Initial Change x +x +x Equilibrium x x x

Transcription

1 Chpter 8 Homework Solutions 8.1 () n(h ) mol/l x 0.05 L 3.60x10-3 mol n(oh - ) 0.15 mol/l x 0.05 L 3.1x10-3 mol After neutrliztion, excess n(h ) 3.6x10-3 mol - 3.1x10-3 mol 4.8x10-4 mol [ H n( H ] V 4 ) 4.8x10 mol 9.6x10 ( ) L ph -log([h ]) -log(9.6x10-3 ).0 3 (b) n(h ) 0.15 mol/l x 0.05 L 3.75x10-3 mol n(oh - ) 0.15 mol/l x L 5.5x10-3 mol After neutrliztion, excess n(oh - ) 5.5x10-3 mol x10-3 mol 1.50x10-3 mol [ OH [ H n( OH ] V 1x10 ] [ OH 14 ) 1x10 ].5x x10 mol ( ) L x10.5x10 13 poh -log([oh - ]) -log(.5x10 - ) 1.60 ph poh (c) n(h ) 0. mol/l x 0.01 L 4.66x10-3 mol n(oh - ) 0.30 mol/l x L 3.00x10-3 mol After neutrliztion, excess n(h ) 4.66x10-3 mol x10-3 mol 1.66x10-3 mol [ H n( H ] V 3 ) 1.66x10 mol ( ) L ph -log([h ]) -log(5.3x10 - ) x10 8. () Clculte the molr concentrtion of KCH3O 1 mol 1 (8.4 g) 0.34 mol L 98.1 g 0.50 L 1 H O(l) Lc (q) ƒ HAc(q) OH (q) Initil Chnge x x x Equilibrium 0.34 x 0.34 x x

2 K b 1.0x10 1.8x x10 10 x 0.34 x 0.34 x 1.9x10-10 x [OH - ] 1.38x10-5 poh -log([oh - ]) -log(1.38x10-5 ) 4.86 ph (b) (3.75 g) 1 mol 97.9 g L mol L1 NH 4 Br NH 4 (q) H O(l) É NH 3 (q) H 3 O (q) Initil Chnge x x x Equilibrium x x x K K K w b 1.0x10 1.8x x x x.14x10-10 x [H ] 1.46x10-5 ph -log([h ]) -log(1.46x10-5 ) 4.83 x () HLc H Lc - K 8.4x10-4 [HLc] x [H ] [Lc - ] x K 8.4x x x x 8.4x10-5 x [H ] 9.x10-3 ph -log([h ]) -log(9.x10-3 ).04 (b) Lc - HO HLc OH - Kb /8.4x x10-11 [Lc - ].0 - x [HLc] [OH - ] x K b 1.x10.0 x.0 11 x x.4x10-11 x [OH - ] 4.9x10-6 poh -log([oh - ]) -log(4.9x10-6 ) 5.3 ph poh (c) Initil: (nhlc)init 0.1 mol/l x 0.0 L 0.0 mol (nlc-)init 0.0 (noh - ) 0.05 mol/l x 0.10 L mol HLc OH - Lc - HO

3 V 0.0 L 0.1 L 0.30 L Finl: (nhlc)fin mol (nlc-)fin mol [HLc]fin mol/0.30 L [Lc - ]fin mol/0.30 L log(8.4x10-4 ) 3.08 ph [ Lc ] log 3.08 log HLc] () Anil HO AnilH HO Kb 4.3x10-10 [Anil] x [AnilH ] [OH - ] x K b 4.3x x x x.15x10-11 x [OH - ] 4.64x10-6 poh -log([oh - ]) -log(4.64x10-3 ) 5.33 ph poh 8.67 (b) AnilH Anil H K 1x10-14 /Kb.3x10-5 [AnilH ] x [Anil] [H ] x K b.3x x x x.3x10-6 x [H ] 1.51x10-3 ph -log([h ]) -log(1.51x10-3 ).8 (c) Initil: (nanil)init 0.05 mol/l x 0.40 L 0.0 mol (nanilh )init 0 nh 0.15 mol/l x 0.10 L mol V 0.40 L 0.10 L 0.50 L Anil H AnilH Finl: (nanil)fin mol [Anil]fin mol/0.50 L (nanilh)fin mol [AnilH ]fin mol/0.50 L b -log(4.3x10-10 ) ph [ Anil] log 4.64 log AnilH ] 0.030

4 8.5 For ech of the solutions below, indicte whether the solution would be buffer. ) YES: n(hac) 1.0 mol, n(oh - ) 0.50 mol. OH - converts prt of HAc to Ac - b) YES: n(ac - ) 1.6 mol, n(h ) 1.0 mol. H converts prt of Ac - to HAc c) NO: HCO3 nd CO3 - re not conjugte cid/conjugte bse of ech other d) YES: n(co3 - ) 1.0 mol, n(h ) 0.50 mol. H converts prt of CO3 - to HCO3 - e) YES: n(co3 - ) 1.0 mol, n(h ) 1.50 mol. H converts ll CO3 - to HCO3 - nd then converts prt of HCO3 - to HCO3 f) NO: n(co3 - ) 1.0 mol, n(h ).50 mol. H converts ll CO3 - to HCO3 -, nd then to HCO3 (nd there is even H left over). We do not hve n cid/bse conjugte pir t the end of the rection. 8.6 The following independent questions re on ph clcultions in solutions of Arsenic Acid (H3AsO4) nd its vrious nions. H3AsO4 is triprotic cid with Acid Dissocition Constnts: K ' 6.0x10-3, K '' 1.0x10-7, K ''' 3.x H AsO H AsO HAsO K ' 6.0x10 K '' 1.0x10 K '' 3.x '. 4 ' ' ) n(h3aso4) 1. mol, n(oh-) 0.8 mol 1st. (only) Rection: H3AsO4 OH- HAsO4- H3AsO4 OH- HAsO4- Initil Chnge Finl Conc [HAsO4-] 0.0 ph ' log. log [H3AsO4] 0.10 b) n(h3aso4) 1. mol, n(oh-) 1.6 mol 1st. Rection: H3AsO4 OH- HAsO4- H3AsO4 OH- HAsO4- Initil Chnge Finl Conc 0 nd. Rection: HAsO4- OH- HAsO4-

5 HAsO4- OH- HAsO4- Initil Chnge Finl Conc [HAsO4 - ] 0.08 ph '' log 7.00 log 7.00 ( 0.30) 6.70 [HAsO4-] 0.16 c) n(aso4-3) 1. mol, n(h) 0.8 mol 1st. (only) Rection: AsO4-3 H HAsO4- AsO4-3 H HAsO4- Initil Chnge Finl Conc [AsO4-3] 0.10 ph ''' log log (-0.30) [HAsO4 - ] 0.0 c) n(aso4-3) 1. mol, n(h) 1.6 mol 1st. Rection: AsO4-3 H AsO4- AsO4-3 H HAsO4- Initil Chnge Finl Conc nd. Rection: HAsO4- H HAsO4- HAsO4- H Initil Chnge Finl Conc HAsO4-

6 [HAsO4 - ] 0.16 ph '' log 7.00 log 7.00 (0.30) 7.30 [HAsO4-] [HAsO4 ] ph '' log 7.00 log (0.18) [HAsO4 ] e) ( ) f) [AsO4-3 ]/[HSO4 - ] 1/ [AsO4 ] ph ''' log log - ( 0.40 ) (-0.40) [HAsO4 ] [HAsO4-] [HAsO4-] g) ph.60 ' log. log [H3AsO4] [H3AsO4] [HAsO4-] log [H3AsO4] [HAsO4-] [H3AsO4] [AsO4-3] [AsO4-3] h) ph ''' log llog [HAsO4 - ] [HAsO4 - ] [AsO4-3] log [HAsO4 - ] [AsO4-3] [HAsO4 - ] [HAsO4 - ] [AsO4-3] HCOOH H HCOO - K 1.8x10-4 -log(1.8x10-4 ) 3.74 ph 4.5 [ HCOO ] [ HCOO ] log 3.74 log HCOOH ] HCOOH ] [ HCOO ] log 0.51 HCOOH ] [ HCOO ] [ HCOOH ] [HCOO - ] 3.4[HCOOH] (Eqn. 1) [HCOOH] [HCOO - ] [HCOOH}init 0.50 (Eqn. 1) Therefore: [HCOOH] 3.4[HCOOH] [HCOOH] 0.50 [HCOOH] 0.50/

7 [HCOO - ] Addition of Strong Bse to HBenz: HBenz NOH N Benz - HO Initil: (nhbenz)init HBenz x VHBenz 0.40 mol/l x 0.50 L 0.0 mol () Number of ml of NOH needed to rech hlf-wy to equiv. point Need nnoh (1/) x 0.0 mol 0.10 mol nnoh 0.10mol VNOH 0.10 L 100mL 1.0mol / L NoH ph Hlf-wy to equivlence point, [Benz - ] [HBenz] -log(k) -log(6.5x10-5 ) 4.19 ph [ Benz ] log log(1) 4.19 HBenz] (b) Number of ml of NOH needed to rech the equiv. point Need nnoh 0.0 mol n NOH V NOH NOH 0.0 mol 1.0 mol / L ph t equivlence point. Hve 100% Benz - (the benzote ion) This is pure bse with Kb /K /6.5x x mol [ Benz ] L 0.0 L It will rect with wter ccording to the eqution: Benz - HO HBenz OH - [ HBenz][ OH ] x x Kb 1.5x10 [ Benz ] [ ] (0.9( ) x OH x x poh x 6 log( ) 5. ph

8 8.9 Al ' '' Al Al [ Al ] [ Al] () [ Al] [ Al ] ph [ Al] ' log.35 log(4.0) Al ] (b) '' [ Al ] [ Al ] '' ph log log ph [ Al] [ Al] [ Al ] 10 [ Al] 6.5 (c) We hve solution contining [Al ] nd [Al]. [Al ] [Al] 0.80 ' [ Al] [ Al] ph.00 log.35 log Al ] Al ] [ Al] l og Al ] [ Al] [ Al ] [Al] 0.45[Al ] 0.80 [Al ] [Al] [Al ] 0.45[Al ] 1.45[Al ] [0.80] [ Al ] 0.55 nd [Al] Lys '.18 Lys '' 8.95 Lys ''' Lys () After dding 0.5 equiv., [Lys ] [Lys ] ' [ Lys ] ph log.18 log(1).18 Avg. Chrge 1.5 Lys ] (b) After dding 1.0 equiv., hve pure Lys ' ' 1 ( ) ( ) ' ph Avg. Chrge 1.0

9 (c) At the isoelectric point, pi, hve pure Lys (with Avg. Chrge 0.0) pi '' '' 1 ( ) ( ) ' (d) At ph 10.53, ph. Therefore, [Lys - ] [Lys] This requires dding.5 equivlents. (e) After dding.5 equiv. of NOH to the protonted form, the solution contins 50% [Lys] nd 50% [Lys - ]. Therefore, the verge chrge is -1/ gf(s) g (q) F - (q) Ks 6.4x10-9 [g ][F - ] () [g ] s [F - ] s Ks 6.4x10-9 s(s) 4s 3 s 3 1.6x10-9 s 1.17x10-3 (b) [g ] 0.1 s 0.1 [F - ] s Ks 6.4x10-9 (0.1)(s) 0.4s s 1.6x10-8 s 1.6x10-4 (c) [g ] s [F - ] 0. s 0. Ks 6.4x10-9 (s)(0.) 0.04s s 1.6x10-7