2.303 [ OH ] [ H ] CT

Transcription

1 CEE Octber 2011 FIRST EXAM Clsed bk, ne page f ntes allwed. Answer all questins. Please state any additinal assumptins yu made, and shw all wrk. Yu are welcme t use a graphical methd f slutin if it is apprpriate. Miscellaneus Infrmatin: R = cal/mle = J/mle Abslute zer = C 1 jule = calries 1230 calries = 1 Triple Whpper 1. (50%) Yu have been asked t prepare a buffer at p The chices are an acetate buffer with a C T f 5mM and carbnate buffer with a C T f 8 mm. Which f the tw will have a higher buffer intensity at the desired p (i.e., at p 5.30) under each f the fllwing cnditins? In answering this please shw the calculated buffer intensity fr bth under each cnditin. Assume a clsed system. a. 25C, I = 0 b. 100C, I = 0 Preferred Apprach recgnize that this is a simple buffer intensity prblem nce yu have the crrect pa s fr part b yu must crrect pa (and p w ) fr temperature, and repeat a. 25C, I = 0 first calculate fr the acetate buffer recall that fr a mnprtic acid, the buffer intensity is calculated frm: [ O ] [ ] CT 0 1 and since this is a buffer prblem, we can nrmally drp the hydrxide and hydrgen in terms:

2 and substituting in fr the alphas, we get: CT C T and fr the particular p f 5.3, this reduces t: [ ] 1 [ ] C T C T S that at 5 mm C T, we have: 2.303C * C T T (5x10 ) next fr the carbnate buffer recall that fr a diprtic acid, the buffer intensity is calculated frm: [ O ] [ ] CT 0 1 CT 1 2 and since the p is clsest t the first p, we can drp the secnd alpha term as well as the hydrxide and hydrgen in terms: and substituting in fr the alphas, we get: CT 0 1

3 2.303 C T and since [ + ]>> 2, then we can simplify: [ ] [ ] [ ] 1 [ ] C T [ ] 1 [ ] 1 and fr the particular p f 5.3, this reduces t: C T C T S that at 8 mm C T, we have: 2.303C * C T T (8x10 ) S, at , the acetate buffer is strnger than the carbnate buffer b. 100C, I = 0 determine enthalpy change fr the imprtant dissciatin reactins. First the acetate reactin: C 3 COO = + + C 3 COO - i f C3COO C3COO ( ) 0.05 cal / mle

4 then re-estimate a lg T2T RT T cal / mle lg x10 cal/ mle lg lg = 1.75 x 10-5 Nw the carbnate reactin (nte that nly the 1 is needed) 2 CO 3 = + + CO 3 - i f CO3 2CO ( 167.0) 1.82 cal / mle then re-estimate a lg T2T RT T cal / mle lg x10 cal/ mle lg lg = 8.28 x 10-7 D the same fr w

5 i O f 13.36cal / mle 2O ( ) then re-estimate w lg lg lg T2 T RT2T lg cal / mle x10 cal / mle w 100 = 9.29 x and nw: fllw methd in part a this time it s the Carbnate that has the higher Beta. ( vs ) 2. (40%) What is the cmplete cmpsitin f a 1-liter vlume f water cntaining 10-2 M f ammnium chlride (N 4 Cl) and 10-2 M f sdium bisufide (NaS)? Apprximate values ( 0.2 lg units) will suffice. Apprach * prepare a lgc vs p diagram fr ammnia system (C T =0.01 M) and the acetic acid system (C T = 0.01M) superimpsed ver it. * write the PBE and find a slutin * read ff cncentratins frm the graph This is a gd prblem fr the graphical slutin (n acid/base cnjugates added, nr any strng acids r bases). The first task is then t prepare the species lines n ur usual lg C vs p axes (see belw)

6 Recall that we re adding ammnium catin (N 4 + ) and the partially deprtnated bisulfide (S - ). These are simple slutins f tw unrelated acids/bases. Therefre we dn t have any acid/cnjugate base mixtures, nr d we have an acids r bases that have been partly titrated with a strng acid r base. This means we are free t use the PBE, and in fact, shuld use the PBE (an ENE wn t give us a clean r identifiable intersectin). Thus, the PBE is: [ 2 S] + [ + ] = [O - ] + [N 3 ] + [S -2 ] And if we presume that + and O- are insignificant, we get: [ 2 S] = [N 3 ] + [S -2 ] And ging a step further we might presume that the sulfide is small as well. This gives us the PBE intersectin [ 2 S] = [N 3 ]

7 O - N 4 - S - N Lg C S S p p 8.15 [ + ] 7.1 x 10-9 lg [ 2 S] [ 2 S] 7.0 x 10-4 lg [S - ] -2.0 [S - ] 1 x 10-2 lg [S -2 ] -7.8 [S -2 ] 2 x 10-8 lg [N + 4 ] -2.0 [N + 4 ] 1 x 10-2 lg [N 3 ] -3.1 [N 3 ] 8 x 10-4 lg [O - ] [O - ] 1.4 x 10-6 Check assumptins: [ 2 S] >> [ + ] >> , YES [N 3 ] >> [O - ] >> , again YES

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9 3. (10%) True/False. Mark each ne f the fllwing statements with either a "T" r an "F". a. T Water has an unusually high biling pint, given its mlecular weight. b. F The Brnsted-Lwry definitin f an acid is a substance that turns red litmus blue. c. T ardness is nrmally defined as the sum f all divalent catins d. F Organic frms f carbn are thse in the IV xidatin state. e. T Mass defects are directly prprtinal t nuclear binding energy f. F g. F h. F i. T The alkalinity minus the acidity is equal t ne-half the C T (ttal carbnates) The reactivity f neutral species is unaffected by changes in inic strength. Increases in inic strength cause an increase in the pa f an acid, if the fully-prtnated frm f the acid is an uncharged species. The standard assumptin used fr calculating the p f a strng acid is that [A-] >> [A]. j. T The value f plus 1 must never equal unity fr a diprtic acid.

10 Selected Acidity Cnstants (Aqueus Slutin, 25 C, I = 0) NAME FORMULA pa Perchlric acid ClO4 = + + ClO STRONG ydrchlric acid Cl = + + Cl - -3 Sulfuric acid 2SO4= + + SO (&2) ACIDS Nitric acid NO3 = + + NO ydrnium in 3O + = + + 2O 0 Trichlracetic acid CCl3COO = + + CCl3COO Idic acid IO3 = + + IO3-0.8 Bisulfate in SO4 - = + + SO4-2 2 Phsphric acid 3PO4 = + + 2PO (&7.2,12.3) -Phthalic acid C64(COO)2 = + + C64(COO)COO (&5.51) Citric acid C35O(COO)3= + + C35O(COO)2COO (&4.77,6.4) ydrfluric acid F = + + F Aspartic acid C26N(COO)2= + + C26N(COO)COO (&9.82) m-ydrxybenzic acid C64(O)COO = + + C64(O)COO (&9.92) p-ydrxybenzic acid C64(O)COO = + + C64(O)COO (&9.32) Nitrus acid NO2 = + + NO2-4.5 Acetic acid C3COO = + + C3COO Prpinic acid C25COO = + + C25COO Carbnic acid 2CO3 = + + CO (&10.33) ydrgen sulfide 2S = + + S (&13.9) Dihydrgen phsphate 2PO4 - = + + PO ypchlrus acid OCl = + + OCl Bric acid B(O)3 + 2O = + + B(O)4-9.2 (&12.7,13.8) Ammnium in N4 + = + + N ydrcyanic acid CN = + + CN p-ydrxybenzic acid C64(O)COO - = + + C64(O)COO Phenl C65O = + + C65O m-ydrxybenzic acid C64(O)COO - = + + C64(O)COO Bicarbnate in CO3 - = + + CO Mnhydrgen PO4-2 = + + PO phsphate Bisulfide in S - = + + S Water 2O = + + O Ammnia N3 = + + N2-23 Methane C4 = + + C3-34

11 Species f G f kcal/mle kcal/mle Ca +2 (aq) CaCO 3 (s), calcite CaO (s) C(s), graphite 0 0 CO 2 (g) CO 2 (aq) C 4 (g) CO 3 (aq) CO - 3 (aq) CO -2 3 (aq) C 3 COO C 3 COO -, acetate (aq) (g) 0 0 F (aq) F - (aq) Fe +2 (aq) Fe +3 (aq) Fe(O) 3 (s) NO - 3 (aq) N 3 (g) N 3 (aq) N + 4 (aq) NO 3 (aq) O 2 (aq) O 2 (g) 0 0 O - (aq) O (g) O (l) PO -3 4 (aq) PO -2 4 (aq) PO - 4 (aq) PO 4 (aq) SO S - (aq) S(g) S(aq)

12 Lg C T p