CHE230 Environmental Chemistry 2010
Tutorial Problem 5A
1) Using information provided by the predominance area diagram below,
indicating aluminium speciation in solutions of different pH and fluoride
concentrations, answer the following questions:
0
1
A lF 4
2
3
pF
A lF 3
A l(O H )3
4
A l(O H )4
A lF 2
5
A lF
A l(O H )2
6
Al
7
2
3
4
5
6
7
8
9
10
11
12
13
14
pH
a) Estimate the overall formation constant (β4) for the formation of [AlF4]
from [Al3+]
The stepwise formation constants for [AlF]2+,[AlF2]+,[AlF3] and [AlF4]- can
be read from the horizontal lines on the diagram. For example, the lowest
horizontal line represents
Al3+ + F- ⇌ [AlF]2+
Where K1 = [AlF]2+/[Al3+][F-] = 1/[F-]
∴ K1 = 10pF
Since only F- is involved in the reaction at each horizontal line, this is true for
all four lines. So
β4 = K1K2 K3 K4 = 106+5+3.8+2.6 = 1017.4 = 2.5*1017
b) Name the second most predominant form of aluminium dissolved in
water with pH =4 and pF= 5.2.
pF= 5.2, which is just below the AlF2 and above AlF line.
Therefore, AlF2 (diflouroaluminium(III)) is closest region and second most
predominant form
c) Derive the equation of the line representing the equilibrium between
[AlF3] and [Al(OH)3]
For OH- complexes,
log(β2)= (14 - 4.5)*2= 19
log(β3)= 19 +(14-6.5) = 26.5
Therefore, 14.8-3pF = 26.5-3pOH
pF = ((14.8-26.5)/3)+pOH = -3.9 +pOH
pF = (-3.9 + 14) – pH = 10.1– pH
d) A water sample from an industrial process is found to contain 10-4.5M of
H+ , 1ppb of dissolved aluminium and 80% of the total dissolved
aluminium in the form of [AlF4]-. Calculate the pF of this water sample.
pH=4.5
[AlF4]- is 80% of total dissolved aluminum
Since AlF3 is next closet region, it is ~ 20% of total dissolved aluminum
AlF4/AlF3 = K4([F-]
Log( k4)=2.6
[F-]= 0.8/0.2*10^-2.6 = 0.010048M
pF = -log (0.010048M) = 1.998
CHE230 Environmental Chemistry 2010
Tutorial Problem 5B
Data for 25oC
Al3+(aq) + OH-(aq) <====> [AlOH]2+(aq)
Al3+(aq) + 2OH-(aq) <====> [Al(OH)2]+(aq)
Al3+(aq) + 3OH-(aq) <====>[Al(OH)3] (aq)
Al3+(aq) + 4OH-(aq) <====> [Al(OH)4]-(aq)
Al(OH)3(s) <===> Al3+(aq) + 3OH-(aq)
K=5x108
β2 =6x1018
log β3 = 26.3
pβ4= -32.3
Ksp=2x10-32
1) A wastewater sample with pH=5 is in equilibrium with Al(OH)3(s)
a) Calculate the total concentration of dissolved Al3+(aq) in this water (in
M).
Al3+(aq) + 3OH-(aq) <====>[Al(OH)3] (aq)
Therefore, Ksp=[ Al3+][ OH-]3
[ Al3+]=Ksp/[ OH-]3=2*10-32/(10-9)8=2*10-5M
log β3 = 26.3, so Ksp=2*10-32
b) Calculate the total concentration of dissolved aluminium in this water,
taking into account OH- complex formation (in M).
Total Al concentration = [ Al3+] + [Al(OH)2]++ [Al(OH)3]+ [Al(OH)4]-
=[ Al3+] + [ Al3+][ OH-]k + [ Al3+][ OH-]2β2+[ Al3+][ OH-]3β3+[ Al3+][ OH-]4β4
=[ Al3+] (1+[ OH-]k + [ OH-]2β2+[ OH-]3β3+[ OH-]4β4)
=2*10-5M (1+ (10-9*5*108)+ ((10-9)2*6*1018)+ ((10-9)3*1026.3)+ ((10-9)4*1032.3)
=1.54*10-4M
c) Name the predominant form of aluminium dissolved in this water
Based on the result above, [Al(OH)2]+=1.2*10-4M (dihydroxyaluminum (III))
is the dominant form
d) Calculate the ratio of Al3+(aq) to [Al (OH)2]+(aq) in this wastewater.
Al3+(aq) + 2OH-(aq) <====> [Al(OH)2]+(aq) β2 =6x1018
Therefore, [Al3+]/[Al(OH)2]+= [ OH-]2β2=((10-9)2*6*1018)=6
e) Calculate the mass of Al(OH)3(s) that would precipitate out, if the pH of a
1.0L sample of this wastewater were raised from 5 to 10.
Mass precipitated = Mass not in solution
At pH=5, total aluminium is solution =1.54*10-4M (part B)
At pH=10,
Total Al concentration =[ Al3+] (1+[ OH-]k + [ OH-]2β2+[ OH-]3β3+[ OH-]4β4)
=[ Al3+] (1+ (10-4*5*108)+ ((10-4)2*6*1018)+ ((10-4)3*1026.3)+ ((10-4)4*1032.3)
=[ Al3+] (2.015*1016)
Al(OH)3(s) <===> Al3+(aq) + 3OH-(aq)
Ksp=2x10-32
[ Al3+]=Ksp/[ OH-]3=2x10-32/(10-4)3=2*10-20
Therefore, total Al concentration = (2*10-20) (2.015*1016)=4.03*10-4M
[total Al concentration]pH10 - [total Al concentration]pH5
= 4.03*10-4M - 1.54*10-4M
= 2.49*10-4M
2.49*10-4M*1.00L=2.49*10-4 mol
2.49*10-4 mol * 78.01g/mol = 0.019 g of Al(OH)3 would dissolve
0.019 g of Al(OH)3 would dissolve in solution if pH is increased from 5 to 10
(increased solubility)