# 6 October 2000

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.      (45%) How much phosphoric acid (H3PO4), in moles/L, would you have to add to a 10-3 M solution of Mono Potassium Phosphate (KH2PO4) to reach a pH of 2.80?  Calculate this for each of the two temperatures below.  Please ignore ionic strength effects. (Ignore ionic strength effects)

a. 25°C

b. 100°C.

#### Preferred Approach

·         recognize that this may be a simple buffer problem

·         then use buffer equation, solving for CHA

·         finally correct pKa for temperature, and repeat

·         check assumptions

a.

Check Assumptions:

[H+] << [K+]

10-2.8 << 10-3.0    NO!!!

next step: return to underlying equations and try again:

must use a charge balance equation, rather than proton balance

[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO4-2] + 3[PO4-3]

since we are at low pH, several terms will probably be insignificant (e.g., [OH-], [HPO4-2], [PO4-3]), so we get:

[H+] + [K+] = [H2PO4-]

and substituting in for the concentrations of [H+] and [K+], we get:

[H2PO4-] = 10-2.8 + 10-3  = 10-2.59

and re-arranging the equation for the pK, we have:

and using the simplified CT equation:

CHA + CA = CT = [H3PO4] + [H2PO4-] = 10-3.24 + 10-2.59 = 10-2.50

And finally the

CHA + CA = CT = 10-2.50

CHA  = 10-2.50 - CA

CHA  = 10-2.50 – 10-3.00

CHA  = 10-2.67 = 0.00215 M

Now check all assumptions.

b.

determine enthalpy change for the reaction:

H3PO4 = H+ + H2PO4-

then re-estimate Ka

K100 = 3.72 x 10-3

and now:

Check Assumptions:

[H+] << [K+]

10-2.8 << 10-3.0    NO!!!

next step: as above, you need to return to underlying equations and try again:

again you must use a charge balance equation, rather than proton balance

[H+] + [K+] = [OH-] + [H2PO4-] + 2[HPO4-2] + 3[PO4-3]

since we are at low pH, several terms will probably be insignificant (e.g., [OH-], [HPO4-2], [PO4-3]), so we get:

[H+] + [K+] = [H2PO4-]

and substituting in for the concentrations of [H+] and [K+], we get:

[H2PO4-] = 10-2.8 + 10-3  = 10-2.59

and re-arranging the equation for the pK, we have:

and using the simplified CT equation:

CHA + CA = CT = [H3PO4] + [H2PO4-] = 10-2.96 + 10-2.59 = 10-2.44

And finally the

CHA + CA = CT = 10-2.44

CHA  = 10-2.44 - CA

CHA  = 10-2.44 – 10-3.00

CHA  = 10-2.57 = 0.00267 M

Now check all assumptions.

2. (45%) What is the complete composition of a 1-liter volume of water to which you have added 10-3 M of sodium bicarbonate (NaHCO3), 10-3.5 M of propionic acid, and 10-2.5 M of potassium fluoride (KF)?  Approximate values (± 0.2 log units) will suffice..

#### Approach

·         prepare a logC vs pH diagram for carbonate system (CT=0.001 M) with the propionate system (CT = 0.003M) and the fluoride (CT = 0.030M) system superimposed over it.

·         write the PBE and find a solution

·         read off concentrations from the graph

the PBE is:

[HF] +[H2CO3] + [H+] = [OH-] + [Pr-] + [CO3-2]

from the graph below, we can conclude that the solution lies at:

[H2CO3] = [Pr-]

 pH =6.7 [H+] = 2.0 x 10-7 log [H2CO3] = -3.5 [H2CO3] = 3.2 x 10-4 log [HCO3-] = -3.2 [HCO3-] = 6.3 x 10-4 log [CO3-2] = -6.8 [CO3-2] = 1.6 x 10-7 log [HPr] = -5.3 [HPr] = 5.0 x 10-6 log [Pr-] = -3.5 [Pr-] = 3.2 x 10-4 log [OH-] = -5.3 [OH-] = 5.0 x 10-6 log [HF] = -5.9 [HF] = 1.3 x 10-6 log [F-] = -2.5 [F-] = 3.2 x 10-3

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

a. ___ T__ Water has an unusually high boiling point based on its molecular weight.

b. ___ F__ Werner Stumm was the 13th president of the United States.

c. ___ T__ Chemical potentials can be used to calculate the gibbs free energy of a reaction

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 produce heat

f. ___ T__ Sodium Acetate is not a strong base.

g. ___ T__ Ionic strength affects the equilibria of acid/base reactions.

h. ___ F__ Increases in ionic strength always increase the activity of negatively charged ions.

i.  ___ T__ The standard assumption used for calculating the pH of buffer solutions is that the [H+] and the [OH-] are negligible.

j.  ___ F__ The value of ao plus a1 must always equal unity.

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