### 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

### 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 OH- are insignificant, we get:

[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 OH- are insignificant, we get:

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