CEE 680

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