CEE
680
|
23 October 2008 |
FIRST EXAM
Closed book, one page of
notes allowed.
Answer
all questions. Please state any
additional assumptions you made, and show all work. You are welcome to use a graphical method of
solution if it is appropriate.
Miscellaneous Information:
R = 1.987 cal/mole°K = 8.314 J/mole°K
Absolute zero = -273.15°C
1 joule = 0.239 calories
1015 joules = 1
crown joule
1. (25%) Calculate the pKa for the ammonia-ammonium ion system at 90°C.
determine enthalpy change for the
reaction:
NH4+ = H+
+ NH3
then re-estimate Ka
Or:
K90 = 2.45 x 10-8
2. (65%) Determine the complete composition of a 1-liter volume of water to which you have added
a. 10-3 M of carbonic acid (H2CO3), and 10-2 M of sodium fluoride (NaF)
b. 10-3 M of sodium bicarbonate (NaHCO3), and 10-2.5 M of sodium monohydrogen phosphate (Na2HPO4)
Approximate values (±
0.2 log units) will suffice.
Approach: part a
· prepare a logC vs pH diagram for carbonate system (CT=0.001 M) and the fluoride system (CT = 0.01M) superimposed over it.
· write the PBE and find a solution
· read off concentrations from the graph
This is a good problem for
the graphical solution (no acid/base conjugates added, nor any strong acids or
bases). The first task is then to
prepare the species lines on our usual log C vs pH
axes (see below)
Recall that we’re adding carbonic acid (H2CO3)
and the deprotonated fluoride. These are simple solutions of two unrelated
acids/bases. Therefore we don’t have any
acid/conjugate base mixtures, nor do we have an acids or bases that have been
partly titrated with a strong acid or base.
This means we are free to use the PBE, and in fact, should use the PBE
(an ENE won’t give us a “clean” or identifiable intersection).
Thus, the PBE is:
[HF] + [H+] = [OH-]
+ [HCO3-] + 2[CO3-2]
And if we presume that H+
and
[HF] = [HCO3-] + 2[CO3-2]
And recognizing that
carbonate will be small
[HF] = [HCO3-]
|
pH ≈ 5.3 |
[H+] ≈ 5 x 10-6 |
|
log [HF]
≈ -4.1 |
[HF]
≈ 8 x 10-5 |
|
log [F-]
≈ -2.0 |
[F-]
≈ 1 x 10-2 |
|
|
|
|
log [H2CO3]
≈ -3.0 |
[H2CO3]
≈ 1 x 10-3 |
|
log [HCO3-]
≈ -4.1 |
[HCO3-]
≈ 8 x 10-5 |
|
log [CO3-2]
≈ -9.1 |
[CO3-2]
≈ 8 x 10-10 |
|
log [OH-]
≈ -8.7 |
[OH-]
≈ 2 x 10-9 |
|
|
|
|
log [Na+]
= -2.0 |
[Na+] = 1 x
10-2 |
Check assumptions:
[HF] >> [H+]
10-4.1 >>
10-5.3, YES
[HCO3-]
>> [OH-]
10-4.1 >>
10-8.7,
again YES
[HCO3-]
>> 2[CO3-2]
10-4.1 >>
10-7.-8.8,
again YES
Approach: part b
* prepare a logC vs pH diagram for carbonate system (CT=0.001 M) and the phosphate system (CT = 10-2.5 M) superimposed over it.
* write the PBE and find a solution
* read off concentrations from the graph
Again, this is a good problem
for the graphical solution (no acid/base conjugates added, nor any strong acids
or bases). The first task is then to
prepare the species lines on our usual log C vs pH
axes (see below)
Recall that we’re adding partially deprotonated
bicarbonate (HCO3-) and the partially deprotonated
monhydrogen phosphate (HPO4-2). These are simple solutions of two unrelated
acids/bases. Therefore we don’t have any
acid/conjugate base mixtures, nor do we have an acids or bases that have been
partly titrated with a strong acid or base.
This means we are free to use the PBE, and in fact, should use the PBE
(an ENE won’t give us a “clean” or identifiable intersection).
Thus, the PBE is:
2[H3PO4]
+ [H2PO4-] + [H2CO3] +
[H+] = [OH-] + [PO4-3] + [CO3-2]
And if we presume that H+
and
2[H3PO4]
+ [H2PO4-] + [H2CO3] =
[PO4-3] + [CO3-2]
And recognizing that the
extreme pH species are likely to be small
[H2PO4-] =
[CO3-2]
|
pH ≈ 9.0 |
[H+] ≈ 1 x 10-9 |
|
log [H3PO4]
≈ -11.2 |
[H3PO4]
≈ 6 x 10-12 |
|
log [H2PO4-]
≈ -4.3 |
[H2PO4-]
≈ 5 x 10-5 |
|
log [HPO4-2]
≈ -2.5 |
[HPO4-2]
≈ 3 x 10-3 |
|
log [PO4-3]
≈ -5.7 |
[PO4-3]
≈ 2 x 10-6 |
|
|
|
|
log [H2CO3]
≈ -5.7 |
[H2CO3]
≈ 2 x 10-6 |
|
log [HCO3-]
≈ -3.0 |
[HCO3-]
≈ 1 x 10-3 |
|
log [CO3-2]
≈ -4.3 |
[CO3-2]
≈ 5 x 10-5 |
|
log [OH-]
≈ -5.0 |
[OH-]
≈ 1 x 10-5 |
Check assumptions:
[H2PO4-]
>> [H+]
10-4.3 >>
10-7.0, YES
[H2PO4-]
>> 2[H3PO4]
10-4.3 >>
10-10.9,
YES
[H2PO4-]
>> [H2CO3]
10-4.3 >>
10-5.7, YES
and
[CO3-2]
>> [OH-]
10-4.3 >>
10-5.0, yes,
close, but OK
[CO3-2]
>> [PO4-3]
10-4.3 >>
10-5.7,
again YES
3. (10%) True/False. Mark each one of the following statements with either a "T" or an "F".
|
a. |
F |
Water has an unusually low
boiling point based on its molecular weight. |
|
b. |
T |
Mark Benjamin is the
author of your textbook. |
|
c. |
F |
Chemical potentials are independent
of concentration |
|
d. |
T |
Strong acids will almost
completely donate their exchangeable hydrogen ions to the surrounding solvent
molecules. |
|
e. |
F |
Reactions with a large
negative DG will always produce heat |
|
f. |
T |
Sodium Acetate is not a
strong base. |
|
g. |
T |
Ionic strength affects the
equilibria of acid/base reactions. |
|
h. |
T |
Increases in ionic
strength cause a decrease in the pKa of an acid, if
the fully-protonated form of the acid is an
uncharged species. |
|
i. |
F |
The standard assumption
used for calculating the pH of buffer solutions is that all positive ions are
negligible. |
|
j. |
T |
The value of ao plus a1 must always equal unity
for a monoprotic acid. |
Selected
Acidity Constants (Aqueous
Solution, 25°C, I = 0)
|
NAME |
FORMULA |
pKa |
|
|
Perchloric acid |
HClO4 = H+ +
ClO4- |
-7 STRONG |
|
|
Hydrochloric acid |
HCl = H+ + Cl- |
-3 |
|
|
Sulfuric acid |
H2SO4= H+ + HSO4- |
-3 (&2) ACIDS |
|
|
Nitric acid |
HNO3 = H+ + NO3- |
-0 |
|
|
Hydronium ion |
H3O+ = H+ + H2O |
0 |
|
|
Trichloroacetic acid |
CCl3COOH = H+ + CCl3COO- |
0.70 |
|
|
Iodic acid |
HIO3 = H+ + IO3- |
0.8 |
|
|
Bisulfate ion |
HSO4- = H+ +
SO4-2 |
2 |
|
|
Phosphoric acid |
H3PO4 = H+ + H2PO4- |
2.15 (&7.2,12.3) |
|
|
o-Phthalic acid |
C6H4(COOH)2 = H+ + C6H4(COOH)COO- |
2.89 (&5.51) |
|
|
Citric acid |
C3H5O(COOH)3= H+ + C3H5O(COOH)2COO- |
3.14 (&4.77,6.4) |
|
|
Hydrofluoric acid |
HF = H+
+ F- |
3.2 |
|
|
Aspartic acid |
C2H6N(COOH)2= H+ + C2H6N(COOH)COO- |
3.86 (&9.82) |
|
|
m-Hydroxybenzoic acid |
C6H4(OH)COOH = H+ + C6H4(OH)COO- |
4.06 (&9.92) |
|
|
p-Hydroxybenzoic acid |
C6H4(OH)COOH = H+ + C6H4(OH)COO- |
4.48 (&9.32) |
|
|
Nitrous acid |
HNO2 = H+ + NO2- |
4.5 |
|
|
Acetic acid |
CH3COOH = H+ + CH3COO- |
4.75 |
|
|
Propionic acid |
C2H5COOH = H+ + C2H5COO- |
4.87 |
|
|
Carbonic acid |
H2CO3 = H+ + HCO3- |
6.35 (&10.33) |
|
|
Hydrogen sulfide |
H2S = H+ + HS- |
7.02 (&13.9) |
|
|
Dihydrogen phosphate |
H2PO4- = H+ + HPO4-2 |
7.2 |
|
|
Hypochlorous acid |
HOCl = H+ + OCl- |
7.5 |
|
|
Boric acid |
B(OH)3 + H2O = H+ + B(OH)4- |
9.2 (&12.7,13.8) |
|
|
Ammonium ion |
NH4+ = H+ + NH3 |
9.24 |
|
|
Hydrocyanic acid |
HCN = H+
+ CN- |
9.3 |
|
|
p-Hydroxybenzoic acid |
C6H4(OH)COO- = H+ + C6H4(O)COO-2 |
9.32 |
|
|
Phenol |
C6H5OH = H+ + C6H5O- |
9.9 |
|
|
m-Hydroxybenzoic acid |
C6H4(OH)COO- = H+ + C6H4(O)COO-2 |
9.92 |
|
|
Bicarbonate ion |
HCO3- = H+ + CO3-2 |
10.33 |
|
|
Monohydrogen phosphate |
HPO4-2 = H+ + PO4-3 |
12.3 |
|
|
Bisulfide ion |
HS- = H+ + S-2 |
13.9 |
|
|
Water |
H2O = H+ + OH- |
14.00 |
|
|
Ammonia |
NH3 = H+ + NH2- |
23 |
|
|
Methane |
CH4 = H+ + CH3- |
34 |
|
|
Species |
kcal/mole |
kcal/mole |
|
Ca+2(aq) |
‑129.77 |
‑132.18 |
|
CaC03(s), calcite |
‑288.45 |
‑269.78 |
|
CaO (s) |
‑151.9 |
‑144.4 |
|
C(s), graphite |
0 |
0 |
|
CO2(g) |
‑94.05 |
‑94.26 |
|
CO2(aq) |
‑98.69 |
‑92.31 |
|
CH4 (g) |
‑17.889 |
‑12.140 |
|
H2CO3 (aq) |
‑167.0 |
‑149.00 |
|
HCO3- (aq) |
‑165.18 |
‑140.31 |
|
CO3-2 (aq) |
‑161.63 |
‑126.22 |
|
CH3COO-, acetate |
‑116.84 |
‑89.0 |
|
H+ (aq) |
0 |
0 |
|
H2 (g) |
0 |
0 |
|
Fe+2 (aq) |
‑21.0 |
‑20.30 |
|
Fe+3 (aq) |
‑11.4 |
‑2.52 |
|
Fe(OH)3 (s) |
‑197.0 |
‑166.0 |
|
NO3- (aq) |
‑49.372 |
‑26.43 |
|
NH3 (g) |
‑11.04 |
‑3.976 |
|
NH3 (aq) |
‑19.32 |
‑6.37 |
|
NH4+ (aq) |
‑31.74 |
‑19.00 |
|
HNO3 (aq) |
‑49.372 |
‑26.41 |
|
O2 (aq) |
‑3.9 |
3.93 |
|
O2 (g) |
0 |
0 |
|
OH- (aq) |
‑54.957 |
‑37.595 |
|
H2O (g) |
‑57.7979 |
‑54.6357 |
|
H2O (l) |
‑68.3174 |
‑56.690 |
|
PO4-3 (aq) |
-305.30 |
-243.50 |
|
HPO4-2 (aq) |
-308.81 |
-260.34 |
|
H2PO4- (aq) |
-309.82 |
-270.17 |
|
H3PO4 (aq) |
-307.90 |
-273.08 |
|
SO4-2 |
‑216.90 |
‑177.34 |
|
HS- (aq) |
‑4.22 |
3.01 |
|
H2S(g) |
‑4.815 |
‑7.892 |
|
H2S(aq) |
‑9.4 |
‑6.54 |
