CEE 680 8 Mrch 018 FIRST EXAM Closed ook, one pge of notes llowed. Answer ny 4 of the following 5 questions. Plese stte ny dditionl ssumptions you mde, nd show ll work. Miscellneous Informtion: R = 1.987 cl/mole = 8.314 J/mole Asolute zero = -73.15C 1 joule = 0.39 clories 10 F = the new 30 F 1. (5%) Use the grphicl solution to determine the ph nd complete solution composition for 1 liter of pure wter to which you ve dded 10 - moles of Phenol. Grph pper is ttched to this exm for this purpose. Prepre grph with p = 9.9 PBE: which reduces to: [Phente - ] + [OH - ] = [H + ] [Phente - ] = [H + ] Equlity (intersection) occurs t ph ~ 6.0 Species pc C (Molr) H + 6.0 1.0 x 10-6 OH - 8.0 1.0 x 10-8 Phenol.0 1.0 x 10 - Phente - 6.0 1.0 x 10-6 1
0-1 - H + OH - Phenol Phente -3-4 -5-6 Log C -7-8 -9-10 -11-1 -13-14 0 1 3 4 5 6 7 8 9 10 11 1 13 14 ph. (5%) Determine the ph nd solution composition of mixture of 10 - moles of Phenol plus 10-3 moles of Disodium o-phthlte [NC6H4(COO - )] in 1 liter of wter. Plese use grphicl solution for this one too. Prepre grph with phenol system (p = 9.9) nd o-phthlic cid system (p1 =.89; p = 5.51). PBE: which reduces to: [Phente - ] + [OH - ] = [H + ] + [Hphthlte-] + [H Phthlte] [Phente - ] = [Hphthlte-] Equlity (intersection) occurs t ph ~ 7.
Species pc C (Molr) H + 7. 6.3 x 10-8 OH - 6.8 1.6 x 10-7 Phenol.0 1.0 x 10 - Phente - 4.7.0 x 10-5 Phthlte - 3.0 1.0 x 10-3 HPhthlte - 4.7.0 x 10-5 HPhthlte 9.0 1.0 x 10-9 PBE Solution 0-1 - -3 H + OH - Phenol Phente H Phthlte HPhthlte - Phthlte - -4-5 -6 Log C -7-8 -9-10 -11-1 -13-14 0 1 3 4 5 6 7 8 9 10 11 1 13 14 ph 3
3. (5%) Determine the complete solution composition of:. solution of 10 - moles of Sodium Hypochlorite (NOCl) in 1 Liter of wter. solution of 10 - moles of the conjugte cid [i.e., Hypochlorous Acid (HOCl)] in 1 Liter of wter. But this time use n lgeric solution. You cn ignore ionic strength effects (i.e., ssume infinite dilution). Rememer to mke simplifying ssumptions For this cse, we hve n cid nd then its conjugte se. The p is 7.5, nd the pc is.. solution of 10 - moles of Sodium Hypochlorite (NOCl) in 1 Liter of wter note tht from the ttched tle: p = 7.5, so tht p = 14-7.5 = 6.5 pc = A. Mke sic solution ssumption, which leds to: [ OH ] 4 C [OH - ] = 5.6076 x 10-5 poh = 4.51 ph = 9.7488 ut esier to mke the comined sic solution nd wek se ssumption, which leds to: [ OH ] C [OH - ] = 5.634 x 10-5 poh = 4.500 ph = 9.7500 some textooks use p = 7.6, insted of 7.5. In this cse we get: p = 14-7.6 = 6.4 A. Mke sic solution ssumption, which leds to: [ OH ] 4 C [OH - ] = 6.897 x 10-5 poh = 4.013 4
ph = 9.7986 ut esier to mke the comined sic solution nd wek se ssumption, which leds to: [ OH ] C [OH - ] = 6.3096 x 10-5 poh = 4.000 ph = 9.8000. solution of 10 - moles of the conjugte cid [i.e., Hypochlorous Acid (HOCl)] in 1 Liter of wter. B. Mke cidic solution ssumption, which leds to: [ H ] 4 C [H + ] = 1.7767 x 10-5 ph = 4.75038 ut esier to mke the comined cidic solution nd wek cid ssumption, which leds to: [ H ] C [H + ] = 1.7783 x 10-5 ph = 4.75000 nd gin, if you use p 7.6, you would get: A. Mke cidic solution ssumption, which leds to: [ H ] 4 C [H + ] = 1.5836 x 10-5 5
ph = 4.80034 ut esier to mke the comined cidic solution nd wek cid ssumption, which leds to: [ H ] C [H + ] = 1.5849 x 10-5 ph = 4.80000 4. (5%) Repet prolem #3, ut this time dd 0.1 M of NCl s well s the 10 - moles of NOCl to your liter of wter. Also this time, do not ignore ionic strength effects. HOClOH HOCl HOClOH OCl OCl OCl HOClOH HOCl OH OCl OCl HOClOH HOCl HOCl OH OCl OCl OCl HOClOH OCl OCl OH OCl OH And with the Guntelerg pproximtion, oth ctivity corrections re identicl so tht does not chnge. However, the w will chnge [ H ] [ OH ] H [ H w ][ OH OH ] w w H OH OH OH Use the Guntelerg pproximtion: 6
log f 0.5z I 1 I log(fy )= 0.1013 fy = fy+ = 0.758357 or w w { H }[ OH ] ( 0.758) 1. 318 p w 0.1 p 13.88 w w Since ph = pw-poh, nd since the poh is unchnged y the increse in ionic strength, then the increse in ionic strength drops the ph y 0.4 log units ph = 9.75-0.1 ph = 9.63 Alterntively, you could simply strt with the p nd modify tht. You would get the sme nswer: log(fy )= 0.1013 fy = fy+ = 0.758357 p' = 7.379873 ph = 9.75-0.1 ph = 9.63 7
5. (5%) True/Flse. Mrk ech one of the following sttements with either T or n F, whichever is most ccurte. T The ph of pure wter in equilirium with tmospheric CO is elow 6. T Equilirium constnts cn e clculted from the rtio of the forwrd nd ckwrd rte constnts c. F The p of n cid is lwys equl to the p of its conjugte se d. F ph = 7 t the end-point of titrtion e. F Nitric cid lwys completely dontes its proton to the solvent it is dissolve in, regrdless of the nture of tht solvent f. T The principle of electroneutrlity is lwys oserved in queous solutions g. F Non-cronte hrdness is equl to the mgnesium concentrtion minus the sulfte concentrtion h. F Increses in ionic strength hve no effect on species with zero chrge. i. F Hypochlorite is very strong se in wter j. T The vlue of o plus 1 must lwys equl 1 for ny monoprotic cid system. 8
Selected Acidity Constnts (Aqueous Solution, 5 C, I = 0) NAME FORMULA p Perchloric cid HClO4 = H + + ClO4 - -7 STRONG Hydrochloric cid HCl = H + + Cl - -3 Sulfuric cid HSO4= H + + HSO4 - -3 (&) ACIDS Nitric cid HNO3 = H + + NO3 - -0 Hydronium ion H3O + = H + + HO 0 Trichlorocetic cid CCl3COOH = H + + CCl3COO - 0.70 Iodic cid HIO3 = H + + IO3-0.8 Bisulfte ion HSO4 - = H + + SO4 - Phosphoric cid H3PO4 = H + + HPO4 -.15 (&7.,1.3) o-phthlic cid C6H4(COOH) = H + + C6H4(COOH)COO-.89 (&5.51) Citric cid C3H5O(COOH)3= H + + C3H5O(COOH)COO - 3.14 (&4.77,6.4) Hydrofluoric cid HF = H + + F - 3. Asprtic cid CH6N(COOH)= H + + CH6N(COOH)COO- 3.86 (&9.8) m-hydroxyenzoic cid C6H4(OH)COOH = H + + C6H4(OH)COO- 4.06 (&9.9) p-hydroxyenzoic cid C6H4(OH)COOH = H + + C6H4(OH)COO- 4.48 (&9.3) Nitrous cid HNO = H + + NO - 4.5 Acetic cid CH3COOH = H + + CH3COO - 4.75 Propionic cid CH5COOH = H + + CH5COO - 4.87 o-phthlte C6H4(COOH)COO - = H + + C6H4(COO-) 5.51 Cronic cid HCO3 = H + + HCO3-6.35 (&10.33) Hydrogen sulfide HS = H + + HS - 7.0 (&13.9) Dihydrogen phosphte HPO4 - = H + + HPO4-7. Hypochlorous cid HOCl = H + + OCl - 7.5 Boric cid B(OH)3 + HO = H + + B(OH)4-9. (&1.7,13.8) Ammonium ion NH4 + = H + + NH3 9.4 Hydrocynic cid HCN = H + + CN - 9.3 p-hydroxyenzoic cid C6H4(OH)COO - = H + + C6H4(O)COO- 9.3 Phenol C6H5OH = H + + C6H5O - 9.9 m-hydroxyenzoic cid C6H4(OH)COO - = H + + C6H4(O)COO- 9.9 Bicronte ion HCO3 - = H + + CO3-10.33 Monohydrogen HPO4 - = H + + PO4-3 1.3 phosphte Bisulfide ion HS - = H + + S- 13.9 Wter HO = H + + OH - 14.00 Ammoni NH3 = H + + NH - 3 Methne CH4 = H + + CH3-34 9
Species o H f o G f kcl/mole kcl/mole C + (q) -19.77-13.18 CCO 3 (s), clcite -88.45-69.78 CO (s) -151.9-144.4 C(s), grphite 0 0 CO (g) -94.05-94.6 CO (q) -98.69-9.31 CH 4 (g) -17.889-1.140 H CO 3 (q) -167.0-149.00 - HCO 3 (q) -165.18-140.31 - CO 3 (q) -161.63-16. HOCl (q) -8.90-19.10 OCl- (q) -5.60-8.80 CH 3 COOH -116.79-95.5 CH 3 COO -, cette -116.84-89.0 H + (q) 0 0 H (g) 0 0 HF (q) -77.3-71.63 F - (q) -80.15-67.8 Fe + (q) -1.0-0.30 Fe +3 (q) -11.4 -.5 Fe(OH) 3 (s) -197.0-166.0 - NO 3 (q) -49.37-6.43 NH 3 (g) -11.04-3.976 NH 3 (q) -19.3-6.37 + NH 4 (q) -31.74-19.00 HNO 3 (q) -49.37-6.41 O (q) -3.9 3.93 O (g) 0 0 OH - (q) -54.957-37.595 H O (g) -57.7979-54.6357 H O (l) -68.3174-56.690-3 PO 4 (q) -305.30-43.50 - HPO 4 (q) -308.81-60.34 - H PO 4 (q) -309.8-70.17 H 3 PO 4 (q) -307.90-73.08 - SO 4-16.90-177.34 HS - (q) -4. 3.01 H S(g) -4.815-7.89 H S(q) -9.4-6.54 Guntelerg Approximtion: log f 0.5z I 1 I 10
0-1 - -3-4 -5 Log C T -6-7 -8-9 -10-11 -1-13 -14 0 1 3 4 5 6 7 8 9 10 11 1 13 14 ph 11
0-1 - -3-4 -5 Log C T -6-7 -8-9 -10-11 -1-13 -14 0 1 3 4 5 6 7 8 9 10 11 1 13 14 ph 1