### CEE 680

12 October 2006

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

1 parsec = 19,173,511,600,000 miles

### 1.                  (50%) What is the pH of a 10-3 M solution of Sodium Dihydrogen Phosphate (NaH2PO4) and 10-2.7 M Sodium Monohydrogen Phosphate (Na2HPO4)?Calculate this for each of the three conditions below.

a. 25°C, I = 0

b. 100°C, I = 0

c. 25°C, I = 0.25

#### Preferred Approach

·         recognize that this is a simple base problem

·         then adopt an appropriate set of assumptions, and solve for [H+]

·         finally correct pKa (and pKw) for temperature, ionic strength, and repeat

·         check assumptions

#### a.  25°C, I = 0

Ø      both H+ and OH- are small compared to the other anions and cations

And then use the buffer equation:

So, considering that we’re dealing with the two amphoteric phosphate species H2PO4- and HPO4-2, we have a system centered on the second Ka for the phosphate system.

pH = 7.5

Check Assumptions:

[H2PO4-] + 2[HPO4-2] >> [OH-]

10-3 + 2x10-2.7 >> 10-6.5            YES!!

[Na+] >> [H+]

2x10-3 + 10-2.7 >> 10-7.5            YES!!

#### b.  100°C, I = 0

determine enthalpy change for the reaction:

H2PO4- = H+ + HPO4-2

then re-estimate Ka

Or:

K100 = 8.89 x 10-8

and now:

Make the usual buffer assumptions:

Ø      both H+ and OH- are small compared to the other anions and cations

And then use the buffer equation:

pH = 7.35

Check Assumptions:

[H2PO4-] + 2[HPO4-2] >> [OH-]

10-3 + 2x10-2.7 >> 10-6.65            YES!!

[Na+] >> [H+]

2x10-3 + 10-2.7 >> 10-7.35            YES!!

### Now, strictly speaking, you would also need to adjust Kw for the higher temperature as well.

#### c.  25°C, I = 0.25

determine activity coefficients for the species in the reaction:

H2PO4- = H+ + HPO4-2

And for

H2O = H+ + OH-

### For this level, use the simple Davies equation for the charged species, and assume no change in the activity of theuncharged species:

So for the singly-charged species:

And for the doubly-charged species:

then re-estimate Ka and Kw under this condition, i.e., the conditional K’s

and:

and now:

Make the same buffer assumptions:

Ø      both H+ and OH- are small compared to the other anions and cations

And then use the buffer equation:

pH = 6.93

Check Assumptions:

[H2PO4-] + 2[HPO4-2] >> [OH-]

10-3 + 2x10-2.7 >> 10-6.79            YES!!

[Na+] >> [H+]

2x10-3 + 10-2.7 >> 10-6.93            YES!!

### 2.                  (40%) What is the complete composition of a 1-liter volume of water to which you have added 10-1.5 M of ammonium nitrate (NH4NO3) and 10-2.5 M of the mono-sodium salt of p-Hydroxybenzoic Acid?Approximate values (± 0.2 log units) will suffice.

#### Approach

·         prepare a logC vs pH diagram for carbonate system (CT=0.001 M) and the acetic acid 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 ammonium cation (NH4+) and the singly-deprotonated hydroxyl-benzoate (I’ll call this: HBz-).  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:

[H2Bz] + [H+] = [OH-] + [Bz-2] + [NH3]

And if we presume that H+ and OH- are insignificant, we get:

[H2Bz] = [Bz-2] + [NH3]

from the graph above, its easy to see that  [NH3] is always above [Bz-2], so that the PBE solution lies at:

[H2Bz] = [NH3]

 pH ≈ 6.3 [H+] ≈ 5.0 x 10-7 log [H2Bz] ≈ -4.4 [H2CO3] ≈ 4.0 x 10-5 log [HBz-] ≈ -2.5 [HCO3-] ≈ 3.2 x 10-3 log [Bz-2] ≈ -5.4 [CO3-2] ≈ 4.0 x 10-6 log [NH4+] ≈ -1.5 [HAc] ≈ 3.2 x 10-2 log [NH3] ≈ -4.4 [Ac-] ≈ 4.0 x 10-5 log [OH-] ≈ -7.7 [OH-] ≈ 2.0 x 10-8

Check assumptions:

[H2Bz] >> [H+]

10-4.4 >> 10-6.3,  YES

[NH3] >> [OH-] + [Bz-2]

10-4.4 >> 10-7.7 + 10-5.4,  closer, but still YES

### 3.                  (10%) True/False.Mark each one of the following statements with either a "T" or an "F".

a. F____ An increase in ionic strength will affect hydrogen ion activity, but not hydrogen ion concentration.

b. T____ Titration curves (pH vs g or f) always exhibit a local minimum slope at a pH equal to the pKa of the acid/base being titrated.

c. T____ Electrons do not usually exist in a free state in water.

d. F____ Strong acids will always cause the pH to drop below 2.

e. T____ Alkalinity is a measure of acid neutralizing capacity.

f.  F____ Sodium Acetate is an amphoteric substance.

g. F____ The buffer intensity of a solution of a pure acid is at its lowest at a pH near the pKa of the acid.

h. F____ Positive DH values indicate that the reaction is exothermic

i.  F____ The standard assumption used for calculating the pH of an acidic solution is that the [H+] is negligible.

j.  F____ For a triprotic acid, the value of ao plus a1 must always equal one.

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 HF (aq) -77.23 -71.63 F- (aq) -80.15 -67.28 Fe+2 (aq) ‑21.0 ‑20.30 Fe+3 (aq) ‑11.4 ‑2.52 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