CHAPTER 17 ADDITIONAL ASPECTS OF ACID BASE EQUILIBRIA PRACTICE EXAMPLES 1A (D) Orgnize the solution round the lnced chemicl eqution, s we hve done efore. Initil: 0.500M 0M 0M Eqution: HF(q) HO(l) HO (q) F (q) Chnges: xm xm xm Equil: (0.500 x) M xm xm [ HO ][ F] ( x) ( x) x HF] x 66. 10 [ 0. 500 0. 500 ssuming x 0.500 x 0500. 6. 6 10 0. 018M One further cycle of pproximtions gives: x (0.500 0.018) 6.6 10 0.018 M [H O ] Thus, [HF] = 0.500 M 0.018 M 0.8 M Recognize tht 0.100 M HCl mens HO initil 0.100 M, since HCl is strong cid. Eqution: HF(q) H O(l) Initil: 0.500M 0.100M 0 M Chnges: xm xm xm Equil: (0.500 x) M (0.100 x) M xm x HO (q) F (q) [ HO ][ F] ( x)( 0100. x) 0100. x 66. 10 [HF] 0. 500 x 0. 500 ssuming x 0.100 6.610 0.500 = =. 10 M = F The ssumption is vlid. 0.100 HF = 0.500 M 0.00M = 0.97 M [H O ] = 0.100 M x = 0.100 M 0.00 M = 0.10 M 786
Chpter 17: Additionl Aspects of Acid Bse Equiliri 1B (M) From Exmple 17-6 in the text, we know tht [ HO ] [ CHO ] 1. 10 Min 0.100 M HCH O. We se our clcultion, s usul, on the lnced chemicl eqution. The concentrtion of H O from the dded HCl is represented y x. Eqution: HCHO (q) HO(l) HO (q) CHO (q) Initil: 0.100M 0M 0M Chnges: 0.00010M 0.00010M 0.00010M From HCl: x M Equil: 0.100 M (0.00010 x) M 0.00010M [ HO ][ CHO ] (0.00010 x) 0. 00010 = 18. 10 5 [ HC H O ] 0100. 5 1.810 0.100 0.00010 x 0.018 M x 0.018 M 0.00010 M 0.018 M 0.00010 0.018 mol H O 1mol HCl 1 Lsoln 1000 ml 1 drop V1 M HCl = 1L 1mol H O 1 mol HCl 1L 0.050 ml 1.00 L 0.drops Since 0. drops corresponds to 1.5 ml of 1 M solution, we see tht the volume of solution does indeed remin pproximtely 1.00 L fter ddition of the 1 M HCl. A (M) We gin orgnize the solution round the lnced chemicl eqution. Eqution: HCHO q H O(l) CHO q H O q Initil: 0.100M 0.150M 0M Chnges: xm xm xm Equil: (0.100 x) M (0.150 x) M xm 1.810 x [CHO ][H O ] (0.150 x)( x) 0.150 x [HCHO ] 0.100 0.100 ssuming x 0.100 0.100 1.810 x = = 1.10 M = HO, <<0.100, thus our ssumption is vlid 0.150 x CHO = 0.150 M 0.0001 M = 0.150M 787
Chpter 17: Additionl Aspects of Acid Bse Equiliri B (M) This time, solid smple of wek se is eing dded to solution of its conjugte cid. We let x represent the concentrtion of cette ion from the dded sodium cette. Notice tht sodium cette is strong electrolyte, thus, it completely ph 5.00 5 dissocites in queous solution. HO = 10 = 10 = 1.0 10 M= 0.000010 M Eqution: HCHOq HO(l) CHO q HO q Initil: 0.100 M 0M 0M Chnges: 0.000010 M 0.000010 M 0.000010 M From NAc: x M Equil: 0.100 M (0.000010 x) M 0.000010 M H O C H O 0.0000100.000010 = = =1.8 10 HC H O 0.100 x 5 5 1.810 0.100 0.000010 x 0.18 M x 0.18 M 0.000010 M = 0.18 M 0.000010 1mol CH 1mol NC O HO 0.18 mol C H O 1mol NC H O 8.0g NC H O mss of NCHO = 1.00 L 1L =15g NC H O A B (M) A strong cid dissocites essentilly completely, nd effectively is source of HO.NCHO lso dissocites completely in solution. The hydronium ion nd the cette ion rect to form cetic cid: HO (q) CHO (q) HCHO (q) HO(l) All tht is necessry to form uffer is to hve pproximtely equl mounts of wek cid nd its conjugte se together in solution. This will e chieved if we dd n mount of HCl equl to pproximtely hlf the originl mount of cette ion. (M) HCl dissocites essentilly completely in wter nd serves s source of hydronium ion. This rects with mmoni to form mmonium ion: NH q HO q NH q HO(l). Becuse uffer contins pproximtely equl mounts of wek se (NH ) nd its conjugte cid (NH ), to prepre uffer we simply dd n mount of HCl equl to pproximtely hlf the mount of NH (q) initilly present. 788
Chpter 17: Additionl Aspects of Acid Bse Equiliri A (M) We first find the formte ion concentrtion, rememering tht NCHO is strong electrolyte, existing in solution s N g q nd CHO q..1g NCHO 1000 ml 1 mol NCHO 1 mol CHO [CHO ] 0.679 M 500.0 ml soln 1L 68.01g NCHO 1 mol NCHO As usul, the solution to the prolem is orgnized round the lnced chemicl eqution. Eqution: HCHO q H O(l) CHO q H O q Initil: 0. M 0.679 M 0 M Chnges: x M xm xm Equil: (0. x) M (0.679 x) M xm HO CHO 0.679 0.679 0. 1.8 10 = x x x = =1.810 x= HCHO 0. x 0. 0.679 This gives HO =1.110 M. The ssumption tht x 0. is clerly correct. ph log HO log 1.110.9.9 B (M) The concentrtions of the components in the 100.0 ml of uffer solution re found vi the dilution fctor. Rememer tht NCHO is strong electrolyte, existing in solution s N g q nd CHO q g. 6.0 ml 7.0 ml [HC H O ] 0.00 M 0.16 M [C H O ] 0.00 M 0.070 M 100.0 ml 100.0 ml As usul, the solution to the prolem is orgnized round the lnced chemicl eqution. Eqution: HC HO q HO(l) CHO q HO q Initil: 0.16 M 0.070 M 0 M Chnges: x M x M x M Equil: ( 016. x) M 0.070 xgm x M 5 1.810 x [H O ][C H O ] x (0.070 x ) 0.070 [HC H O ] 0.16 0.16 5 18. 10 016. 5 x 1. 10 M=[H O 5 ]; ph = log [ H O ] log 1. 10 51. 0. 070 Note tht the ssumption is vlid: x 0.070 0.16. Thus, x is neglected when dded or sutrcted 789
Chpter 17: Additionl Aspects of Acid Bse Equiliri 5A (M) We know the initil concentrtion of NH in the uffer solution nd cn use the ph to find the equilirium [OH - ]. The rest of the solution is orgnized round the lnced chemicl eqution. Our first gol is to determine the initil concentrtion of NH. poh 500. 5 poh = 1.00 ph = 1.00 9. 00 500. [ OH ] 10 10 10. 10 M Eqution: NH q H O(l) NH q OH q Initil: 0.5 M x M 0 M Chnges: 5 1.0 10 M 5 1.0 10 M 5 1.0 10 M Equil: 5 (0.5 1.0 10 ) M ( x 10. 10 ) M 10. 10 5 M 5 5 5 [NH ][OH ] 5 ( x 1.0 10 )(1.0 10 ) 1.0 10 1.810 5 [NH ] 0.5 1.010 0.5 x 5 5 0.51.810 Assume x 1.010 x= = 0.6 M = initil NH 5 concentrtion 1.010 0.6 mol NH 1 mol NH SO 1.1 g NH SO mss NH SO = 0.500 L 1L soln mol NH 1mol NH SO Mss of (NH ) SO = 1 g 5B (M) The solution is composed of.05 g NC H O H O dissolved in 00.0 ml of 0.50 M HCl. NC H O H O, strong electrolyte, exists in solution s N (q) nd C H O - (q) ions. First we clculte the numer of moles of NC H O H O, which, sed on the 1:1 stoichiometry, is lso equl to the numer of moles of C HO - tht re relesed into solution. From this we cn clculte the initil [C H O - ] ssuming the solution's volume remins t 00. ml. moles of NCHO H O (nd moles of C H O - ).05 g NCHO HO = 0.moles NC H O H O moles C H O 1 mole NCHO HO 16.08 g NC H O H O 0. mol CHO [CHO ] 0.810 M 0.00 L soln (Note: [HCl] is ssumed to remin unchnged t 0.50 M) We orgnize this informtion round the lnced chemicl eqution, s efore. We recognize tht virtully ll of the HCl hs een hydrolyzed nd tht hydronium ion will rect to produce the much weker cetic cid. 790
Chpter 17: Additionl Aspects of Acid Bse Equiliri Eqution: HCHO (q) HO(l) CHO (q) HO (q) Initil: 0 M 0.810 M 0.50 M Form HAc: 0.50 M 0.50 M 0.50 M 0.50 M 0.560 M 0 M Chnges: x M x M x M Equil: (0.50 x) M (0.560 x) M x M [HO ][CHO ] (0.560 x ) 5 0.560 1.8 10 [HCHO ] 0.50 x 0.50 5 1.810 0.50 6 x 8.0 10 M = [HO ] 0.560 (The pproximtion ws vlid since x << oth 0.50 nd 0.560 ) 6 ph = log [H O ] log 8.010 5.09 5.1 6A (D) () For formic cid, p log( 18. 10 ) 7.. The Henderson-Hssellch eqution provides the ph of the originl uffer solution: [CHO ] 0 ph = p log. log. 50 7 5. [ HCHO ] 0550. () The dded cid cn e considered completely rected with the formte ion to produce formic cid. Ech mole/l of dded cid consumes 1 M of formte ion nd forms 1 M of formic cid: CHO (q) HO (q) HCHO (q) HO(l). neut = / w 5600. Thus, CHO 0.50 M 0.0050 M = 0.5 M nd HCHO = 0.550 M 0.0050 M = 0.555 M. By using the Henderson-Hssellch eqution [CHO ] 0.5 ph = p log =.7 log =.5 [HCHO ] 0.555 (c) Added se rects completely with formic cid producing, n equivlent mount of formte ion. This continues until ll of the formic cid is consumed. Ech 1 mole of dded se consumes 1 mol of formic cid nd forms 1 mol of formte ion: HCHO OH CHO HO. neut = / w 1.8 10 10. Thus, CHO = 0.50 M 0.0050 M = 0.55 M HCHO = 0.550 0.0050 M = 0.55 M. With the Henderson-Hssellch eqution we find CHO 0.55 ph = p log =.7 log =.55 HCHO 0.55 791
Chpter 17: Additionl Aspects of Acid Bse Equiliri 6B (D) The uffer cited hs the sme concentrtion s wek cid nd its nion, s does the uffer of Exmple 17-6. Our gol is to rech ph = 5.0 or ph 5.0 6 HO = 10 = 10 = 9.10 M. Adding strong cid ch O h, of course, produces t the expense of CHO. Thus, dding H drives the rection to the left. HCHO Agin, we use the dt round the lnced chemicl eqution. Eqution: HC H O q H O(l) CHO qg H O q Initil: 0.50 M 0.560 M 8.0 10 6 M Add cid: y M Form HAc: y M y M y M 0.50 yg M 0.560 y x M x g M 0 M g Chnges: M x M Equil: 0.50 y xg M 0.560 y xg M 9.10 6 M 6 6 [HO ][C HO ] 9.10 (0.560 y x) 5 9.10 (0.560 y) 1.810 [HCHO ] 0.50 y x 0.50 y (Assume tht x is negligile compred to y) 5 1.8 10 0.50 y 0.560 0.8 = 0.8 1.9 y= 0.560 y y= 9. 10 1.9 1.00 = 0.06 6 x 0.50 y = 0.76 0.560 y = 0.5. g M Notice tht our ssumption is vlid: 0.06 mmol H O 1mL HNO q V 00.0 ml uffer = 1. ml of 6.0 M HNO HNO 1 ml uffer 6.0 mmol HO Insted of the lgeric solution, we could hve used the Henderson-Hssellch eqution, since the finl ph flls within one ph unit of the p of cetic cid. We let z indicte the increse in HCH O, nd lso the decrese in CHO ph p log CHO 0.560 z = =.7 log HC H O 0.50 =5.0 0.560 z 5.0.7 L O = 10 = 1.95 NM QP z 0.50 z 0.560 0.560 0.88 z= 1.950.50 zg= 0.88 1.95 z z= 1.951.00 =0.0 M This is, nd should e, lmost exctly the sme s the vlue of y we otined y the I.C.E. tle method. The slight difference is due to imprecision rising from rounding errors. 7A (D) () The initil ph is the ph of 0.150 M HCl, which we otin from HO strong cid solution. of tht [H O 0150. mol HCl 1 mol H O ] 0150. M, 1Lsoln 1 mol HCl ph = log [H O ] log ( 0150. ) 08. 79
Chpter 17: Additionl Aspects of Acid Bse Equiliri () To determine HO nd then ph t the 50.0% point, we need the volume of the solution nd the mount of H O left unrected. First we clculte the mount of hydronium ion present nd then the volume of se solution needed for its complete neutrliztion. mount H O 0.150 mmol HCl 1 mmol H O = 5.00 ml 1 ml soln 1 mmol HCl =.75mmol H O V 1 mmol OH 1 mmol NOH 1 ml titrnt =.75 mmol H O 1 mmol H O 1 mmol OH 0.50 mmol NOH cid 15.0 ml titrnt At the 50.0% point, hlf of the H O (1.88 mmol H O ) will remin unrected nd only hlf (7.50 ml titrnt) of the titrnt solution will e dded. From this informtion, nd the originl 5.00-mL volume of the solution, we clculte HO nd then ph. 7B (c) (d) (D) () 1.88 mmol H O left 5.00 ml originl 7.50 ml titrnt ph = log 0.0578 = 1.8 HO = = 0.0578 M Since this is the titrtion of strong cid y strong se, t the equivlence point, the ph = 7.00. This is ecuse the only ions of pprecile concentrtion in the equivlence point solution re N (q) nd Cl - (q), nd neither of these species undergoes detectle hydrolysis rections. Beyond the equivlence point, the solution ph is determined lmost entirely y the concentrtion of excess OH - (q) ions. The volume of the solution is 0.00mL 1.00mL = 1.00 ml. The mount of hydroxide ion in the excess titrnt is clculted nd used to determine OH, from which ph is computed. 0.50 mmol NOH mount of OH = 1.00 ml = 0.50 mmol OH 1mL 0.50 mmol OH OH = = 0.006098 M 1.00 ml poh = log 0.006098 =.15; ph = 1.00.15 = 11.785 The initil ph is simply the ph of 0.0081 M BOHg, which we otin from OH for the solution. g mol B OH mol OH OH = 0.0081 = 0.016 M 1L soln 1mol BOHg poh = log OH = log 0.016 = 1.790; ph = 1.00 poh = 1.00 1.790 = 1.1 79
Chpter 17: Additionl Aspects of Acid Bse Equiliri () To determine OH nd then ph t the 50.0% point, we need the volume of the solution nd the mount of OH unrected. First we clculte the mount of hydroxide ion present nd then the volume of cid solution needed for its complete neutrliztion. 0.0081 mmol B OH mmol OH mount OH = 50.00 ml = 0.81 mmol OH 1mL soln 1mmol B OH 1mmol HO 1mmol HCl 1mL titrnt cid 1mmol OH 1mmol HO V = 0.81mmol OH =.8mL titrnt 0.050 mmol HCl At the 50.0 % point, hlf (0.06 mmol OH ) will remin unrected nd only hlf (16. ml titrnt) of the titrnt solution will e dded. From this informtion, nd the originl 50.00-mL volume of the solution, we clculte OH nd then ph. 0.06mmol OH left OH = = 0.0061 M 50.00 ml originl 16. ml titrnt poh = log 0.0061 =.1; ph = 1.00 poh = 11.79 (c) Since this is the titrtion of strong se y strong cid, t the equivlence point,ph = 7.00. The solution t this point is neutrl ecuse the dominnt ionic species in solution, nmely B (q) nd Cl - (q), do not rect with wter to detectle extent. 8A (D) () Initil ph is just tht of 0.150 M HF ( p = log 6.6 10 =.18 c h ). 0.150 mmol HF [Initil solution contins 0.00 ml =.00 mmol HF] 1mL Eqution : HF q H O(l) H O q F q Initil : 0.150M 0M 0M Chnges : xm x M x M Equil: (0.150 x) M x M x M HO F = = = 6.610 HF 0.150 x 0.150 - x x x x 0.150 6.6 10 9.9 10 M x 0.05 0.150. The ssumption is invlid. After second cycle of pproximtion, HO = 9.6 10 M ; ph = log 9.6 10 =.0 79
Chpter 17: Additionl Aspects of Acid Bse Equiliri () 8B When the titrtion is 5.0% complete, there re (0.5.00=) 0.75 mmol F for every.00 mmol HF tht were present initilly. Therefore, (.00-0.75=).5 mmol HF remin untitrted. We designte the solution volume (the volume holding these.00 mmol totl) s V nd use the Henderson-Hssellch eqution to find the ph. ph p log F 0.75 mmol / V = =.18 log =.70 HF.5 mmol / V (c) At the midpoint of the titrtion of wek se, ph = p =.18. (d) At the endpoint of the titrtion, the ph of the solution is determined y the conjugte se hydrolysis rection. We clculte the mount of nion nd the volume of solution in order to clculte its initil concentrtion. 0.150 mmol HF 1 mmol F mount F = 0.00 ml =.00 mmol F 1 ml soln 1 mmol HF 1mmol OH 1 ml titrnt volume titrnt =.00 mmol HF = 1.0 ml titrnt 1mmol HF 0.50 mmol OH.00mmol F F = = 0.098M 0.00mL originl volume 1.0mL titrnt We orgnize the solution of the hydrolysis prolem round its lnced eqution. Eqution : F q HO(l) HF q OH q Initil: 0.098M 0M 0M Chnges : xm xm xm Equil : 0.098 M M M x x x 1 HF OH 1.010 x x x 6.6 10 0.098 x 0.098 w 11 = = = =1.5 10 = F 11 6 x 0.09 1.510 1. 10 M = [OH ] The ssumption is vlid (x 0.09). 6 poh = log 1.10 = 5.9; ph = 1.00 poh = 1.00 5.9 = 8.08 (D) () The initil ph is simply tht of 0.106 M NH. NH q HO(l) NH q OH q Initil: 0.106 M 0 M 0 M Chnges: x M x M x M Equil: 0.106 xgm x M x M Eqution: 795
Chpter 17: Additionl Aspects of Acid Bse Equiliri NH OH x x x = = =1.810 5 NH 0.106 x 0.106 x 0.1061.810 1. 10 M = [OH - ] The ssumption is vlid (x << 0.106). poh = log 0.001 =.85 ph = 1.00 poh = 1.00.85 = 11.15 () When the titrtion is 5.0% complete, there re 5.0 mmol NH for every 100.0 mmol of NH tht were present initilly (i.e., there re 1. mmol of NH in solution),.98 mmol NH remin untitrted. We designte the solution volume (the volume holding these 5.0 mmol totl) s V nd use the sic version of the Henderson-Hssellch eqution to find the ph. 1. mmol NH poh = p log =.7 log V =.6 NH.98 mmol V ph = 1.00.6 = 9.7 (c) At the midpoint of the titrtion of wek se, poh = p =.7 nd ph = 9.6 (d) At the endpoint of the titrtion, the ph is determined y the conjugte cid hydrolysis rection. We clculte the mount of tht ction nd the volume of the solution in order to determine its initil concentrtion. mount NH mount NH = 5.0 mmol NH 0.106 mmol NH 1mmol NH = 50.00 ml 1 ml soln 1 mmol NH 1 mmol HO 1 ml titrnt titrnt =5.0mmol NH 1 mmol NH 0.5 mmol HO V =.6mL titrnt 5.0 mmol NH NH = = 0.070 M 50.00 ml originl volume.6 ml titrnt We orgnize the solution of the hydrolysis prolem round its lnced chemicl eqution. Eqution: NH q HO(l) NH q HO q Initil: 0.070 M 0 M 0 M Chnges: x M x M x M Equil: 0.070 xgm x M x M 796
Chpter 17: Additionl Aspects of Acid Bse Equiliri NH H O w 1.0 10 = = = NH 1.8 10 1 5 10 x x x =5.610 = 0.070 x 0.070 x 0.070 5.6 10 6.10 M=[H O ] 10 6 The ssumption is vlid ( x << 0.070). c h 6 ph = log 6.10 = 5.0 9A (M) The cidity of the solution is principlly the result of the hydrolysis of the cronte ion, which is considered first. Eqution: CO q H O(l) HCO q OH q Initil: 1.0 M 0 M 0 M Chnges: x M x M x M Equil: 1.0 x g M x M x M 1 w 1.010 HCO OH = = =.1 10 = x x x 11 (HCO ) CO x.7 10 1.0 1.0 x x M Now we consider the hydrolysis of the icronte ion. 1.0.1 10 1.5 10 M 0.015 M [OH ] The ssumption is vlid ( 1.0 ). Eqution: HCO q HO(l) HCOq OH q Initil: 0.015 M 0 M 0.015 M Chnges: y M y M y M Equil: 0.015 ygm y M 0.015 yg M 1 y y w 1.0 10 HCO OH 8 0.015 = F HCO = =. 10 = = 7 H I. 10 HCO 0.015 x g 0.015y y 0.015 = 8 The ssumption is vlid (y << 0.015) nd y = HCO =.10 M. Clerly, the second hydrolysis mkes negligile contriution to the cidity of the solution. For the entire solution, then poh = log OH = log 0.015 = 1.8 ph = 1.00 1.8 = 1.18 g 9B (M) The cidity of the solution is principlly the result of hydrolysis of the sulfite ion. Eqution: SO q HO(l) HSO q OH q Initil: 0.500 M 0 M 0 M Chnges: x M x M x M Equil: 0.500 xgm x M x M 797
Chpter 17: Additionl Aspects of Acid Bse Equiliri 1 HSO 7 OH w 1.010 = = =1.6 10 x x x 8 HSO x 6. 10 [SO ] 0.500 0.500 x The ssumption is vlid ( x << 0.500). 7 0.500 1.6 10.8 10 M = 0.0008 M = [OH ] Next we consider the hydrolysis of the isulfite ion. Eqution: HSO q H O(l) H SO q OH q Initil: 0.0008 M 0 M 0.0008 M Chnges: y M y M y M Equil: 0.0008 y g M y M 0.0008 y 1.010 = = =7.710 1 w HSO 1.10 1 HSO OH y 0.0008 y 0.0008 y 1 =7.7 10 = = = HSO 0.0008 y 0.0008 1 The ssumption is vlid (y << 0.0008) nd y = HSO = 7.7 10 M. Clerly, the second hydrolysis mkes negligile contriution to the cidity of the solution. For the entire solution, then poh = log OH = log 0.0008 =.55 ph = 1.00.55 = 10.5 g g M y A. (D) INTEGRATIVE EXAMPLE From the given informtion, the following cn e clculted: ph of the solution =.716 therefore, [H ] = 1.9 10 - ph t the hlfwy point = p ph =.60 = p p = -log therefore =.50 10-5 FP = 0 C T T = -i m f f f f T -1 1.86 C / m mollity 798
Chpter 17: Additionl Aspects of Acid Bse Equiliri mollity = # moles solute kg solvent = # moles solute 0.500 kg - 0.00750 kg To determine the numer of moles of solute, convert 7.50 g of unknown cid to moles y using its molr mss. The molr mss cn e clculted s follows: ph =.716 [H ] = 1.9 10 - HA H - A 7.50 g Initil MM 0.500 L 0 0 Chnge x x x Equilirium 7.50 g MM 1.9 10 0.500 L 1.9 10-1.9 10 - -5.50 10 = - [H ][A ] [HA] (1.9 10 ).50 10 7.50 MM 1.9 10 0.500 MM =100. g/mol -5 1 mol # moles of solute = 7.50 g 0.077 mol 100. g Mollity = # moles solute g solvent 0.077 mol = 0.15 0.500 g - 0.00750 g m T = -i m f f f f T -1 1.86 C / m 0.15 m T -0.8 C FP = 0 C T = 0 C 0.8 C = 0.8 C f 799
Chpter 17: Additionl Aspects of Acid Bse Equiliri B. (M) By looking t the titrtion curve provided, one cn deduce tht the titrnt ws strong cid. The ph efore titrnt ws dded ws sic, which mens tht the sustnce tht ws titrted ws se. The ph t the end of the titrtion fter excess titrnt ws dded ws cidic, which mens tht the titrnt ws n cid. Bsed on the titrtion curve provided, the equivlence point is t pproximtely 50 ml of titrnt dded. At the hlfwy point, of pproximtely 5 ml, the ph = p. A ph ~8 is otined y extrpoltion the hlfwy point. ph = 8 = p = 10-8 = 1 10-8 = w / = 1 10-6 ~50 ml of 0. M strong cid (1 10 - mol) ws needed to rech the equivlence point. This mens tht the unknown contined 1 10 - mol of wek se. The molr mss of the unknown cn e determined s follows: 0.800 g 1 10 mol 80 g/mol The Common-Ion Effect EXERCISES 1. (M) () Note tht HI is strong cid nd thus the initil HO = HI = 0.089 M Eqution : HCH5O HO CH5O HO Initil : 0.75 M 0M 0.089M Chnges : x M x M x M Equil : 0.75 x M M 0.089 x M x CH5O HO 5 0.089 0.089 5 = x x x = 1.10 = x.010 M HC H O 0.75 x 0.75 5 The ssumption tht x 0.089 M is correct. HO = 0.089 M () OH w 1.0 10 = = HO 0.089 1 =1.110 1 M (c) 5 CHO 5 = x =.0 10 M 800
Chpter 17: Additionl Aspects of Acid Bse Equiliri I = HI = 0.089 M (d) int. (M) () The NH Cl dissocites completely, nd thus, NH = Cl = 0.10 M int int Eqution: NH (q) HO(l) NH (q) OH (q) Initil: 0.16 M 0.10 M 0 M Chnges: x M x M x M Equil: 0.16 xgm 0.10 xgm x M NH OH 0.10 0.10 = = = 1.8 10 ; =.9 10 M x x 5 x 5 x NH 0.16 x 0.16 Assumed x 0.10 M, clerly vlid ssumption. OH = x =.910 5 M () NH = 0.10 x = 0.10M (c) Cl = 0.10 M 1 1.0 10 10 (d) HO = =.10 M 5.9 10. (M) () We first determine the ph of 0.100 M HNO. Eqution HNO (q) HO(l) NO (q) HO (q) Initil : 0.100 M 0M 0M Chnges : x M xm xm Equil : 0.100 x M xm xm NO HO x = = 7.10 = HNO 0.100 x Vi the qudrtic eqution roots formul or vi successive pproximtions, x = 8.110 M = H O. Thus ph = log 8.1 10 =.09 When 0.100 mol NNO is dded to 1.00 L of 0.100 M HNO, solution with NO = 0.100 M = HNO is produced. The nswer otined with the Henderson- Hssellch eqution, is ph = p = log 7. 10 =.1 cused ph chnge of 1.05 units. c h. Thus, the ddition hs 801
Chpter 17: Additionl Aspects of Acid Bse Equiliri () NNO contriutes nitrte ion, NO, to the solution. Since, however, there is no moleculr HNO q g in equilirium with hydrogen nd nitrte ions, there is no equilirium to e shifted y the ddition of nitrte ions. The [H O ] nd the ph re thus unffected y the ddition of NNO to solution of nitric cid. The ph chnges re not the sme ecuse there is n equilirium system to e shifted in the first solution, wheres there is no equilirium, just chnge in totl ionic strength, for the second solution.. (M) The explntion for the different result is tht ech of these solutions hs cette ion present, CHO, which is produced in the ioniztion of cetic cid. The presence of this ion suppresses the ioniztion of cetic cid, thus minimizing the increse in HO. All three solutions re uffer solutions nd their ph cn e found with the id of the Henderson-Hssellch eqution. () ph p log CHO 0.10.7 = =.7 log =.7 HO = 10 = 1.8 10 M HC H O 1.0 HO 1.8 10 M % ioniztion = 100% 100% 0. 018% HC H O 1.0M () (c) 5. (M) () ph = p log =.7 log =.7 CHO 0.10 HCHO 0.10.7 5 HO =10 =1.810 M HO 5 1.8 10 M % ioniztion = 100% 100% 0. 018% HC H O 0.10M ph = p log =.7 log = 5.7 CHO 0.10 HCHO 0.010 5.7 6 HO =10 =1.810 M HO 6 1.8 10 M % ioniztion = 100% 100% 0. 018% HC H O 0.010M The strong cid HCl suppresses the ioniztion of the wek cid HOCl to such n extent tht negligile concentrtion of H O is contriuted to the solution y HOCl. Thus, HO = HCl = 0.05 M 80
Chpter 17: Additionl Aspects of Acid Bse Equiliri () This is uffer solution. Consequently, we cn use the Henderson-Hssellch p = log 7. 10 =.1; eqution to determine its ph. NO 0.100 M ph = p log =.1 log =.0 HNO 0.0550 M (c) HO =10 =.0 10.0 M This lso is uffer solution, s we see y n nlysis of the rection etween the components. HO q, from HCl CHO q, from NC H O HCHO q HO(l) Eqution: In soln: 0.055 M 0.0768 M 0 M Produce HAc: 0.055 M 0.055 M 0.055 M Initil: 0M 0.0 M 0.055 M Now the Henderson-Hssellch eqution cn e used to find the ph. 5 p = log 1.8 10 =.7 c h CHO 0.0M ph = p log =.7 log =.1 HC H O 0.055M.1 5 HO =10 =.9 10 M 6. (M) () Neither B g q nor Cl g q hydrolyzes to mesurle extent nd hence they hve no effect on the solution ph. Consequently, OH y the B OH g solute. 0.006 mol B OH mol OH OH = = 0.01 M 1 L soln 1mol B OH is determined entirely () We use the Henderson-Hssellch eqution to find the ph for this uffer solution. mol NH 1 mol NH SO NH = 0.15 M NH SO = 0.60 M p = 9.6 for NH. NH 0.86 M ph = p log = 9.6 log = 9.15 poh = 1.00 9.15 =.85 NH 0.60 M OH =10 =1.10.85 5 M 80
Chpter 17: Additionl Aspects of Acid Bse Equiliri (c) This solution lso is uffer, s nlysis of the rection etween its components shows. Eqution: NH q, from NH Cl OH q, from NOH NH q HO(l) In soln: 0.6 M 0.196 M 0 M Form NH : 0.196 M 0.196 M 0.196 M Initil: 0.068 M 0M 0.196 M Buffer Solutions NH 0.196 M ph = p log = 9.6 log = 9.7 poh = 1.00 9.7 =.8 0.068 M NH.06 5 7. (M) HO =10 =8.7 10 M. We let S = CHO Eqution: HCHO q H O(l) CHO q H O q Initil : 0.66M S M 0M 5 5 5 Chnges : 8.7 10 M 8.7 10 M 8.7 10 M 5 5 Equil : 0.66 M S 8.7 10 M 8.7 10 M int 5 5 HO CHO S 8.7 10 8.7 10 5 c h 8.7 10 S = L =1.8 10 = HCHO O ; S =0.76 M 0.66 0.66 NM QP 5 To determine S, we ssumed S 8.710 M, which is clerly vlid ssumption. Or, we could hve used the Henderson-Hssellch eqution (see elow). p = log 1.8 10 =.7 c h.06 =.7 log ; =.1; CHO =.1 0.66 = 0.77 M CHO CHO HCHO HCHO The difference in the two nswers is due simply to rounding. 8. (E) We use the Henderson-Hssellch eqution to find the required [NH ]. 5 p = log 1.8 10 =.7 c h p p ph log NH = 1.00 = 1.00.7 = 9.6 = 9.1 = 9.6 NH NH 0.1 = 10 = 0.7 NH = 0.7 NH = 0.7 0.7 M = 0.5M NH 80
Chpter 17: Additionl Aspects of Acid Bse Equiliri 9. (M) HCHO 7 5 q HO(l) CHO 7 5 (q) HO (q) Initil: 0.01 M 0.0 M 0 M Chnges: x M x M x M Equil: 0.01 xm 0.0 xgm x M () Eqution: H O C H O 0.0 0.0 = = 6. 10 = =. 10 M 7 5 5 x x x 5 x HC7H5O 0.01 x 0.01 To determine the vlue of x, we ssumed x 0.01 M, which is n ssumption tht clerly is correct. 5 5 HO =.10 M ph= log.10 =.6 () c h Eqution: NH q HO(l) NH (q) OH (q) Initil : 0.08 M 0.15 M 0 M Chnges : x M x M x M Equil : 0.08 x M 0.15 x M x M g NH OH x 0.15 x = L =1.8 10 = NH O NM QP 0.08 x 0.15x 0.08 x =.810 5 5 To determine the vlue of x, we ssumed x 0.15, which clerly is vlid ssumption. OH =.8 10 5 M; poh = log.8 10 5 =.; ph = 1.00. = 9.68 10. (M) Since the mixture is uffer, we cn use the Henderson-Hssellch eqution to determine of lctic cid. 1.00 g NC H O 5 1000 ml 1 mol NC H O 1 mol C H O 5 5 CHO = =0.089 M 5 100.0 ml soln 1 L soln 11.1 g NC H O 1 mol NC H O CH5O 0.089 M HCH5O 0.0500 M 5 5 ph =.11 = p log = p log = p 0.51 p =.11 0.51 =.86; = 10 = 1. 10.86 M 11. (M) () () 0.100 M NCl is not uffer solution. Neither ion rects with wter to detectle extent. 0.100 M NCl 0.100 M NH Cl is not uffer solution. Although wek cid, NH, is present, its conjugte se, NH, is not. 805
Chpter 17: Additionl Aspects of Acid Bse Equiliri (c) (d) (e) (f) 1. (M) () () (c) 0.100 M CH NH nd 0.150 M CH NH Cl is uffer solution. Both the wek se, CH NH, nd its conjugte cid, CH NH, re present in pproximtely equl concentrtions. 0.100 M HCl 0.050 M NNO is not uffer solution. All the NO hs converted to HNO nd thus the solution is mixture of strong cid nd wek cid. 0.100 M HCl 0.00 M NCHO is uffer solution. All of the HCl rects with hlf of the C H O to form solution with 0.100 M HCHO, wek cid, nd 0.100 M C H O, its conjugte se. 0.100 M HCH O nd 0.15 M NCH5O is not uffer in the strict sense ecuse it does not contin wek cid nd its conjugte se, ut rther the conjugte se of nother wek cid. These two wek cids (cetic, =1.810 5, nd propionic, =1.510 5 ) hve pproximtely the sme strength, however, this solution would resist chnges in its ph on the ddition of strong cid or strong se, consequently, it could e rgued tht this system should lso e clled uffer. Rection with dded cid: HPO HO HPO HO Rection with dded se: H PO OH HPO H O We ssume initilly tht the uffer hs equl concentrtions of the two ions, HPO = HPO HPO HPO ph = p log = 7.0 0.00 = 7.0 (ph t which the uffer is most effective). HPO 0.150 M ph = 7.0 log = 7.0 log = 7.0 0.8 = 7.68 HPO 0.050 M 1. (M) 1 g 1mol C H NH Cl 1mol C H NH moles of solute = 1.15 mg 1000 mg 19.6 g 1mol C H NH Cl CHNH 6 5 =8.8710 mol C H NH 6 mol C H NH = 8.87 10.18 L soln 6 6 5 6 5 =.7910 6 5 6 5 6 5 Eqution: C H NH q H O(l) C H NH q OH q 6 M 6 5 6 5 Initil: 0.105 M.79 10 6 M 0 M Chnges: x M x M x M 6 Equil: 0.105 xgm c.79 10 xhm x M 806
C6H5NH OH 10 = L =7. 10 = CHNHO NM 6 5 QP c 6.79 10 xx 0.105 x Chpter 17: Additionl Aspects of Acid Bse Equiliri 7.10 0.105 x =.7910 x x; 7.810 7.10 x=.79 10 x x 10 6 11 10 6 6 10.79 10 7. 10 7.8 10 11 = 0; x x x.79 10 6 x 7.810 11 0 c.79 10 7.78 10.110 x = = 6 1 10 c h 6 poh = log 7.510 = 5.1 ph = 1.00 5.1 = 8.88 1. (M) We determine the concentrtion of the ction of the wek se. 15. (M) 1 mmol C6H5NH Cl 1 mmol C6H5NH 8.50 g 6 5 6 5 h 6 =7.510 M= OH 19.6 g 1 mmol C H NH Cl [C H NH ] 0.087 M 1 L 750 ml 1000 ml In order to e n effective uffer, ech concentrtion must exceed the ioniztion constnt c =7.10 h 10 y fctor of t lest 100, which clerly is true. Also, the rtio of the two concentrtions must fll etween 0.1 nd 10: CHNH 6 5 M LCHNHO = 0.087 = 0.07. NM 6 5 QP 0.15M Since oth criteri re met, this solution will e n effective uffer. 5 () First use the Henderson-Hssellch eqution. p = 1.00.7 = 9.6 to determine NH p = log 1.8 10 =.7, in the uffer solution. NH NH NH ph = 9.5 = p log = 9.6 log ; log = 9.5 9.6 = 0.19 NH NH NH NH 0.19 NH 0.58M =10 =1.55 NH = = =0.17M NH 1.55 1.55 We now ssume tht the volume of the solution does not chnge significntly when the solid is dded. 1 L soln 0.17 mol NH 1molNH SO mssnh SO = 5 ml 1000 ml 1 L soln molnh 1.1 g NH SO =.8g NH SO 1mol NH SO 807
Chpter 17: Additionl Aspects of Acid Bse Equiliri () We cn use the Henderson-Hssellch eqution to determine the rtio of concentrtions of ction nd wek se in the ltered solution. NH NH NH ph = 9.0 = p log = 9.6 log log = 9.0 9.6 = 0.0 NH NH NH NH 0.0 0.58 = 10 = 1.1 = 0.19 1.1x 0.58 x 0.06 M NH 0.17 M x M The reson we decided to dd x to the denomintor follows. (Notice we cnnot remove component.) A ph of 9.0 is more cidic thn ph of 9.5 nd therefore the conjugte cid s dnh i concentrtion must increse. Additionlly, mthemtics tells us tht for the concentrtion rtio to decrese from 1.55 to 1.1, its denomintor must increse. We solve this expression for x to find vlue of 0.06 M. We need to dd NH to increse its concentrtion y 0.06 M in 100 ml of solution. NH 0.06 mol NH SO mss = 0.100 L 1 L mol NH 1mol NH SO 1mol NH SO 1.1g NH SO = 0.1g NH SO Hence, we need to dd 0. g 16. (M) () () 1mol HC H O n =.00 g HC H O = 0.016mol HC H O 7 5 HC7H5O 7 5 7 5 1.1g HC7H5O 1mol NCHO 7 5 1mol CHO 7 5 =.00 g NCHO CHO 7 5 7 5 1.1g NC 7 H 5 O 1mol NC 7 H 5 O = 0.019 mol C7H5O n CHO 7 5 0.019 mol C H O /0.7500 L ph = p log = log6.10 5 log 7 5 HC H O 0.016 mol HC H O /0.7500 L 7 5 7 5 =.0 0.0718 =.1 To lower the ph of this uffer solution, tht is, to mke it more cidic, enzoic cid must e dded. The quntity is determined s follows. We use moles rther thn concentrtions ecuse ll components re present in the sme volume of solution. 0.019 mol C7H5O 0.019 mol C H O.00 =.0 log log xmol HC H O xmol HC H O 7 5 7 5 7 5 = 0.0 0.019mol C H O 0.019 x mol HC H O 0.6 7 5 0.0 = 10 = 0.6 x = = 0.0 mol HC7H5O (required) 7 5 808
Chpter 17: Additionl Aspects of Acid Bse Equiliri HC H O tht must e dded = mount required mount lredy in solution 7 5 HC H O tht must edded = 0.0mol HC H O 0.016mol HC H O 7 5 7 5 7 5 HC7H5O tht must edded = 0.006 mol HC7H5O 1.1g HC H O dded mss HCHO 7 5 = 0.006mol HCHO 7 5 1mol HC H O 7 5 7 5 =0.7g HC H O 7 5 17. (M) The dded HCl will rect with the mmoni, nd the ph of the uffer solution will decrese. The originl uffer solution hs NH = 0.58 M nd NH = 0.17 M. We first clculte the [HCl] in solution, reduced from 1 M ecuse of dilution. [HCl] 0.55mL dded =1M = 0.066M We then determine p for mmonium ion: 100.6mL 5 p = log 1.8 10 =.7 p = 1.00.7 = 9.6 c h Eqution: NH q H O q NH q H Ol Buffer: 0.58 M 0 M 0.17 M Added: 0.066 M Chnges: 0.066 M 0.066 M 0.066 M Finl: 0.19 M 0 M 0. M ph p log NH 0.19 = L =9.6 log NH 0. =9.16 N M O QP 18. (M) The dded NH will rect with the enzoic cid, nd the ph of the uffer solution will increse. Originl uffer solution hs [C 7 H 5 O ] = 0.019 mol C 7 H 5 O /0.750 L = 0.0185 M nd HC7H5O = 0.016 mol HC7H5O /0.7500 L = 0.019 M. We first clculte the NH in solution, reduced from 15 M ecuse of dilution. 0.5 ml NH dded = 15 M = 0.0070 M 750.5 ml 5 For enzoic cid, p = log 6.10 =.0 c h 7 5 7 5 Buffer: 0 M 0.019 M 0 M 0.0185 M Added: 0.0070 M Chnges: 0.0070M 0.0070 M 0.0070 M 0.0070 M Finl: 0.000 M 0.019 M 0.0070 M 0.055 M Eqution: NH q HCHO q NH q CHO q CHO 7 5 0.055 ph = p log =.0 log =. HC H O 0.019 7 5 809
Chpter 17: Additionl Aspects of Acid Bse Equiliri 19. (M) The p s of the cids help us choose the one to e used in the uffer. It is the cid with p within 1.00 ph unit of.50 tht will do the trick. p =.7 for HCHO, p =.7 for HCHO, nd p 1 =.15 for HPO. Thus, we choose HCHO nd NCHO to prepre uffer with ph =.50. The Henderson-Hssellch eqution is used to determine the reltive mount of ech component present in the uffer solution. ph =.50 =.7 log log =.50.7 = 0. CHO = 10 = 0.58 HCHO 0. CHO CHO HCHO HCHO This rtio of concentrtions is lso the rtio of the numer of moles of ech component in the uffer solution, since oth concentrtions re numer of moles in certin volume, nd the volumes re the sme (the two solutes re in the sme solution). This rtio lso is the rtio of the volumes of the two solutions, since oth solutions eing mixed contin the sme concentrtion of solute. If we ssume 100. ml of cid solution, V cid = 100. ml. Then the volume of slt solution is V = 0.58 100. ml = 58 ml 0.100 M NCHO slt 0. (D) We cn lower the ph of the 0.50 M HCHO 0.560 M CH O uffer solution y incresing HCH O or lowering CHO. Smll volumes of NCl solutions will hve no effect, nd the ddition of NOH(q) or NC H O q g will rise the ph. The ddition of 0.150 M HCl will rise HCH O nd lower CHO through the rection HO (q) CHO (q) HCHO(q) HO(l) nd ring out the desired lowering of the ph. We first use the Henderson-Hssellch eqution to determine the rtio of the concentrtions of cette ion nd cetic cid. ph log CHO = 5.00 =.7 LHCH O O NM QP CHO CHO 0.6 log = 5.00.7 = 0.6; = 10 = 1.8 HCHO HCHO Now we compute the mount of ech component in the originl uffer solution. mount of C H O 0.560 mmol C H O = 00. ml 1 ml soln = 168mmol C H O 0.50mmol HC H O mount of HC H O = 00. ml 1mL soln = 75.0mmol HC H O 810
Chpter 17: Additionl Aspects of Acid Bse Equiliri Now let x represent the mount of H O dded in mmol. 168 x 1.8 = ; 168 x = 1.875 x= 15 1.8x 168 15 =.8x 75.0 x 168 15 x= =1mmol HO.7 Volume of 0.150 M HCl = 1 mmol H O 1mmol HCl 1mmol HO = 80mL 0.150M HCl solution 1mL soln 0.150 mmol HCl 1. (M) () The ph of the uffer is determined vi the Henderson-Hssellch eqution. ph p log CHO 5 0.100M = L =.89 log =.89 HC H O O 0.100M NM 5 QP The effective ph rnge is the sme for every propionte uffer: from ph =.89 to ph = 5.89, one ph unit on either side of p for propionic cid, which is.89. () To ech liter of 0.100 M HCH5O 0.100M NCH5O we cn dd 0.100 mol OH efore ll of the HCH5O is consumed, nd we cn dd 0.100 mol HO efore ll of the CH5O is consumed. The uffer cpcity thus is 100. millimoles (0.100 mol) of cid or se per liter of uffer solution.. (M) () () The solution will e n effective uffer one ph uniton either side of the p of methylmmonium ion, CHNH, =. 10 for methylmine, p = logc. 10 h =.8. For methylmmonium ction, p = 1.00.8 = 10.6. Thus, this uffer will e effective from ph of 9.6 to ph of 11.6. The cpcity of the uffer is reched when ll of the wek se or ll of the conjugte cid hs een neutrlized y the dded strong cid or strong se. Becuse their concentrtions re the sme, the numer of moles of se is equl to the numer of moles of conjugte cid in the sme volume of solution. 0.0500 mmol mount of wek se = 15 ml = 6.5 mmol CHNH or CHNH 1mL Thus, the uffer cpcity is 6.5 millimoles of cid or se per 15 ml uffer solution. 811
Chpter 17: Additionl Aspects of Acid Bse Equiliri. (M) () The ph of this uffer solution is determined with the Henderson-Hssellch eqution. CHO 8.5 mmol/75.0 ml ph = p log = log 1.810 log HCHO 15.5 mmol/75.0 ml =.7 0.6 =.8 1 [Note: the solution is not good uffer, s CHO =1.1 10, which is only ~ 600 times ] () mmol OH Amount of dded OH =0.5mmol B OH =0.50mmol OH 1mmol B OH The OH dded rects with the formic cid nd produces formte ion. Eqution: HCHO (q) OH (q) CHO (q) HO(l) Buffer: 15.5mmol 0 M 8.5mmol Add se: 0.50mmol Rect: 0.50 mmol 0.50 mmol 0.50 mmol Finl: 15.0 mmol 0 mmol 9.0 mmol CHO 9.0 mmol/75.0 ml ph = p log = log 1.810 log HCHO 15.0 mmol/75.0 ml =.7 0. =.5 (c) Amount of dded H O = 1.05 ml cid 1 mmol HCl 1 mmol H O 1 ml cid 1 mmol HCl The HO dded rects with the formte ion nd produces formic cid. Eqution: CHO (q) HO (q) HCHO (q) HO(l) Buffer : 8.5mmol 0 mmol 15.5mmol Add cid : 1mmol Rect : 8.5mmol 8.5mmol 8.5mmol Finl : 0 mmol.5mmol.0 mmol The uffer's cpcity hs een exceeded. The ph of the solution is determined y the excess strong cid present..5 mmol HO = = 0.059 M; 75.0mL 1.05mL ph = log 0.059 = 1. = 1 mmol H O 81
Chpter 17: Additionl Aspects of Acid Bse Equiliri. (D) For NH, p = log 1.810 () 5 c h For NH p p 1.68g NH 1 mol NH NH = = 0.197 M 0.500 L 17.0 g NH, = 1.00 = 1.00.7 = 9.6.05 g NH SO 1 mol NH SO mol NH NH = = 0.1 M 0.500 L 1.1 g NH SO 1mol NH SO ph p log NH = L =9.6 log NH O NM QP 0.197 0.1 M M =9.6 () The OH qg rects with the NH q g to produce n equivlent mount of NH qg. g NOH mol NOH mol OH OH = 0.88 1 1 = 0.0 M i 0.500L 0.00 g NOH 1mol NOH Eqution: NH q OH q NH q HO(l) Initil : 0.1M 0M 0.197 M Add NOH : 0.0 M Rect : 0.0 M 0.0 M 0.0 M Finl : 0.079M 0.0000M 0.1M ph p log NH 0.1M = L =9.6 log NH O 0.079 M NM QP =9.7 (c) Eqution: NH q H O q NH q H O(l) Initil : 0.197 M 0M 0.1M Add HCl : x M Rect : xm xm xm 9 x x Finl : 0.197 M 110 M 0 0.1 M ph p log NH 0.197 x = 9.00 = L = 9.6 log NH O 0.1 x log 0.197 x 0.1 x g g NM M = 9.00 9.6 = 0.6 M QP 0.197 x 0.1 x g g M M M 0.6 = 10 = 0.55 M 0.197 x = 0.55 0.1 x = 0.068 0.55 x 1.55 x = 0.197 0.068 = 0.19 x = 0.19 1.55 = 0.08 M 0.08 mol HO 1mol HCl 1000 ml HCl volume HCl = 0.500L =.5mL 1L soln 1mol H O 1 mol HCl 81
Chpter 17: Additionl Aspects of Acid Bse Equiliri 5. (D) () We use the Henderson-Hssellch eqution to determine the ph of the solution. The totl solution volume is 5 6.00 ml 6.00 ml = 100.00 ml. p = 1.00 p = 1.00 log 1.8 10 = 9.6 NH 6.00mL 0.00 M NH = 100.00 ml mmol NH = 7.0 100.0 ml = 0.070 M NH 6.00 ml 0.00 M NH = 100.00 ml mmol NH = 1.8 100.0 ml = 0.18 M [NH ] 0.070 M ph = p log 9.6 log 9.01 9.00 [NH ] 0.18M () The solution hs OH =10.99 =1.010 5 M The Henderson-Hssellch eqution depends on the ssumption tht: 5 NH 1.8 10 M NH If the solution is diluted to 1.00 L, NH =7.010 M, nd NH =1.8 10 M. These concentrtions re consistent with the ssumption. However, if the solution is diluted to 1000. L, NH NH =1.810 5 =7.10 6 M, nd M, nd these two concentrtions re not consistent with the ssumption. Thus, in 1000. L of solution, the given quntities of NH nd NH will not produce solution with ph = 9.00. With sufficient dilution, the solution will ecome indistinguishle from pure wter (i.e.; its ph will equl 7.00). (c) (d) The 0.0 ml of dded 1.00 M HCl does not significntly ffect the volume of the solution, ut it does dd 0.0 ml 1.00 M HCl = 0.0 mmol H O. This dded HO rects with NH, decresing its mount from 7.0 mmol NH to 7.00 mmol NH, nd incresing the mount of NH from 1.8 mmol NH to 1.0 mmol NH, s the rection: NH H O NH H O 7.00 mmol NH / 100.0mL ph =9.6 log =8.99 1.0 mmol NH / 100.0 ml We see in the clcultion of prt (c) tht the totl volume of the solution does not ffect the poh of the solution, t lest s long s the Henderson-Hssellch eqution is oeyed. We let x represent the numer of millimoles of H O dded, 81
Chpter 17: Additionl Aspects of Acid Bse Equiliri through 1.00 M HCl. This increses the mount of NH nd decreses the mount of NH, through the rection NH H O NH H O 7.0 x 7.0 x ph = 8.90 = 9.6 log ; log = 8.90 9.6 = 0.6 1.8 x 1.8 x Inverting, we hve: x 7.0 x 1.8 = 10 0.6 =.9; 1.8 =.9 7.0 = 16.5.9 16.5 1.8 x = = 1.1 mmol HO 1.00.9 x x x 1mmol HCl 1mL soln vol 1.00 M HCl = 1.1mmol H O = 1.1mL 1.00 M HCl 1mmol H O 1.00 mmol HCl 6. (D) () 1.0g NCHO 1mol NCHO 1mol CHO CHO = =0.88M CHO 0.00 L soln 8.0g NCHO 1mol NCHO Eqution: HCHO HO CHO HO Initil : 0M 0M 0.00M Add CHO 0M 0.88M 0M Consume HO 0.00 M 0.00 M 0.00 M Buffer : 0.00 M 0.88 M 0 M Then use the Henderson-Hssellch eqution to find the ph. ph p log CHO 0.88M = L =.7 log =.7 0.16 =.90 HC H O O 0.00M NM QP g. () We first clculte the initil OH due to the dded B OH OH g g B OH = 1.00 0.00L g g 1mol B OH 171.g B OH mol OH 1mol B OH g = 0.089 M Then HC H O is consumed in the neutrliztion rection shown directly elow. Eqution: HCHO OH CHO HO Initil: 0.00 M 0.089 M 0.88 M Consume OH : -0.089 M -0.089 M 0.089 M Buffer: 0.161 M ~ 0M 0.7 M 815
Chpter 17: Additionl Aspects of Acid Bse Equiliri Then use the Henderson-Hssellch eqution. ph p log CHO 0.7 M = L =.7 log =.7 0.1 = 5.05 HC H O O 0.161M NM QP (c) (d) g cn e dded up until ll of the HC H O g g B OH B OH HC H O B C H O H O is consumed. 0.00 mol HC H O 1mol B OH moles of B OH = 0.00 L = 0.000 mol B OH 1L soln mol HC H O 171.g B OH mss of B OH = 0.000 mol B OH = 5.1 g B OH 1mol B OH 0.6 g BOHg is too much for the uffer to hndle nd it is the excess of OH - originting from the B(OH) tht determines the poh of the solution. g B OH mol B OH g 1 g mol OH OH = 0.6 0.00 L soln c h g =1.10 171.g B OH 1mol B OH g poh = log 1. 10 = 1.85 ph = 1.00 1.85 = 1.15 M OH Acid Bse Indictors 7. (E) () The ph color chnge rnge is 1.00 ph unit on either side of p Hln. If the ph color chnge rnge is elow ph = 7.00, the indictor chnges color in cidic solution. If it is ove ph = 7.00, the indictor chnges color in lkline solution. If ph = 7.00 flls within the ph color chnge rnge, the indictor chnges color ner the neutrl point. Indictor HIn Bromphenol lue 1. 10 Bromcresol green.110 5 Bromthymol lue 7.9 10 8,-Dinitrophenol 1.10 Chlorophenol red 1.0 10 6 Thymolphthlein 1.0 10 10 HIn p ph Color Chnge Rnge Chnges Color in?.85.9 (yellow) to.9 (lue) cidic solution.68.7 (yellow) to 5.7 (lue) cidic solution 7.10 6.1 (yellow) to 8.1 (lue) neutrl solution.89.9 (colorless) to.9 (yellow) cidic solution 6.00 5.0 (yellow) to 7.0 (red) cidic solution 10.00 9.0 (colorless) to 11.0 (lue) sic solution () If romcresol green is green, the ph is etween.7 nd 5.7, proly out ph =.7. If chlorophenol red is ornge, the ph is etween 5.0 nd 7.0, proly out ph = 6.0. 816
Chpter 17: Additionl Aspects of Acid Bse Equiliri 8. (M) We first determine the ph of ech solution, nd then use the nswer in Exercise 7 () to predict the color of the indictor. (The wekly cidic or sic chrcter of the indictor does not ffect the ph of the solution, since very little indictor is dded.) 1 mol H O HO = 0.100 M HCl = 0.100 M; ph = log 0.100 M = 1.000 1 mol HCl,-dinitrophenol is colorless. () () (c) Solutions of NCl(q) re ph neutrl, with ph = 7.000. Chlorphenol red ssumes its neutrl color in such solution; the solution is red/ornge. Eqution: NH (q) HO(l) NH (q) OH (q) Initil: 1.00 M 0 M 0 M Chnges: x M x M x M 1.00 x M x M x M Equil: NH OH 5 x x = L =1.8 10 = x =. 10 M= OH NH O NM QP 1.00 x 1.00 (x << 1.00 M, thus the pproximtion ws vlid). c h poh = log. 10 =.8 ph = 1.00.8 = 11.6 Thus, thymolphthlein ssumes its lue color in solutions with ph 11.6. (d) From Figure 17-8, sewter hs ph = 7.00 to 8.50. Bromcresol green solution is lue. 9. (M) () () In n Acid Bse titrtion, the ph of the solution chnges shrply t definite ph tht is known prior to titrtion. (This ph chnge occurs during the ddition of very smll volume of titrnt.) Determining the ph of solution, on the other hnd, is more difficult ecuse the ph of the solution is not known precisely in dvnce. Since ech indictor only serves to fix the ph over quite smll region, often less thn.0 ph units, severl indictors crefully chosen to spn the entire rnge of 1 ph units must e employed to nrrow the ph to 1 ph unit or possily lower. An indictor is, fter ll, wek cid. Its ddition to solution will ffect the cidity of tht solution. Thus, one dds only enough indictor to show color chnge nd not enough to ffect solution cidity. 817
Chpter 17: Additionl Aspects of Acid Bse Equiliri 0. (E) () We use n eqution similr to the Henderson-Hssellch eqution to determine the reltive concentrtions of HIn, nd its nion, In, in this solution. In In In ph = p HIn log ;.55 =.95 log ; log =.55.95 = 0.0 HIn HIn HIn In x 0 x x x HIn 100 x 1.0 0.0 = 10 = 0.0 = = 0 0.0 = = 9% In nd 71% HIn () 1. (E) () () (c) (d) (e) When the indictor is in solution whose ph equls its p (.95), the rtio In / HIn = 1.00. And yet, t the midpoint of its color chnge rnge (out ph = 5.), the rtio In /[HIn] is greter thn 1.00. Even though HIn In t this midpoint, the contriution of HIn to estlishing the color of the solution is out the sme s the contriution of In. This must men tht HIn (red) is more strongly colored thn In (yellow). 0.10 M OH is n lkline solution nd phenol red will disply its sic color in such solution; the solution will e red. 0.10 M HC H O is n cidic solution, lthough tht of wek cid, nd phenol red will disply its cidic color in such solution; the solution will e yellow. 0.10 M NH NO is n cidic solution due to the hydrolysis of the mmonium ion. Phenol red will disply its cidic color, tht is, yellow, in this solution. 0.10 M HBr is n cidic solution, the queous solution of strong cid. Phenol red will disply its cidic color in this solution; the solution will e yellow. 0.10 M NCN is n lkline solution ecuse of the hydrolysis of the cynide ion. Phenol red will disply its sic color, red, in this solution. (f) An equimolr cetic cid potssium cette uffer hs ph = p =.7 for cetic cid. In this solution phenol red will disply its cid color, nmely, yellow.. (M) () ph = log0.05 g = 0.688 The indictor is red in this solution. () The totl volume of the solution is 600.0 ml. We compute the mount of ech solute. mount HO = 50.0 ml 0.05 M = 71.8 mmol HO mount NO = 50.0mL 0.500 M = 15mmol NO mmol mmol HO = 71.8 = 0.10 M NO = 15 = 0.08M 600.0mL 600.0mL The H O nd NO rect to produce uffer solution in which HNO = 0.10 M nd NO = 0.08 0.10 = 0.088 M. We use the 818
Chpter 17: Additionl Aspects of Acid Bse Equiliri Henderson-Hssellch eqution to determine the ph of this solution. p = log 7. 10 =.1 c h NO 0.088 M ph = p log =.1 log =.01 The indictor is yellow in this HNO 0.10 M solution. (c) The totl volume of the solution is 750. ml. We compute the mount nd then the concentrtion of ech solute. Amount OH = 150mL 0.100 M = 15.0 mmol OH This OH rects with HNO in the uffer solution to neutrlize some of it nd leve 56.8 mmol ( = 71.8 15.0) unneutrlized. mmol mmol HNO = 56.8 = 0.0757 M NO = 1515 g = 0.187 M 750. ml 750. ml We use the Henderson-Hssellch eqution to determine the ph of this solution. ph p log NO 0.187 M = L =.1 log =.5 HNO O 0.0757 M NM QP The indictor is yellow in this solution. g. (d) We determine the OH due to the dded B OH OH g g B OH = 5.00 0.750 L g g 1mol B OH 171. g B OH mol OH 1mol B OH g = 0.0778 M This is sufficient OH to rect with the existing HNO nd leve n excess OH = 0.0778 M 0.0757 M = 0.001M. poh = log 0.001 =.68. ph = 1.00.68 = 11. The indictor is lue in this solution.. (M) Moles of HCl = C V = 0.0050 M 0.01000 L =.050 10 moles Moles of B(OH) t endpoint = C V = 0.0110 M 0.01790 L =.005 10 moles. Moles of HCl tht rect with B(OH) = moles B(OH) Moles of HCl in excess.050 10 moles.010 10 moles =.05 10 6 moles Totl volume t the equivlence point = (10.00 ml 17.90 ml) = 7.90 ml [HCl] excess = 6.0510 mole HCl 0.0790 L = 1.5 10 M; ph = log(1.5 10 ) =.8 () () The pproximte p HIn =.8 (generlly 1 ph unit) This is reltively good indictor (with 1 % of the equivlence point volume), however, p Hin is not very close to the theoreticl ph t the equivlence point (ph = 7.000) For very ccurte work, etter indictor is needed (i.e., romthymol lue (p Hin = 7.1). Note:,-dinitrophenol works reltively well here ecuse the ph ner the equivlence point of strong cid/strong se 819
Chpter 17: Additionl Aspects of Acid Bse Equiliri titrtion rises very shrply ( 6 ph units for n ddition of only drops (0.10 ml)).. Solution (): 100.0 ml of 0.100 M HCl, [H O ] = 0.100 M nd ph = 1.000 (yellow) Solution (): 150 ml of 0.100 M NC H O of HC H O = 1.8 10 5 of C H O = 5.6 10 10 C H O (q) H O(l) -10 = 5.6 10 HC H O (q) OH (q) Initil 0.100 M 0 M ~ 0 M Chnge x x x Equil. 0.100 x x x Assume x is smll: 5.6 10 11 = x ; x = 7.8 10 6 M (ssumption vlid y inspection) [OH ] = x = 7.8 10 6 M, poh = 5.1 nd ph = 8.87 (green-lue) Mixture of solution () nd (). Totl volume = 50.0 ml n HCl = C V = 0.1000 L 0.100 M = 0.0100 mol HCl n C HO = C V = 0.1500 L 0.100 M = 0.0150 mol C H O HCl is the limiting regent. Assume 100% rection. Therefore, 0.0050 mole C H O is left unrected, nd 0.0100 moles of HC H O form. [C H O ] = n 0.0050 mol = V 0.50 L = 0.00 M [HC H O ] = n 0.0100 mol = V 0.50 L = 0.000 M -5 HC H O (q) H O(l) = 1.8 10 C H O (q) H O (q) Initil 0.000 M 0.00 M ~ 0 M Chnge x x x Equil. 0.000 x 0.00 x x 1.8 10-5 = x(0.00 x) 0.000 x x(0.00) 0.000 x =.6 10 5 (proof 0.18 % < 5%, the ssumption ws vlid) [H O ] =.6 10 5 ph =. Color of thymol lue t vrious phs: ph ph ph ph 1..8 8.0 9.8 Red Ornge Yellow Green Blue Solution () RED Solution (c) YELLOW Solution () GREEN 80
Chpter 17: Additionl Aspects of Acid Bse Equiliri Neutrliztion Rections 5. (E) The rection (to second equiv. pt.) is: HPO q OH q HPO q HO(l). The molrity of the HPO solution is determined in the following mnner. 0.0 mmol OH 1mmol HPO 1.15mL OHsoln 1mL OHsoln mmol OH HPO molrity = = 0.1508M 5.00mL H PO soln 6. (E) The rection (first to second equiv. pt.) is: NH PO q NOH q N HPO q H O(l) solution is determined in the following mnner. H PO. The molrity of the H PO 0.1885mmol NOH 1mmol H PO 18. 67 ml NOH soln 1mLNOHsoln 1mmol NOH molrity = 0.00mL H PO soln =0.1760M 7. (M) Here we must determine the mount of H O or OH in ech solution, nd the mount of excess regent. 0.0150 mmol H SO mmol H O mount HO = 50.00mL =1.50mmol H O 1mL soln 1mmol H SO (ssuming complete ioniztion of H SO nd HSO in the presence of OH ) mount 0.085 mmol NOH 1mmol OH OH = 50.00 ml = 1.9 mmol OH 1mL soln 1mmol NOH Titrtion rection : OH q HO q HO(l) Initil mounts : 1.9mmol 1.50 mmol After rection : 0.mmol 0 mmol Result: 0.mmol OH OH = =.10 M 100.0 ml soln poh = log. 10 =.7 ph = 1.00.7 = 11.6 81
Chpter 17: Additionl Aspects of Acid Bse Equiliri 8. (M) Here we must determine the mount of solute in ech solution, followed y the mount of excess regent. HO = 10 = 0.00 M.50 0.00 mmol H O 1mmol HCl mmol HCl = 100.0 ml 1mL soln 1mmol H O =0.mmol HCl 0.0010 mmol OH mmol NOH = 100.0 ml 1mL soln Result:.00 poh = 1.00 11.00 =.00 OH = 10 = 1.0 10 M 1mmol NOH =0.10mmol NOH 1mmol OH Titrtion rection : NOH q HCl q NCl q H O(l) Initil mounts : 0.10 mmol 0. mmol 0 mmol After rection : 0.00 mmol 0. mmol 0.10 mmol 0. mmol HCl 1mmol H O 00.0 ml soln 1mmol HCl HO = =1.110 M ph = log 1.1 10 =.96 Titrtion Curves 9. (M) First we clculte the mount of HCl. The relevnt titrtion rection is HCl q OH q Cl q H O(l) 0.160 mmol HCl mount HCl = 5.00 ml 1 ml soln =.00 mmol HCl =.00 mmol HO present Then, in ech cse, we clculte the mount of OH tht hs een dded, determine which ion, OH g q or H O g, q is in excess, compute the concentrtion of tht ion, nd determine the ph. 0. mmol OH () mount OH = 10.00 ml =. mmol OH ; HO is in excess. 1 ml soln F HG 1 mmol H O.00 mmol H O. mmol OH 1 mmol OH [H O ] = 500. ml originlly 10.00 ml titrnt ph = log 0.051 = 1.6 g I J = 0.051 M 8