CEE November 2011 SECOND EXAM. Answer all questions. Please state any additional assumptions you made, and show all work.

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1 EE 680 November 0 SEOND EXAM losed book, two pages of notes allowed. Answer all questions. Please state any additional assumptions you made, and show all work.. arbonate System. (50% for A & B) wo different drinking water supplies are used to provide a total plant flow of 5 MGD. Water # is a badly polluted surface water that has elevated levels of ammonia. Water # is a relatively pristine groundwater. he two are characterized as follows: Water Flow (MGD) alinity Ammonia p (mg/l as ao 3 ) (mg-n/l) # # 5 50 ~0 8.0 A. Water # is pre-treated with sodium hypochlorite to oxidize the ammonia to nitrogen gas prior to blending with water #. What is the p of water # after sodium hypochlorite (NaOl) addition. Assume the reaction with chlorine is stoichiometric (see equation below) and assume there are no other reactions occurring. N 3NaOl N 3Na 3l 3 O 4 ( g ) here are several ways to solve this. I based my alkalinity change on the amount of ammonia to be removed, but you could also base it on the requisite chlorine dose Ammonia = B A = (3 ) (3 0) = eq M he amount of ammonia lost is equal to the amound present in the raw water Ammonia = mg-n/l = /4,000= 0043 M Note thas sodium hypochlorite solutions usually come with some NaO, but for purposes of this problem, let s assume that the solution added is pure NaOl Although we call this ammonia, it is really present as ammonium ion (N 4 ) at all ps below 9.

2 Note that a high level of precision is needed to get an accurate p. For this reason, I will retain many more significant figures than I would normally. And so: alinity = eq/m * 0043 M = 0043 equ And now the final alkalinity is: f = i = = equ L So now we calculate the for this water (which does not change with chlorine addition) Recall that: = O α α α = { α = and for p =6.5, we have (using 6.35 and 0.33 as the two pa s): α = α = 8. 66x0 5 = 0.x (8.66x0 5 ) = 0.7x0 3 M Now we expect that with a drop in alkalinity, the p will go well below the first p, and the apha s can be simplied for the case where p<p α = = α = =

3 = W { And if we assume that α will be very small = W { = W ( W ) = 0 Solving this with the quadratic formula get us: p = 4.35 B. What will the p of the blended water be immediately after mixing water # (remember that this has just been treated with sodium hypochlorite) and water #? his is a closed system problem. herefore the total carbonate concentrations ( 's) must be determined and treated as conservative. Likewise the alkalinities are conservative, and then the final p can be determined from the blended and alkalinity. for either water: = O α α first, I would determine the alpha's at the two p's. Recall the general equations for a diprotic acid are: 3

4 α = α = his results in the following values (assuming ps of 6.3 and 0.3: p α α x so, for water #: = {(-.4/50,000) - ( ) ( )}/(0993 *.04x0-8 ) = 007 M But we didn t have to calculate this again, as it was already determined in part B and for water #: = {(50/50,000) - ( ) (0-8.0 )}/( *0757) = M now we need to calculate the blended alkalinities and total carbonates: = (0*(-.4) 5*50)/5/50,000 = 079 equ/l = (0*007 5*05035)/5 = 0638 M Now calculate the p from the earlier equation for total carbonates, and making any one of the following three sets of simplifying assumptions:. is large compared to O and 4

5 5 O = α α which becomes: α α ( ) ) ( α α Now use the quadratic equation: ( ) ( ) ( ) 4 ± = which is simplified to: ( ) ( ) ( ) ( ) ( ) ( ) 4 ± = and ( ) is = and ( ) is 079 (0638) = ( ) ( ) * x x x x ± = =

6 or p = O 3 - >> O 3 -, p >> p and is large compared to O and = O α α which becomes: α ( α ) But if p>>p, then we can ignore the first terms in the denominator of the alpha quotient. So this equation simplifies to: Now we can solve directly for : ( - ) 0 = = 4.75x0 or p =

7 or, 3. Make no assumptions and solve for the exact solution. his gives: p = omplexation (40% total for both parts) hloride forms few strong complexes. Mercury is one exception. he following two part problem concerns complexes of this metal-ligand combination. A. (0%) Attached is an accurate graph of alpha values (vs log l-) for the Mercuric hloride system. Using this graph determine the complete mercury speciation in sea water where free chloride is about M and total mercury is about 0-9 M. Assume the only mercury species are free g and the various mercuric chloride complexes (i.e., gl x ). B. (0%) Now determine the complete compsition of a 0-7 M solution of sodium chloride to which you have added 0-7 M of gl. Ignore all other mercury complexes except for the chloride ones (i.e., gl x ). 7

8 ...0 α 0 α α Alpha α α Log L A. Draw a verticle line at L =M or logl = hen read off the alpha values and determine concentrations. Below is an expanded view of the graph to aide in finding the best values. For the purposes of your exam, you only needed to draw the line over the full sized graph and locate the points as best you can by eye. he correct values are presented in the table below. 8

9 n-bar (equ) Alpha n-bar Log L Graph Variable Value Species onc (M) L l - α 7 gl 7x0 - α gl 3 4.7x0-0 α gl 4 4.7x0-0 B. Draw the two n-bar lines (equilibrium line and mass balance line). Find the intersection of the two lines and draw a vertical at that point. hen read off the alpha values and determine concentrations. Below is an expanded view of the graph to aide in finding the best values. For the purposes of your exam, you only needed to draw the line over the full sized graph and locate the points as best you can by eye. he correct values are presented in the table below. n mb L L x L L = = = M 7 7 9

10 Alpha n-bar (mass balance) n-bar (equ) n-bar Log L Alpha n-bar (mass balance) n-bar (equ).0.5 n-bar Log L 0

11 Graph Variable Value Species onc (M) L l -.7x0-7 α 0 g.5x0-8 α gl 4.0x0-8 α gl 4.5x Multiple hoice. (0%) Answer all 0 of the following questions. Indicate which of the options is the best choice. Question #6 was a bit ambiguous, so I didn t consider this one in the grade. he sum of total acidity and total organic carbon on any given sample is equal to a. the UV absorbance b. twice the total carbonate c. the value one d. half of the hardness e. none of the above. alinity is said to be conservative when: a. the system being studied is open to the atmosphere b. the system being studied is isolated in the subsurface c. the system being studied is at alkaline ps d. all of the above e. none of the above 3. Phenolphthalein a. is a hexadentate ligand b. is rarely used because noone can spell it c. complexes calcium forming an insoluble salt d. is the drug of choice for malaria e. changes from colorless to red as p increases 4. O 3 *: a. is composed mostly of aqueous O b. is conservative in closed systems c. is am ampholyte d. all of the above e. none of the above

12 5. A ligand atom: a. is always charged b. forms coordinate covalent bonds with metals c. is almost never dissolved d. only forms outer-sphere complexes e. none of the above 6. he ligand number: a. is usually 6 or less b. is related to the molecular weight of the central atom c. is a function of the size of the ligand d. all of the above e. none of the obove 7. he buffer intensity of the acetate/acetic acid system: a. is independent of the p b. is independent of the total acetate ( ) c. is zero when the p is zero. d. is at a minimum when the p is equal to the p of a pure acetate solution e. is at a minimum when the p = p 8. Detergent surfactants are used to: a. help solubilize grease b. complex trace metals c. take hardness cations from the surfactants d. elevate the acidity e. reduce the caloric content 9. EDA a. stands for ethylene dioxo-tetraacetic acid b. is most commonly used as a p buffer c. is a higly potent carcinogen d. all of the above e. none of the above 0. he Irving Williams Series a. is a means of estimating alkalinity b. describes the inverse proportionality of acidity to alkalinity c. includes a number of books, such as he hapman Report, and he Prize d. follows the increase in ligand affinity from Mn(II) to u(ii) e. provides a comprehensive description of ligand structure

13 Selected Acidity onstants (Aqueous Solution, 5, I = 0) NAME FORMULA pa Perchloric acid lo4 = lo SRONG ydrochloric acid l = l - -3 Sulfuric acid SO4= SO (&) AIDS Nitric acid NO3 = NO ydronium ion 3O = O 0 richloroacetic acid l3oo = l3oo Iodic acid IO3 = IO3-0.8 Bisulfate ion SO4 - = SO4 - Phosphoric acid 3PO4 = PO4 -.5 (&7.,.3) itric acid 35O(OO)3= 35O(OO)OO (&4.77,6.4) ydrofluoric acid F = F - 3. Nitrous acid NO = NO Acetic acid 3OO = 3OO Propionic acid 5OO = 5OO arbonic acid O3 = O (&0.33) ydrogen sulfide S = S (&3.9) Dihydrogen phosphate PO4 - = PO4-7. ypochlorous acid Ol = Ol Boric acid B(O)3 O = B(O)4-9. (&.7,3.8) Ammonium ion N4 = N3 9.4 ydrocyanic acid N = N Phenol 65O = 65O m-ydroxybenzoic 64(O)OO - = 64(O)OO- 9.9 acid Bicarbonate ion O3 - = O Monohydrogen PO4 - = PO4-3.3 phosphate Bisulfide ion S - = S- 3.9 Water O = O Methane 4 =

14 Log L 4

15 O - Log p