CEE 680 |
|
Fall 2011 |
Water
Chemistry
Homework #8
1.
Prepare a complete Log C vs pH diagram for a system containing pure water and
an excess of a-aluminum
hydroxide (25ºC and I=0).
(2 points)
Use
the equilibrium constants in Stumm & Morgan’s Table A6.1 (after page 324 in
the 3rd edition) or the values in the tables on pgs 241 and 242 in
the 2nd edition (see table below).
Show how you determined the equations for each of the lines.
Equilibrium Data
|
Species |
Equilibrium |
Log
K |
|
|
Benjamin[1] |
Stumm & Morgan[2] |
||
|
Al(OH)3 |
Al(OH)3 = Al+3 +
3OH- |
-33.23 |
-33.5 |
|
AlOH+2 |
Al+3
+ OH- = Al(OH)+2 |
9.01 |
9.0 |
|
Al(OH)2+ |
Al+3
+ 2OH- = Al(OH)2+ |
17.90 |
18.7 |
|
Al(OH)3o |
Al+3
+ 3OH- = Al(OH)3o |
26.00 |
27.0 |
|
Al(OH)4- |
Al+3
+ 4OH- = Al(OH)4- |
33.00 |
33.0 |
|
Al3(OH)4+5 |
3Al+3
+ 4OH- = Al3(OH)4+5 |
|
42.1 |
|
Al13O4(OH)24+7 |
13Al+3
+ 32OH- = Al13O4(OH)24+7
+ 4H2O |
|
349.3 |
Solution to #1
LogC
vs pH Line Summary
|
Species |
Slope |
Intercept |
|
|
Benjamin[3] |
Stumm & Morgan[4] |
||
|
Al+3 |
-3 |
8.77 |
8.5 |
|
AlOH+2 |
-2 |
3.78 |
3.5 |
|
Al(OH)2+ |
-1 |
-1.33 |
-0.8 |
|
Al(OH)3o |
0 |
-7.23 |
-6.5 |
|
Al(OH)4- |
+1 |
-14.23 |
-14.5 |
|
Al3(OH)4+5 |
-5 |
|
11.6 |
|
Al13O4(OH)24+7 |
-7 |
|
11.8 |
See diagram below
2.
Determine the pH of this solution. (1 point)
Solution to #2
Use ENE
3[Al+3] + 2[AlOH+2] + [Al(OH)2+]
+ 5[Al3(OH)4+5]
+7[ Al13O4(OH)24+7]
+ [H+] = [OH-] + [Al(OH)4-]
Make simplifying assumptions to end up
with:
[H+]
= [OH-]
Which with the help of the solubility
graph gives us:
pH = 7.0
3.
Determine the pH of a 0.1 mM solution of alum (aluminum sulfate). (1 point)
Solution to #3
Use ENE, this time including sulfate
3[Al+3] + 2[AlOH+2] + [Al(OH)2+]
+ 5[Al3(OH)4+5]
+7[ Al13O4(OH)24+7]
+ [H+] = [OH-] + 2[SO4-2] + [Al(OH)4-]
Make simplifying assumptions to end up
with:
3[Al+3]
= 2[SO4-2]
Which with the help of the solubility
graph gives us:
pH = 4.1
you may recognize that at this pH our
implicit assumption that the system is in equilibrium with aluminum hydroxide
solid is on the verge of bein invalid (we will have 2x10-4M total
Al). Its pretty close, and even if we
don’t have the solid phase, this doesn’t change our answer very much because
all of the Al species will retain the same relative concentrations (i.e., Al+3
will be dominant at these low pHs)
