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