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Chem 16 Exm 009 NONE OF THE ULTIPLE CHOICE PROBLES REQUIRES EXTENSIVE OR TIE COSUING CALCULATIONS. IF YOUR ETHOD REQUIRES EXTENSTIVE CALCULATIONS IT IS EITHER WRONG OR HARDER THAN WHAT IS REQUIRED. 1. Which of the cids in the tble is the strongest cid?. Acetic b. Fluorocetic c. Formic d. Hydrocynic e. Hydrofluoric The one with the lrgest K is the strongest.. 14 g HNO is diluted to 50 ml. Wht is the ph?..05 b. 1.5 W.. HNO = 1 14 x16= 6 g/ mol c. 0.0 1mol 14g 6g mol d. 0.05 = = = 0.89 ph = log H log ( 0.89) 0.051 L 0.5L = = e. 0.89. 50.0 ml of 0.5 HBr is mixed with 100. ml of 0.09 HI? Wht is the ph?. 0.46 Strong cid strong cid b. 0.75 0.5mol 50mL 0.09mol 100mL mmol HBr = x = 17.5 mmol; mmol HI = x = 9mmol c. 1.04 L 1 L 1 6.5mmol mol d. 0.6 H = = 0.177 = 0.177 ; ph = log H = log( 0.177) = 0.75 150mL L e..6 4. 50.0 ml of 0.50 HF is mixed with 50. ml of 0.0 HCl? Wht is the ph?. 0.70 b. 0.46 c. 0.9 d. 1. e. 1.00 ( ) ( ) Strong cid HCl wek cid HF ; strong cid determines the ph 50mL H = 0.0 = 0.1 = ph = log H = log ( 0.1) = 1 100mL

Chem 16 Exm 009 5. 75.0 ml of 0.60 HCl is mixed with 00. ml of 0.0 B(OH). Wht is the ph?. 1.10 b. 0.90 c. 1.74 d. 1.6 e. 10.78 Acid-bse neutrliztion: 0.60mol 75mL 0.mol 00mL mmol H = x = 45 mmol; mmol OH = x = 80mmol L 1 L 1 H q OH q H O q ( ) ( ) ( ) I 45 80 C -45-45 45 E 0 5 5mmol OH = = 0.17 ; 75mL poh = log ( 0.17) = 0.90; ph = 14 poh = 1.10 6. Wht is the ph of 0.64 HCN? K = 4.79 x 10-10.. 4.60 b. 5.01 c. 4.85 d. 9.51 e. 4.76 ( ) ( ) ( ) ( ) HCN q H O q HO q CN g I 0.64 - ~0 0 C -x x x E 0.64-x x x HO CN x = 10 10 5 K; = 4.79x10 ; x = 4.79x10 ( 0.64) = 1.75x10 HCN 0.64 x [ ] 5 ( x ) ph = log 1.75 10 = 4.76 7. A 0. solution of wek cid, HB, hs ph of 5.5? Wht is K?. 1.4x10-5 b. 4.5x10-6 c. 6.0x10-11 d. 4.5 x10-10 e. 1.4x10-11 = = = ph 5.5 6 H 10 10 4.47x10 [ HB] ( ) ( ) ( ) ( ) HB q H O q H O q B g I 0. - ~0 0 C -4.47x10 4.47x10 4.47x10 6 6 6 6 E 0. 4.47x10 4.4 6 ( 4.47x10 ) B 11 HO K = ; = 6.0x10 0. 710 x 6 8. Which of the following ions is cidic?. Fe - d. SO 4 b. Cl - e. N c. CN - N - nd Cl re neutrl ions CN nd SO - - 4 Fe like most trnsition metl ions nd Al is cidic. re conjugte bses to wek cids nd re bsic.,,.

Chem 16 Exm 009 9. A 0.10 solution of ech compound is prepred. Which solution is most bsic?. NH K b = 1.8x10-5 b. NH 4 Br Acidic The one with the lrgest K b is the most bsic. c. C 6 H 5 NH K b = 4.17x10-10 d. NCH COO K = 1.7x10-5 (for CH COOH) K b = 10-14 /1.75x10-5 = 5.9x10-10 e. K S K = 1.1x10-1 (for HS - ) K b = 10-14 /1.1x10-1 = 9.1x10-10. A 150 ml solution hs 0.95 HCN nd 0.15 NCN. Wht is the ph? K = 4.8 x 10-10. 9. b. 10.1 c. 8.5 d. 9.7 e. 8.9 The solution consists of wek cid nd its conjugte bse: use buffer eqution. pk 10 ( x ) = log 4.8 10 = 9. B 0.15 ph = pk log = 9. log = 8.5 HB 0.95 11. Wht hppens to the equilibrium percent dissocition of cetic cid s sodium hydroxide is dded to solution of cetic cid. (The frction dissocited = cette ion concentrtion/cid concentrtion.) CH COOH(q) H O(q)! H O (q) CH COO - (q). No chnge Le Chtelier's principle: stress too little H H O OH H O, so b. Increses c. Decreses d. It cn not be determined ( ) rection goes to right using up CHCOOH nd mking more H nd CH COO. So, the percent dissocition increses. 1. Which of the bses is best for prepring buffer with ph of 9? Creful!. mmoni b. methylmine c. trimethylmine d. hydroxylmine e. niline pk = 14 pk = 14 4.76 = 9.4 b pk = 14.4 = 10.66 pk = 14 4.1 = 9.79 pk = 14 7.97 = 6.0 Wnt pk ph = 9 pk = 14 9.8 = 4.6

Chem 16 Exm 009 1. Buffers neutrlize both cids nd bses. When HCl is dded to buffer solution consisting of potssium formte (KHCOO) nd formic cid (HCOOH), which rection (when completed) shows the neutrliztion of the cid?. H (q) Cl - (q) b. H (q) K (q) c. H (q) HCOOH(q) d. H (q) HCOO - (q) e. None of the bove 14. 40 ml of 0.10 LiOH is dded to 100 ml of solution contining 0.5 NCH COO nd 0.50 CH COOH. Wht is the ph of the resulting solution? (K = 1.75 x 10-5 for cetic cid.). 4. b. 4.50 c. 4.56 d. 4.76 e. 4.96 ( ) ( ) ( ) ( ) I 0.1mol 40mL = 4mmol L 50mmol 5mmol C 4 mmol 4 mmol 4mmol E 0 mmol 46 mmol 9mmol pk OH q CH COOH q H O l CH COO q 5 CHCOO 9 = log ( 1.75x10 ) = 4.76; ph = pk log = 4.76 log = 4.56 CHCOOH 46 15. Which of the following is the best nswer regrding HPO 4?. It is only n cid. b. It is only bse. c. It is the conjugte bse to HPO 4 nd is n cid. d. It is the conjugte bse to HPO 4 nd is n cid. e. It is the conjugte cid to HPO 4 nd is bse. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) HPO q HOl HPO q HO q 4 4 HPO q HOl HPO q HO q 4 4 HPO q H O l PO q H O q 4 4 From this eqution, it is the conjugte bse to HPO: 4 HPO 4 ( q) HOl ( ) HPO4 ( q) HO ( q) From this eqution it is n cid: HPO q H O l PO q H O q ( ) ( ) ( ) ( ) 4 4 16. In which of the following rections does the underlined compound ct s Lewis cid? ( ) ( ) ( ) ( ) I. NH q H O l NH q OH q ( ) () ( ) ( ) II. NH q H O l NH q H O q 4 ( ) ( ) 4 III. NH G OH NH G OH where the N is bonded to the G () only I (b) only II (c) only III (d) I nd II (e) I, II, nd III

Chem 16 Exm 009 PROBLES (1 POINTS EACH) I. () The K sp of Y(OH) is 1.0 x 10-4 t 5ºC. ( PT) Write the solubility product expression (K sp =?): ( ) ( )! ( ) ( ) Y OH s Y q OH q Y OH K sp = (b) (5 PT) Wht is the solubility of Y(OH) in pure wter t 5ºC in mol/l? USE CORRECT NUBER OF SIGNIFICANT FIGURES. I C E ( ) ( )! ( ) ( ) Y OH s Y q OH q excess 0 0 -s s s excess s s 1.0x10 s s = 1x10 ; s= 7 ( )( ) 0.5 = 6 1.4x10 / mol L (c) (5 PT) Wht is the solubility of Y(OH) in solution buffered to ph of 10 t 5ºC? poh = 14 10= 4; OH = 10 = 1.00x10 poh 4 I C E ( ) ( )! ( ) ( ) Al OH s Al q OH q excess 0 10 -s s 0 (buffered solution!) excess s 10-4 -4 4 1.0x10 10 s10 ( ) = 1.010 x ; s = = 1.010 x mol/ L 1 10

Chem 16 Exm 009 II. 5 ml of wek cid re titrted with 0.1 NOH s shown in the grphs. () ( pt) Wht volume of bse is dded to rech the equivlence point? 50 ml (b) (4 pt) Wht is the concentrtion of the cid? (Obviously, this mens before the titrtion strts.) At the equivlence point ll the cid is neutrlized, tht is, moles cid = moles bse. 0.1mol 50mL mol bse = x = 6.5mmol = mol cid L 1 6.5mmol Concentrtion cid = = 0.19 5mL (c) ( pt) Wht is K for the cid? Use the lbeled grph. (There is much extr informtion.) B ph = pk log = pk When [ HB] ( ) [ HB] B =, tht is, when hlf the cid is converted to conjugte bse. So, wnt ph fter 5 ml of bse hve been dded. ph 4.484.0 / = 4. 9 4.9 = = = 4.9 5 pk; pk 10 4.1x10 (d) ( pt) Clerly indicte the region where the buffer eqution works to clculte the ph. Buffer eqution works (e) ( pt) Clerly indicte the region where the strong bse pproximtion works to clculte the ph. Strong bse pproximtion works

Chem 16 Exm 009 III. () (6 pt) 5.00 HCl is used to prepre 50. ml solution whose ph is 1.10. How mny ml of concentrted HCl re needed? H = = = ph 1.1 10 10 0.0794 0.0794mol H 0.5L 1L 1000mL x x x =.97 ml L 1 5 mol conc. H 1L (b) (6 pt) You need to prepre buffered solution with ph of 5.10. Acetic cid is used. The K = 1.74 x 10-5. You strt with 00 ml of 0.45 cetic cid. How mny ml of 0.0 sodium cette must be dded to the 00 ml of cid? B mmol B ph = 5.10 = pk log = pk log becuse the volume is the sme. [ HB] mmol HB pk 5 ( x ) = log 1.74 10 = 4.76 mmol B mmol B = = = = mmol HB mmol HB 0.41 log 5.10 4.759 0.41; 10.19 0.45mol 00ml mmol HB = x = 90mmol L 1 ( ) mmol B =.19 mmol HB =.19x90 = 197mmol 197mmol ml x = 658mLbse must be dded 1 0.mmol