CEE 680

18 October 1996

FIRST EXAM

Closed book, one page of notes allowed.

Miscellaneous Information:

R = 1.987 cal/mole°K = 8.314 J/mole°K

Absolute zero = -273.15°C

1 joule = 0.239 calories

1015 newtons = 1 figanewton

(50%) You have an industrial wastewater composed entirely of 0.01 F Sodium Bisulfide (NaHS) and 0.000316 F Propionic Acid (C2H5COOH). What is the equilibrium composition of this mixture?

Approach

• construct Log C vs pH Diagram
• pKa's are 7.02 and 13.9 for sulfide system and 4.87 for propionic acid (from table)
• write PBE to get intersection

PBE

[H2S] + [H+] = [OH-] + [S-2] + [C2H5COO-]

which reduces to:

[H2S] = [C2H5COO-]

Species

Graph

C

pC

H+

3e-9

8.5

OH-

3e-6

5.5

H2S

3e-4

3.5

Na+

1e-2

2.0

HS-

1e-2

2.0

S-2

4e-8

7.4

HPr

6e-8

7.2

Pr-

3e-4

3.5

(20%) What would the pH be if you added 0.005 F HCl to the wastewater in question #1?

Approach

• use previously constructed Log C vs pH Diagram
• write ENE to get intersection

ENE

[Na+] + [H+] = [OH-] + [HS-] + 2[S-2] + [C2H5COO-] + [Cl-]

which reduces to:

[Na+] - [Cl-] = [HS-]

Species

Graph

C

pC

H+

2e-7

6.7

OH-

5e-8

7.3

H2S

8e-3

2.1

Na+

1e-2

2.0

Cl-

5e-3

2.3

HS-

5e-3

2.3

S-2

2e-10

9.7

HPr

5e-6

5.3

Pr-

3e-4

3.5

(30%) What is the pH of pure water at 2°C?

Approach

• recognize that this requires a temperature correction on kw
• first enthalpy of reaction is needed
• then use Van't Hoff equation

H2O = H+ + OH-

Change in Enthalpy = (0) + (-230.0) - (-285.83) = 55.83 kJ

ln(K2/K25) = (55.83 kJ/8.314 J/mole°K)*[(1/298.15°K)-(1/275.15°K)]*1000 J/kJ

ln(K2/K25) = -1.8827

K2 = 1e-14 * exp(-1.8827)

K2 = 1.52e-15

for pure water [H+] = [OH-], so:

pH = 0.5*log(K2) 0.5*(14.82)

pH = 7.41

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

Thermodynamic Properties at Standard Temperature & Pressure@

Species

Gibbs Free Energy of Formation (kJ/mol)

Enthalpy of Formation (kJ/mol)

Entropy of Formation (kJ/ mol)

H2CO3* (aq)

-623.2

-699.6

187.0

CO2 (g)

-394.37

-393.5

213.6

HCO3- (aq)

-586.8

-692.0

91.2

CO3-2 (aq)

-527.9

-677.1

-56.9

H+ (aq)

0

0

0

H2O (l)

-237.18

-285.83

69.91

H2O (g)

-228.57

-241.8

188.72

OH- (aq)

-157.29

-230.0

-10.75

@From Stumm & Morgan, 1996