
#B4
-13
-14
0
1
2
3
4
5
6
7
pH
8
8
9
10
11
12
13
14