CHEMISTRY 16 HOUR EXAM III Mrch 26, 1998 Dr. Finkle - Answer key 1. Phenol, lso known s crbolic cid, is wek monoprotic cid (HA). A 1.0 M solution of phenol hs ph of 4.95. Wht is the K of phenol? 1. 2.0 10 HA º H + A K = {[H ][A ]}/[HA] b. 1.1 10 1.0 x +x +x 10 4.95 c. 9,1 10 x = [H ] = 10 = 1.1 10 M 10 d. 1.2 10 K = (1.1 10 ) /(1.0 1.1 10 ) = 1.2 10 e. 1.0 10 14 2 10 2. Wht is the expression for K for crbonic cid (H CO )? 1 2 3. [HCO 3 ]/[H2CO 3] H2CO 3 º H + HCO3 K1 + 2 b. {[H ][HCO ]}/[H CO ] HCO º H + CO K + 2 2 c. {[H ] [CO 3 ]}/[H2CO 3] + 2 d. {[H ][CO 3 ]}/[HCO 3 ] e. [H ][HCO 3 ] 3 2 3 3 3 2 3. Wht is the ph of 0.30 M NH 3 solution? K b(nh 3) = 1.8 10 NH 3 + H2O º NH 4 + OH K b = {[NH 4 ][OH ]}/[NH 3]. 11.37 0.30 x +x +x 2 2 6 ½ 3 b. 5.27 1.8 10 = x /(0.30 x). x /0.30; x = (5.4 10 ) = 2.3 10 (< (0.1)(0.3)) 3 c. 13.48 [OH ] = 2.3 10 ; poh = 2.63; ph = 14.00 2.63 = 11.37 d. 2.67 e. 9.26 5 4. Wht is the pk for NH 4 if K b(nh 3) = 1.8 10?. 4.74 K = K w/k b = 1.0 10 /1.8 10 = 5.6 10 b. 14.00 10 pk = log(5.6 10 ) = 9.26 c. 9.26 d. 7.00 e. 5.56 14 10
5. NH4 Cl is dded to solution of NH 3. Wht hppens to the ph? NH 3 + H2 O º NH 4 + OH +. The ph increses. Adding NH 4 (from NH4Cl) shifts the bse b. The ph decreses. dissocition equilibrium bove to the left. + c. The ph remins constnt. [OH ] down, [H ] up, ph down. 6. Which of the following sttements is flse bout solution contining 0.50 M HNO 2 nd 0.50 M NNO 2? (c) is flse becuse bse OH does not rect with bse (NO 2 ).. The ph is equl to the pk of HNO 2. (true t equl conc. of conj. bse & cid) b. Any dded strong cid is consumed by rection with NO 2. (true) c. Any dded OH is consumed by rection with NO 2. d. The solution will resist chnges in ph. (true; the solution is buffer) 7. A buffer contins 1.0 M cetic cid (HA) nd 0.20 M sodium cette (NA). Wht is the ph? K (HA) = 1.8 10 A buffer. ph = pk + log([a ]/[HA]. 4.74 pk = log(1.8 10 ) = 4.74 b. 4.04 ph = 4.74 + log(0.20/1.0) = 4.74 + (0.70) = 4.04 c. 0.30 d. 0.48 e. 4.26 8. To 1.0 L of the preceding buffer is dded 0.10 mol of NOH. Wht is the finl ph? Rect the OH with the conj. cid HA:. 4.74 HA + OH º A + H2O b. 4.04 init. 1.0 mol 0.10 mol 0.20 mol c. 0.30 ) 0.10-0.10 +0.10 d. 0.48 finl 0.90 0 0.30 mol e. 4.26 ph = 4.74 + log(0.30/0.90) = 4.74 + (0.48) = 4.26
9. Three populr indictors for cid-bse titrtions re phenolphthlein (color chnge t ph 8.3-10.0), bromthymol blue (color chnge t ph 6.0-7.6) nd methyl red (color chnge t ph 4.2-6.0). Which indictor is best suited for titrtion of NH Cl solution with NOH? 4. phenolphthlein NH 4 is wek cid, OH is strong bse. At the equivlence b. bromthymol blue volume, the dominnt species in solution is the conj. bse NH. 3 c. methyl red The ph will be bsic (> 7.0), so phenolphthlein is the best choice. 10. Wht is the correct K expression for Ag PO? sp 3 4 + 3 + 3. [3Ag ][PO 4 ] Ag3PO 4 (s) º 3Ag + PO4 + 3 3 b. {[Ag ] [PO 4 ]}/[Ag3PO 4 ] The pure solid Ag3PO 4 does not pper in the + 3 c. {[Ag ][PO 4 ]}/[Ag3PO 4] K sp expression. + 3 3 d. [Ag ] [PO 4 ] + 3 3 e. [Ag ][PO 4 ] 17 11. Wht is the molr solubility of AgI? K sp(agi) = 8.5 10 17. 4.2 10 M AgI(s) º Ag + I 6 b. 2.8 10 M +x +x 17 2 c. 9.6 10 M K sp = [Ag+][I ] = 8.5 10 = x 9 17 ½ 9 d. 9.2 10 M x = (8.5 10 ) = 9.2 10 M 3 e. 8.5 10 M 12. Which of the following slts becomes more soluble when cid is dded to solution sturted with the slt? A. AgCl B. CCO C. Mg(OH) 3 2 2. A only. Acid rects with bsic nions such s CO 3 nd OH. The stress of b. B only. lowering the conc. of these ions increses the solubility of the slt. c. C only. d. A nd B. e. B nd C.
13. Led sulfte (PbSO 4 ) nd sulfuric cid (H2SO 4) re both found in led-cid storge btteries, 8 which we ll depend on to strt our crs. Wht is the molr solubility of PbSO 4 (K sp = 1.8 10 ) 2 if the bttery electrolyte contins 10. M SO 4? 2+ 2 2+ 2. 1.8 10 M PbSO 4 (s) º Pb + SO4 K sp = [Pb ][SO 4 ] 4 b. 1.3 10 M +x 10. + x 8 c. 1.3 10 M 1.8 10 = (x)(10. + x). x(10.) 7 8 9 d. 1.8 10 M x = 1.8 10 /10. = 1.8 10 M (< (0.1)(10.)) 9 e. 1.8 10 M 2+ 2 14. 10.0 mls of 0.0040 M Mg re mixed with 10.0 mls of 0.0040 M CO. Will MgCO -6 precipitte? K sp(mgco 3) = 6.8 10 2+ 2 2+ 2 MgCO 3 (s) º Mg + CO3 IP = [Mg ][CO 3 ] 2+ 2. Yes. After mixing & before rection, [Mg ] = [CO ] = 0.0020 M 3 3 3 6 b. No. IP = (0.0020)(0.0020) = 4.0 10. Since IP < K sp, no precipitte forms. c. Cn t be determined from the dt given. 15. Which of the following sttements is flse?. H O is weker cid thn HF. (F is more electronegtive thn O 6 weker HX bond) 2 b. H O is weker cid thn H S (H-O bond is stronger) 2 2 c. HClO is weker cid thn HClO. (more oxygens on Cl 6 stronger cid) 2 3 d. For diprotic cid, K is much greter thn K. (usully K << K ) 2 1 2 1 16. In the clssic Agth Christie mystery novels, potssium cynide (KCN) ws frequently used poison. Apprently the stuff ws lying round in every kitchen cupbord in Englnd. Wht is the 10 ph of 0.10 M solution of KCN? K (HCN) = 4.9 10 CN + H O º HCN + OH K = {[HCN][OH ]}/[CN ] 2 b. 9.31 0.10 x +x +x 14 10 b. 4.69 K b = K w/k = 1.0 10 /4.9 10 = 2.0 10 2 2 6 ½ 3 c. 11.15 2.0 10 = x /(0.10 x). x /0.10; x = (2.0 10 ) = 1.4 10 (< (0.1)(0.1)) - 3 d. 9.36 [OH ] = 1.4 10 ; poh = 2.85; ph = 14.00 2.85 = 11.15 e. 2.85
17. This question counts double. Scchrin, the rtificil sweetener, is wek cid with K = 12 2.1 10. Wht is the ph of 0.50 M solution of scchrin? Let HA be the wek cid scchrin.. 6.00 HA º H + A K = {[H ][A ]}/[HA] b. 7.00 0.50 x +x +x 12 2 2 c. 8.00 K = 2.1 10 = x /(0.50 x). x /0.50 12 ½ 6 d. 11.68 x = (1.05 10 ) = 1.0 10 (< (0.1)(0.5)) 6 e. 5.84 [H ] = 1.0 10 ; ph = 6.00 18. Mrk the sme nswer s #17. 4 19. This question counts double. 50.0 mls of 0.100 M HF (K = 3.5 10 ) is titrted with 0.200 M NOH. Wht is the ph t V = 20.0 mls? A wek cid - strong bse titrtion. First, find V eq.. 7.00 (50.0 mls)(0.100 M HF) = V eq (0.200 M NOH); V eq = 25.0 mls b. 4.06 Hence, this is cse (2) 0 < V < V eq. It is not the specil cse V = V eq/2. c. 3.46 HF + OH º F + H2O d. 2.86 init. 0.0050 0.0040 0 moles (volume conc.) e. 1.00 ) 0.0040 0.0040 +0.0040 moles finl 0.0010 0 0.0040 Since this is buffer, use the H-H eqution: - 4 ph = pk + log([f ]/[HF]); pk = log(3.5 10 ) = 3.46 ph = 3.46 + log(0.0040/0.0010) = 3.46 + (0.60) = 4.06 20. Mrk the sme nswer s #19.