CEE 370

Fall 2006

Exam #1

 

October 16, 2006

 

Closed Book, one sheet of notes allowed

Please answer any 4 of the following 9 questions.  Each is worth 25 points.  Show all work.  Be neat, and box-in your answer.

 

 

 

1.      Calculate the ThOD (in mg/L) of the following solutions: (25 points)

a.      10-3 moles/L of sodium nitrite, NaNO2

 

 

The balanced stoichiometric equation is:

The ThOD is then:

 

b.      3x10-3 moles/L of formic acid, HCOOH

 

 

The balanced stoichiometric equation is:

The ThOD is then:

 

 

c.      10-2 moles/L of ammonium carbonate, (NH4)2CO3

 

 

The balanced stoichiometric equation is:

The ThOD is then:

 

 

2. Perform a charge balance on the following water.  As part of this balance determine the amount of unbalanced charge and its percentage of the total.  What does this number tell you about the accuracy of the chemical analysis? (25 points)

Chemical Substance

Conc. (mg/L)

GFW

Conc. (mM)

Conc. (meq/L)

Na+

12.5

22.9898

0.5437

0.5437

K+

1.1

39.0983

0.0281

0.0281

Ca+2

39.3

40.0800

0.9805

1.9611

Mg+2

3.5

24.3050

0.1440

0.2880

NO3-

8.8

62.0049

0.1419

0.1419

SO4-2

12.4

96.0576

0.1291

0.2582

Cl-

33.9

35.4530

0.9562

0.9562

HCO3-

112.3

61.0171

1.8405

1.8405

 

 

 

3. What is the TDS (in mg/L), and the ionic strength for the water in question #2? (25 points)

 

 

 

 

 

TDS = 224 or 112 mg/L

Ionic Strength = 4.3 mM

 

 

 

 

 

 

4. The diagram below shows a long deep-rock tunnel carrying water.  There is one major inflow (Q1), one minor inflow (Q2), one minor outflow (Q3) and the major outflow (Q4), all but one of which are known.  In addition, the concentration of chloride (a conservative substance) is known at 2 of the 4 inflow/outflow locations.  Calculate the missing flow (Q3) and the missing chloride concentrations (c3 and c4) for this system (25 points)

 

 

 

Location

Q (cfs)

Chloride (mg/L)

1 (main inflow)

15

2

2 (side inflow)

3

6

3 (side outflow)

???

???

4 (main outflow)

17

???

 

Mass Balance on Water across the entire system

 

Mass Balance on chloride for the system:

 

 

Recognize that we’re dealing with a conservative substance.  This means that the chloride doesn’t decay.  For this reason, the concentration will be constant downstream of the last input.  Thus, c3 = c4.

 

 

5. Use the following data to estimate the rate constant for the loss of chlorine dioxide in water.  Assume the reaction is first order in chlorine dioxide (25 points)

 

Time (min)

Concentration (µM)

k (min-1)

ti-1 to ti

t0 to ti

0

100

 

 

50

22

0.03028

 

100

5

0.02996

0.02963

Average:

0.03012

 

 

or

So:

 And we can do this several ways, as there are three sets of paired C-t data.  While there is some varability among these it is rather small, and the rate constant can be said to be about

k = 0.030 min-1 = 5x10-4 sec-1

 

 

 

6. True/False.  Indicate whether the following statements are true (T) or false (F).  (25 points total; 2 points each; 1 free point)

 

1.

T

The Arrhenius equation allows one to adjust rate constants from one temperature to another

2.

F

A triprotic acid has three times the strength as a monoprotic acid

3.

T

A conjugate base is what forms when an acid losses a proton

4.

T

When an organic compound name ends in “ol”, it usually means the compound is an alcohol

5.

T

Catalysts accelerate chemical reactions by lowering the activation energy

6.

F

When Gibbs Free Energy increases, the reaction will tend to go forward

7.

T

Alkynes all have triple bonds

8.

T

Henry’s law describes the relationship between partial pressure and dissolved concentration

9.

F

BTEX is a new type of dog food

10.

T

Changes in ionic strength can cause shifts in chemical equilibria

11

F

An element of high electronegativity will share its bonding electrons equally with an element of low electronegativity

12

F

Biochemical oxygen demand (BOD) differs from theoretical oxygen demand (ThOD) in that BOD only refers to oxygen demand from pharmaceuticals

 

 

 

7. The half life for nonylphenol in groundwater is estimated to be 35 days.  If the concentration of nonylphenol in a groundwater plume is currently 12 ppb, how long will it take for the concentration to drop below 1 ppb?  (25 points)

 

Since I give you a half-life, its reasonable to assume this is a first order process, so

And for 1st order reactions:

Which gives us:

≈125 days

 

8.  The equilibrium constant for the following reaction is 10-10.3.  What percent of the total carbonate is in the form of HCO3- at pH 11.5?  (25 points)

 

HCO3- = CO3-2 + H+

 

The equilibrium expression for this reaction is:

So:

And therefore:

Thus:

 

And recognizing that at such a high pH, virtually all of the carbonate are in the form of bicarbonate and carbonate (i.e., there is essetially no carbonic acid):

Or                                               HCO3-/total carbonate ≈ 6.3%

9. You need to oxidize butyric acid (CH3CH2CH2COOH) completely to carbon dioxide (CO2) by addition of potassium dichromate (K2Cr2O7).  After the reaction is complete, the chromium is in the trivalent form (Cr+3).  How many mg/L of potassium dichromate are needed to completely oxidize 4.6 mg/L of butyric acid? (25 points).

 

This is a redox (reduction-oxidation) reaction, so the electron transfer must be balanced.  This requires that we first separate out the two half reactions,

            First the oxidation half reaction we have carbon becoming oxidized:

 

            Next, dichromate must be the substance that is reduced (i.e., it does the oxidizing of the butyric acid).  We’re also told that an end product is Cr+3.  So the reduction half reaction is:

 

Then combine to balance the overall equation:

3 x   ()

10 x ()

        

Or

        

 

Next calculate the mass requirements

 

Appendix

Selected Chemical Constants

Element

Symbol

Atomic #

Atomic Wt.

Valence

Electronegativity

Aluminum

Al

13

 26.98

3

1.47

Boron

B

 5

 10.81

3

2.01

Calcium

Ca

20

 40.08

2

1.04

Carbon

C

 6

 12.01

2,4

2.50

Chlorine

Cl

17

35.453

1,3,5,7

2.83

Chromium

Cr

24

52.00

many

1.56

Helium 

He

 2

  4.00

0

 

Holmiuum

Ho

67

164.93

3

1.10

Hydrogen

H

 1

  1.01

1

2.20

Magnesium

Mg

12

 24.31

2

1.23

Manganese

Mn

25

 54.94

2,3,4,6,7

1.60

Nitrogen  

N

7

14.01

many

3.07

Oxygen  

O

 8

16.00

2

3.50

Potassium

K

19

 39.10

1

0.91

Sodium  

Na

11

 22.99

1

1.01

Sulfur  

S

16

 32.06

2,4,6

2.44

 

Selected Acidity Constants  (Aqueous Solution, 25°C, I = 0)

   NAME

   FORMULA

 pKa

Hydrochloric acid

HCl = H+ + Cl-

-3

Sulfuric acid

H2SO4= H+ + HSO4-

-3

Nitric acid

HNO3 = H+ + NO3-

-0               

Bisulfate ion

HSO4- = H+ + SO4-2

 2

Phosphoric acid

H3PO4 = H+ + H2PO4-

 2.15

Hydrofluoric acid

HF = H+  + F-

 3.2

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

Hydrogen sulfide

H2S = H+  + HS-

 7.02

Dihydrogen phosphate

H2PO4- = H+  + HPO4-2

 7.2

Hypochlorous acid

HOCl = H+  + OCl-

 7.5

Ammonium ion

NH4+ = H+  + NH3

 9.24

Hydrocyanic acid

HCN = H+  + CN-

 9.3

Phenol

C6H5OH = H+  + C6H5O-

 9.9

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