### CEE 680

17 October 2007

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 10-2 M solution of Sodium Nitrite (NaNO2) to which you have added 5 x 10-3 M Nitric Acid (HNO3)?Calculate this for each of the three conditions below (obviously the ionic strength will never be zero for this solution, but let’s assume that ideal case for part “a” and “b”, anyway).

a. 25°C, I = 0

b. 1°C, I = 0

c. 25°C, I = 0.10

#### 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

First, recognize that nitric is a strong acid.  Also, the addition of 5 mM of nitric acid to 10 mM of sodium nitrite will result in the equivalent of a 5 mM solution of nitrous acid and a 5 mM solution of nitrite.

Next, 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:

So, considering that we’re dealing with the two adjacent nitrite species HNO2 and NO2-, we have a system centered on the Ka for the nitrite system.

pH = 4.5

Check Assumptions:

[NO2-] >> [OH-]

10-2.3  >> 10-9.5            YES!!

[Na+]-[NO3-] >> [H+]

10-2.3 >> 10-4.5            YES!!

#### b.  1°C, I = 0

determine enthalpy change for the reaction:

HNO2- = H+ + NO2-

then re-estimate Ka

Or:

K1 = 1.89 x 10-5

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 = 4.72

Check Assumptions:

[NO2-] >> [OH-]

10-2.3  >> 10-9.28            YES!!

[Na+]-[NO3-] >> [H+]

10-2.3 >> 10-4.72            YES!!

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

#### c.  25°C, I = 0.10

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 the uncharged species:

So for the singly-charged species:

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

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 = 4.28

Check Assumptions:

[NO2-] >> [OH-]

10-2.3  >> 10-9.72            YES!!

[Na+]-[NO3-] >> [H+]

10-2.3 >> 10-4.28            YES!!

### 2.             (40%) What is the complete composition of a 1-liter volume of water to which you have added 10-2 M of ammonium acetate (NH4CH3CO2)?Approximate values (± 0.2 log units) will suffice.

#### Approach

·       prepare a logC vs pH diagram for ammonia system (CT=0.01 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 deprotonated acetate (I’ll call this: Ac-).  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:

[HAc] + [H+] = [OH-] + [NH3]

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

[HAc] = [NH3]

 pH ≈ 7.0 [H+] ≈ 1 x 10-7 log [HAc] ≈ -4.3 [HAc] ≈ 5.0 x 10-5 log [Ac-] ≈ -2.0 [Ac-] ≈ 1 x 10-2 log [NH4+] ≈ -2.0 [NH4+] ≈ 1 x 10-2 log [NH3] ≈ -4.3 [NH3] ≈ 5.0 x 10-5 log [OH-] ≈ -7.0 [OH-] ≈ 1 x 10-7

Check assumptions:

[HAc] >> [H+]

10-4.3 >> 10-7.0,  YES

[NH3] >> [OH-]

10-4.3 >> 10-7.0,  again YES

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

a.  ___ T_ Solutions of pure compounds exhibit their highest buffer intensities at pHs near the compound’s pKa .

b. ___ F_ Solutions of hydrochloric acid are always lower in pH than solutions of acetic acid.

c.  ___ T_ A 10-2 F solution of HCl has a pH of about 2.

d. ___ F_ Strong acids will always have strong conjugate bases.

e.  ___ T_ The sum of alkalinity and acidity equals twice the total inorganic carbon.

f.  ___ T_ Dihydrogen phosphate (H2PO4-) is an amphoteric substance.

g. ___ F_ The buffer intensity of a solution is the inverse of the alkalinity.

h. ___ F_ Positive DH values indicate that the reaction is not spontaneous

i.   ___ T_ The standard assumption used for calculating the pH of an acidic solution is that the [OH-] is negligible.

j.   ___ F_ For a diprotic 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 NO2- (aq) ‑25.00 ‑8.89 NO3- (aq) ‑49.372 ‑26.43 NH3 (g) ‑11.04 ‑3.976 NH3 (aq) ‑19.32 ‑6.37 NH4+ (aq) ‑31.74 ‑19.00 HNO2 (aq) ‑28.49 ‑10.27 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