Homework #6 Chpter 7 Homework Acids nd Bses 18. ) H O(l) H 3O (q) OH (q) H 3 O OH Or H O(l) H (q) OH (q) H OH ) HCN(q) H O(l) H 3O (q) CN (q) H 3 O HCN CN Or HCN(q) H (q) CN (q) H CN HCN c) NH 3(q) H O(l) NH 4 (q) OH (q) 4 NH OH NH 3 For this question you could hve picked ny wek cid or se or just used generl wek cid (HA) nd wek se (B). ) HClO 4 strong cid ) HOCl wek cid c) H SO 4 strong cid (1 st deprotontion only) d) H SO 3 wek cid 5. The lrger the the stronger the cid (tle 7.). HClO 4 > HClO > NH 4 > H O Strongest cid Wekest cid Note: HClO 4 is strong cid, therefore, will hve the highest vlue. H O is the conjugte cid of OH. OH is strong se. Strong ses hve the wekest conjugte cids. (H O(l) OH (q) OH (q) H O(l)) 6. The lrger the the stronger the cid (tle 7.), the stronger the cid the weker the conjugte se, therefore, the strongest se will hve the conjugte cid with the smllest. NH 3 > ClO > H O > ClO 4 Note: HClO 4 is strong cid, therefore, will hve the highest vlue. Acids re clssified s strong cids if they hve greter thn 1. H 3O is one of the wekest strong cids. Therefore, H 3O (H O 1
conjugte se) is one of the wekest strong cid nd H O will e slightly more sic thn ClO 4. 7. The lrger the the stronger the cid. ) HCl strong cid nd H O is wek cid w = (1.0 10 14 ) ) HNO The of HNO (4.0 10 4 ) > w = (1.0 10 14 ) c) HCN The HCN (6. 10 10 ) > of HOC 6H 5 (1.6 10 10 ) 8. The higher the the weker the conjugte se ) Cl conjugte se of strong cid, HCl H O conjugte se of one of the wekest strong cids H 3O H O stronger se ) H O conjugte se of one of the wekest strong cids H 3O NO conjugte se of wek cid NO stronger se c) CN conjugte se of the wek cid HCN ( = 6. 10 10 ) OC 6H 5 conjugte se of the wek cid HOC 6H 5 ( = 1.6 10 10 ) OC 6H 5 stronger se 9. ) CH 3CO H cid H O se CH 3CO conjugte se H 3O conjugte cid Therefore, H O nd CH 3CO re competing for protons. ) CH 3CO is the stronger se (H 3O is strong cid, therefore, forms weker conjugte cid thn CH 3CO H) c) CH 3CO is wek se ecuse it is the conjugte se of wek cid. Conjugte ses of wek cid re lso wek, therefore, the do no fully dissocite in solution. CH 3CO H O(q) CH 3CO H(q) OH (q) 31. ) wek cid ) strong cid c) wek se d) strong se e) wek se f) wek cid g) wek cid h) strong se i) strong cid (1 st deprotontion) 36. The solution is cidic if [H ]>[OH ], the solution is sic if [H ]<[OH ], nd the solution is neutrl if [H ]=[OH ]. The reltionship etween [H ] nd [OH ] is: ) ) 1.010 OH 1.010 7 14 OH 1.010 7 [H ] = [OH ] neutrl solution 8.310 OH 1.010 16 14 OH 1 [H ] < [OH ] sic solution w H OH 1.010 14
c) d) 14 1 OH 1.010 OH 8.310 16 [H ] > [OH ] cidic solution 5.410 OH 1.010 5 14 OH 1.910 10 [H ] > [OH ] cidic solution 37. If the ph < 7 the solution is cidic, if the ph > 7 the solution is sic, nd if the ph = 7 the solution is neutrl. If the poh < 7 the solution is sic, if the poh > 7 the solution is cidic, nd if the poh = 7 the solution is neutrl. The reltionship etween ph nd [H ] is: ph log H The reltionship etween [H ] nd [OH ] is: ) Bsic H 10 10 4.010 ) Bsic ph 7.4 8 4.010 OH 1.010 8 14 OH.510 7 H 10 10 510 ph 15.3 16 510 OH 1.010 OH c) Acidic 16 14 0 H ph 1.0 10 10 10 10 OH 1.010 OH d) Acidic 110 15 14 H 10 10 6.310 ph 3.0 4.10 OH 1.010 4 14 OH 1.610 11 e) Bsic poh OH 10 10 110 5.0 5 H 110 1.010 H 110 f) Acidic 6 14 9 w H OH 1.010 14 3
OH 10 10.510 poh 9.60 10 H 1.510 1.010 10 14 H 4.010 5 43. ) HNO is wek cid The mjor species in solution re HNO nd H O The following equilirium will occur in solution HNO (q) NO (q) H (q) = 4.0 10 4 (Tle 7.) Clculte the concentrtion of H ions t equilirium HNO NO H Initil 0.50 0 0 Chnge x x x Equilirium 0.50x X X NO H 4 4.010 HNO 0.50 x 1.010 4.010 x x x 4 4 4.010 x1.010 0 4 4 4 4 4 4c 4.010 4.010 4.010 x 0.0098 nd 0.010 x is 0.0098 ecuse concentrtion cnnot e negtive. Clculte ph H x 0.0098 ph log H log0.0098. 01 ) HC H 3O is wek cid The mjor species in solution HC H 3O nd H O The following equilirium will occur in solution HC H 3O (q) C H 3O (q) H (q) Clculte the concentrtion of H ions t equilirium HC H 3O C H 3O H Initil 0.50 0 0 Chnge x x x Equilirium 0.50x X X C H O H 3 1.810 5 = 1.8 10 5 (Tle7.) HC H O 0.50 x 3 Since is smll ssume 0.50x = 0.50 5 1.810 0.50 x 0.001 Check ssumption (5% rule) 4
Amount Gined/Lost 0.001 100% 100% 0.84% Originl Concentrtion 0.5 Less thn 5%, therefore, good ssumption Clculte ph H x 0.001 ph log H log0.001. 68 44. ) HOC 6H 5 is wek cid therefore The mjor species in solution re HOC 6H 5 The following equilirium will occur in solution HOC 6H 5(q) OC 6H 5 (q) H (q) = 1.6 10 10 Clculte the concentrtions of H t equilirium HOC 6H 5 OC 6H 5 H Initil 0.50 0 0 Chnge x x x Equilirium 0.50 x x x OC6H 5 H 10 1.610 HOC H 0.50 x 6 5 Since is smll ssume 0.50x = 0.50 1.610 10 x 6.310 6 0.50 Check ssumption (5% rule) 6 Amount Gined/Lost 6.310 100% 100% 0.005% Originl Concentrtion 0.50 Less thn 5%, therefore, good ssumption. 6 H x 6.3 10 Clculte the concentrtion of OH from wter equilirium 14 H OH 1.0 10 Clculte ph w 5 1.5 10 OH 10 OH 6.7 10 1.010 14 ph H 6 log log 6.3 10 5.0 ) HCN is wek cid The mjor species in solution re HCN nd H O The following equilirium will occur in solution HCN(q) CN (q) H (q) = 6. 10 10 (Tle 7.) 5
Clculte the concentrtion of H ion t equilirium HCN CN H Initil 0.50 0 0 Chnge x x x Equilirium 0.50x x x CN H 10 6.10 HCN 0.5 x Since is smll ssume 0.50x = 0.50 6. 10 10 x 1. 10 5 0.50 Check ssumption (5% rule) 5 Amount Gined/Lost 1.10 100% 100% 0.0048% Originl Concentrtion 0.5 Less thn 5%, therefore, good ssumption. Clculte ph 5 H x 1. 10 ph log 5 H log1. 10 4. 9 45. HF is wek cid therefore, the rections tht re occurring in solution re: HF(q) F (q) H (q) Clculte the concentrtions of HF, F, nd H t equilirium (you will lso need to clculte the consentrtion of OH which will e in the soltion from the following equilirium HO(l) OH (q) H (q)) HF F H Initil 0.00 0 0 Chnge x x x Equilirium 0.00x x x F H 4 7.10 HF 0.00 x 1.4410 7.10 x x x 5 4 7.10 x1.4410 0 4 5 4 4 5 4c 7.10 7.10 5.7610 x 0.0035 nd 0.004 x is 0.030 ecuse concentrtion cnnot e negtive. Find the finl concentrtions HF x 0.00 0.00 0.0035 0.017 F H x 0.0035 Clculte the concentrtion of OH from wter equilirium 6
Clculte ph w H OH 0.0035OH 1.0 1 OH.9 10 ph log 1.0 10 10 14 14 H log0.0035. 46 47. ) HA is wek cid. There re more HA molecules (green nd white toms together) thn H ions (single white tom) nd A ions (single green toms). This shows tht the cid did not completely dissocite mking it wek cid. ) Clculte the percentge dissocited In the picture there re 9 prticles tht hve note dissocited nd 1 prticle tht hs prticles dissocited 100% = 1 100% = 10% totl prticles 1 9 Clculte the equilirium constnt The rection of interest is HA(q) H (q) A (q) HA H A Initil 0.0 0 0 Chnge x x x Equilirium 9 0.0 x= x = 1 N V N V x = 1 N V Note from the picture t equilirium the concentrtion of HA is 9 N = 9 N is V N V Avogdro s numer nd is used to convert the 9 prticles to numer of moles of HA. The concentrtion of H nd A re Solve for V 0.0 x = 9 N A V Solve for 0.0 1 N V = 9 N V V = 10 N 0.0 = 50 N = [H ][A ] [HA] = 1 N( 50 N ) 1 N( 50 N ) 9 = N( 50 N ) 1 N V 1 (50)(9) = 0.00 48. HC H ClO is wek cid, therefore, the equilirium estlished is: HC H ClO (q) C H ClO (q) H (q) Find the concentrtion of H t equilirium HC H ClO C H ClO H Initil 0.10 0 0 Chnge x x x Equilirium 0.10x x x 7
CHClO H 3 1.3510 HC H ClO 0.10 x 1.410 1.3510 x x x 4 3 1.3510 x1.410 0 3 4 3 3 4 4c 1.3510 1.3510 5.610 x 0.011 nd 0.013 x is 0.011 ecuse concentrtion cnnot e negtive. H x 0.011 Clculte ph ph log H log0.011 1. 96 49. HIO 3 is wek cid, therefore, the equilirium estlished is: HIO 3(q) IO 3 (q) H (q) Find the concentrtion of H t equilirium HIO 3 IO 3 H Initil 0.010 0 0 Chnge x X x Equilirium 0.010x X x IO 3 H 0.17 HIO 0.010 x 0.0017 0.17x x x 0.17x 0.0017 0 3 4c 0.17 0.17 0.0068 x 0.0094 nd 0.18 x is 0.0094 ecuse concentrtion cnnot e negtive. Clculte ph H x 0.0094 ph log H log0.0094. 03 51. 1.13 g C6H5CO H mol 0.56 g C H CO H 0.0046 mol C H CO H 1 mol C6H5CO H 6 5 1.13 g C6H5CO H 6 5 n 0.0046mol 0.0046 V 1.0L C 6H 5CO H is wek cid, therefore, the equilirium estlished is: H O(l) OH (q) H (q) C 6H 5CO H(q) C 6H 5CO (q) H (q) 8
Clculte the equilirium concnetrtions C 6H 5CO H C 6H 5CO H Initil 0.0046 0 0 Chnge x x x Equilirium 0.0046x X X C6H5CO H 5 6.410 C H CO H 0.0046 x 6 5.910 6.410 x x x 7 5 6.410 x.910 0 5 7 5 5 6 4c 6.410 6.410 1.10 x 5.10 nd 5.810 x is 5. 10 4 ecuse concentrtion cnnot e negtive. Find the finl concentrtions 4 C6H5CO H 0.0046 x 0.0046 5.10 0.0041 6 5 5. 10 4 C H CO H x Clculte the concentrtion of OH from wter equilirium w 4 5. 10 OH OH Clculte ph H OH 1.9 10 ph log H 1.0 10 1.010 11 14 14 8 log 5. 10 4 3. 5. C 6H 5CO H is solid efore it is put into wter. Therefore, two equiliriums re estlished. C 6H 5CO H(s) C 6H 5CO H(q) C 6H 5CO H(q) C 6H 5CO (q) H (q) They wnt to know how much C 6H 5CO H will dissolve. This will e the mount of C 6H 5CO H tht strts in the second equilirium which we will denote s s. C 6H 5CO H C 6H 5CO H Initil S 0 0 Chnge x x x Equilirium sx x x The ph of the solution will give us the equilirium concentrtion of the H ions ph log H ph.80 H 10 10 0.0016 C 6H 5CO H C 6H 5CO H Initil s 0 0 Chnge 0.0016 0.0016 0.0016 Equilirium s0.0016 0.0016 0.0016 4 4 9
C6H5CO H 5 6.410 C H CO H S 0.04 6 5 0.00160.0016 S 0.0016 Convert to grms per 100 ml S = ( 0.04mol ) ( 1L ) 1L 1000mL (1.13g) = 0.0051 g 1mol ml Therefore the wter soluility per 100 ml is 0.51 g per 100 ml 55. ) The first solution contins 0.10 HCl (strong cid) nd 0.10 HOCl (wek cid). Since there is oth strong nd wek cid with equl concentrtions of ech the numer of H ions from the wek cid will e miniml therefore the ph will e completely from the strong cid (ph = log[h ] = log 0.10 = 1.00 This cn e proven using n ICE tle HOCl(q) H (q) OCl (q) = 3.5 0 8 HOCl OCl H Initil 0.10 0 0.10 Chnge x x x Equilirium 0.10x x 0.10x OCl H 8 x 0.10 x 3.510 HOCl 0.10 x Since is smll ssume 0.50x = 0.50 3.510 8 x 3.510 8 x 0.10 0.10 Check ssumption (5% rule) Amount Gined/Lost 3.510 Originl Concentrtion 0.10 Less thn 5%, therefore, good ssumption. Clculte ph 8 5 100% 100% 3.510 % 8 H 0.10 x 0.10 3.5 10 0.10 ph log H log 0.10 1.00 ) The second solution contins 0.050 HNO 3 (strong cid) nd 0.50 HC H 3O (wek cid). Since there is oth strong nd wek cid nd even though the concentrtion of the wek cid is 10x less thn the concentrtion of the strong cid the numer of H ions from the wek cid will e miniml therefore the ph will e completely from the strong cid (ph = log[h ] = log 0.050 = 1.30 This cn e proven using n ICE tle HC H 3O (q) H (q) C H 3O (q) = 1.8 0 5 HC H 3O C H 3O H Initil 0.50 0 0.050 Chnge x x x Equilirium 0.50x X 0.050x 10
CH3O H 5 x 0.050 x 1.810 HC H O 0.50 x 3 Since is smll ssume 0.50x = 0.50 nd 0.05x=0.05 1.810 5 x 1.810 4 x 0.05 0.50 Check ssumption (5% rule) Note:0.05<0.50 therefore is the ssumption works for 0.05x then it will work form 0.50x 4 Amount Gined/Lost 1.810 100% 100% 0.36% Originl Concentrtion 0.05 Less thn 5%, therefore, good ssumption. Clculte ph H 4 0.050 x 0.050 1.8 10 0.050 ph log H log 0.50 1.30 59. HOCN is wek cid, therefore, the following equilirium is estlished: HOCN(q) OCN (q) H (q) HOCN OCN H Initil 1.00 10 0 0 Chnge x x x Equilirium 1.00 10 x x x The ph of the solution t equilirium will llow us to clculte the concentrtion of H ions t equilirium =x ph logh ph.77 H 10 10 0. 0017 Updting the ICE tle gives HOCN OCN H Initil 1.00 10 0 0 Chnge 0.0017 0.0017 0.0017 Equilirium 0.0083 0.0017 0.0017 OCN H HOCN 0.0017 0.0017 0.0083 3.510 63. The lrger the the stronger the se. Therefore the strongest se will hve the lrgest. NH 3 > C 5H 5N > H O > NO 3 Strongest se Wekest se Note: HNO 3 is strong cid. The stronger the cid the weker the conjugte se. Therefore, NO 3 is very wek se. Note: H O is the conjugte se of (H 3O ). H 3O is one of the wekest strong cids. Therefore, H O is only slightly more sic thn NO 3 4 11
64. The smller the the stronger the conjugte cid. HNO 3 > C 5H 5NH > NH 4 > H O Strongest cid Wekest cid Note: HNO 3 is strong cid. Note: C 5H 5N nd NH 3 re oth wek ses, therefore, will hve wek conjugte cids. The of NH 3 > k of C 5H 5N resulting in C 5H 5NH eing stronger cid thn NH 4. Note: H O is the conjugte cid of OH. OH is strong se. Strong ses hve the wekest conjugte cids. (H O(l) OH (q) OH (q) H O(l)) 68. ) C(OH) is strong se (fully dissocites) mol C( OH ) OH 0.00040 C( OH ) 1 mol C( OH ) 0.00080 OH poh log OH log 0.00080 3.10 ph poh 14 ph 14 poh 14 3.10 10.90 ) OH is strong se (fully dissocites) OH 56.11 g mol 5 g OH 0.45 mol OH 1 mol OH 1 mol OH 56.11 g OH 1 mol OH n 0.45mol OH 0.45 V 1L poh log OH log 0.45 0.35 ph poh 14 ph 14 poh 14 0.35 13.65 c) NOH is strong se (fully dissocites) g 40.00 N5OH mol 1 mol NOH 1 mol OH 40.00 g NOH 1 mol NOH 150.0 g NOH 3.750 mol OH n 3.750mol OH 3.750 V 1L poh log OH log3.750 0.5740 ph poh 14 ph 14 poh 14 0.5740 14.5740 1
69. B(OH) is strong se (fully dissocites) Clculte the concentrtion of OH 14 ph poh poh 14 ph 14 10.50 3.50 poh logoh poh 3.50 4 OH 10 10 3. 10 Clculte the concentrtion of B(OH) 1 mol B OH B OH mol OH 3. 10 OH 1.6 10 4 4 73. The presence of nitrogen most often results in sic molecule. Nitrogen hs n unpired electron, therefore, onds esily to H ions. 93. The lrger the ph the more sic solution is ) HI strong cid HF wek cid NF (wek se) NF(s) N (q) F (q) N H O NOH H (NOH is strong se) N is neutrl F H O HF OH F is sic NI (neutrl) NI(s) N (q) I (q) N H O NOH H (NOH is strong se) N is neutrl I H O HI OH (HI is strong cid) I is neutrl HI < HF < NI < NF Incresing ph ) NH 4Br (wek cid) NH 4Br(s) NH 4 (q) Br (q) NH 4 H O NH 3 H 3O NH 4 is cidic Br H O HBr OH (HBr is strong cid) Br is neutrl HBr strong cid Br (neutrl) Br(s) (q) Br (q) H O OH H (OH is strong se) is neutrl Br H O Br is neutrl NH 3 wek se HBr OH (HBr is strong cid) 13
14
HBr < NH 4Br < Br < NH 3 Incresing ph c) C 6N 5NH 3NO 3 (wek cid) C 6N 5NH 3NO 3(s) C 6N 5NH 3 (q) NO 3 (q) C 6N 5NH 3 H O C 6N 5NH H 3O C 6N 5NH 3 is cidic NO 3 H O HNO 3 OH (HNO 3 is strong cid) NO 3 is neutrl NNO 3 (neutrl) NNO 3(s) N (q) NO 3 (q) N H O NOH H (NOH is strong se) N is neutrl NO 3 H O HNO 3 OH (HNO 3 is strong cid) NO 3 is neutrl NOH strong se HOC 6H 5 wek cid ( = 1.6 10 10 ) OC 6H 5 (wek se) OC 6H 5(s) (q) OC 6H 5 (q) H O OH H (OH is strong se) is neutrl OC 6H 5 H O HOC 6H 5 OH OC 6H 5 is sic C 6H 5NH wek se HNO 3 Strong cid There re wek cid C 6N 5NH 3NO 3 (C 6H 5NH 3 ) nd HOC 6H 5 Compre the vlues to determine the weker cid HOC 6H 5 = 1.6 10 10 The vlue of C 6H 5NH 3 is not given therefore, it must e clculted C 6H 5NH (q) = 3.8 10 10 14 w 1.010.610 10 3.810 Therefore, C 6N 5NH NO 3 is stronger cid thn HOC 6H 5 There re wek ses C 6N 5NH nd OC 6H 5 (OC 6H 5 ) Compre the vlues to determine the weker se C 6N 5NH = 3.8 10 10 The vlue of OC 6H 5 is not given therefore, it must e clculted HOC 6H 5(q) = 1.6 10 10 14 w 1.010 6.10 10 1.610 Therefore, OC 6H 5 is stronger se thn C 6N 5NH HNO 3< C 6N 5NH 3NO 3 < HOC 6H 5 < NNO 3< C 6H 5NH < OC 6H 5 < NOH 5 5 15
95. See chrt for prolem 85. ) Sr(NO 3) (slt formed form strong cid nd strong se) Sr(NO 3) Sr(NO 3) (s) Sr (q) NO 3 (q) Sr H O Sr(OH) H (Sr(OH) is strong se) Sr is neutrl NO 3 H O HNO 3 OH (HNO 3 is strong cid) NO 3 is neutrl Therefore, Sr(NO 3) is neutrl ) C H 5NH 3CN C H 5NH 3CN(s) C H 5NH 3 (q) CN (q) C H 5NH 3 H O C H 5NH H 3O C H 5NH 3 is cidic CN H O HCN OH CN is sic Therefore, need to compre nd vlues of rections We re not given the of C H 5NH 3, therefore, we need to clculte it C H 5NH (q) = 5.6 10 4 14 w 1.010 1.810 4 5.610 11 We re not given the of CN, therefore, we need to clculte it HCN(q) = 6. 10 10 14 w 1.010 1.610 10 6.10 Since < C H 5NH 3CN is sic c) C 5H 5NHF C 5H 5NHF(s) C 5H 5NH (q) F (q) C 5H 5NH H O C 5H 5N H 3O C 5H 5NH is cidic F H O HF OH F is sic Therefore, need to compre nd vlues of rections We re not given the of C 5H 5NH, therefore, we need to clculte it C H 5N(q) = 1.7 10 9 14 w 1.010 5.910 9 1.710 We re not given the of F, therefore, we need to clculte it HF(q) = 7. 10 4 14 w 1.010 1.410 4 7.10 Since > C 5H 5NHF is cidic 5 6 11 16
d) NH 4C H 3O NH 4C H 3O (s) NH 4 (q) C H 3O (q) NH 4 H O NH 3 H 3O NH 4 is cidic C H 3O H O HC H 3O OH C H 3O is sic Therefore, need to compre nd vlues of rections NH 4 = 5.6 10 10 We re not given the of C H 3O therefore we need to clculte it HC H 3O (q) = 1.8 10 5 14 w 1.010 10 5.610 5 1.810 = NH 4C H 3O is neutrl e) NHCO 3 NHCO 3(s) N (q) HCO 3 (q) N H O N(OH) H (NOH is strong se) N is neutrl HCO 3 H O H CO 3 OH HCO 3 H O CO 3 H 3O Since H CO 3 is polyprotic cid must look t nd of the two possile rections. HCO 3 = 4.8 10 11 We re not given the of HCO 3 therefore we need to clculte it H CO 3(q) = 4.3 10 7 14 w 1.010.310 7 4.310 Since < HCO 3 is sic Therefore, NHCO 3 is sic 8 97. ) Cl NO Cl(s) (q) Cl (q) H O OH H (OH is strong se) is neutrl Cl H O HCl OH (HCl is strong cid) Cl is neutrl Therefore, Cl is neutrl Therefore, the solution will e neutrl ph=7.00. For neutrl solutions the concentrtion of H nd OH equls 1.0 10 7. 17
) F F(s) (q) F (q) H O OH H (OH is strong se) is neutrl F H O HF OH F is sic Therefore, F is sic Interested in the following rection F (q) H O(l) HF(q) OH(q) = [HF][OH ] [F ] Prolem we do not know HF(q) =7. 10 4 = w 4 = 1.4 10 11 Clculte the concentrtion of OH ions F HF OH Initil 1.0 0 0 Chnge x x x Equilirium 1.0x x x = 1.0 10 14 7. 10 11 OH HF 1.410 F 1.0 x Since is smll ssume tht 1.0x = 1.0 11 1.410 1.0 6 x3.710 Check ssumption (5% rule) Amount Gined/Lost 3.710 Originl Concentrtion 1.0 Less thn 5%, therefore, good ssumption. Clculte OH concentrtion 6 OH x 3.7 10 Clculte H concentrtion [H ][OH ] = 1.0 10 14 [H ] = 1.0 10 14 =.7 10 9 3.7 10 6 6 4 100% 100% 3.710 % Clculte ph 9 ph log H log.710 8.57 18
101. ) NO NO (s) (q) NO (q) H O OH H (OH is strong se) is neutrl NO H O HNO OH NO is sic Therefore, NO is sic Interested in the following rection NO (q) H O(l) HNO (q) OH (q) HNO OH NO Prolem we don t know now HNO (q) = 4.0 10 4 14 w 1.010.510 4 4.010 11 Clculte the concentrtion of OH ions NO HNO OH Initil 0.1 0 0 Chnge x x x Equilirium 0.1x X x HNO OH NO 11.5 10 0.1 x Since is smll ssume tht 0.1x = 0.1 11.510 0.1 6 x1.710 Check ssumption (5% rule) 6 Amount Gined/Lost 1.710 100% 100% 0.0014% Originl Concentrtion 0.1 Less thn 5%, therefore, good ssumption. Clculte OH concentrtion 6 OH x 1.710 Clculte ph 6 poh log OH log 1.7 10 5.77 ph poh 14 ph 14 poh 14 5.77 8.3 19
) NOCl NOCl(s) N (q) OCl (q) N H O NOH H (NOH is strong se) N is neutrl OCl H O HOCl OH OCl is sic Therefore, NOCl is sic Interested in the following rection OCl (q) H O(l) HOCl(q) OH (q) HOCl OH OCl Prolem we don t know HOCl(q) = 3.5 10 8 14 1.010.910 8 3.510 Clculte the concentrtion of OH ions OCl HOCl OH Initil 0.45 0 0 Chnge x x x Equilirium 0.45x X x 7 7 HOCl OH.910 OCl 0.45 x Since is smll ssume tht 0.45x = 0.45 7.910 0.45 4 x3.610 Check ssumption (5% rule) 4 Amount Gined/Lost 3.610 100% 100% 0.08% Originl Concentrtion 0.45 Less thn 5%, therefore, good ssumption. Clculte OH concentrtion 4 OH x 3.610 Clculte ph poh ph poh 14 4 log OH log 3.6 10 3.44 ph 14 poh 14 3.44 10.56 0
c) NH 4OCl 4 ( slt from wek se nd strong cid) NH 4OCl 4(s) NH 4 (q) OCl 4 (q) NH 4 H O NH 3 H 3O NH 4 is cidic OCl 4 H O HOCl 4 OH (HOCl 4 strong cid) OCl 4 is neutrl Therefore, NH 4OCl 4 is cidic Interested in the following rection NH 4 (q) NH 3(q) H (q) NH H w 3 10 5.610 NH 4 Clculte the concentrtion of H ions NH 4 NH 3 H Initil 0.40 0 0 Chnge x x x Equilirium 0.40x x X NH H NH Since is smll ssume tht 0.40x = 0.40 3 10 5.6 10 0.40 x 4 10 5.610 0.40 5 x1.510 Check ssumption (5% rule) 5 Amount Gined/Lost 1.510 100% 100% 0.0038% Originl Concentrtion 0.40 Less thn 5%, therefore, good ssumption. Clculte H concentrtion 5 H x 1.4 10 Clculte ph ph H 5 log log 1.5 10 4.8 11. Becuse the solution hs ph of 4.6 we know tht the solution is cidic, therefore, NOH, NH 3, nd NCN cnnot e the solute. Becuse the ul is right it mens tht there re lot of ions in solution. Therefore, strong electrolyte is need. Only HCl, NOH, nd NH 4Cl re strong electrolytes. NOH ws lredy eliminted ecuse it ws sic. HCl could light the ul right ecuse it is strong cid ut the ph of 1.0 solution of HCl is ph log H log 1.0 0 1
Confirm the solution is mde from NH 4Cl y compre ctul to clculted vlues Determine the of the wek cid. HA(q) A (q) H (q) = unknown A H HA HA A H Initil 1.0 0 0 Chnge x x x Equilirium 1.0x x x Determine x x = [H ] ph = log[h ] 4.6 = log[h ] [H ] =.5 10 10 5.5 10.5 10 5 A H 6.310 5 HA 1.0.510 This of NH 4 is 5.6 10 10 which is close to the clculted vlue. Therefore the solution must contin NH 4Cl 10