Dr. Stewrd s CHM15 Em # Review Spring 014 (Ch. 16. 17) KEY 1. For ech of the following cid / bse rections, determine if ech of the following will hve more products or rectnts (K HSO 4 = 1.10, K b NH = 1.810 5, K HF = 6.810 4 ): NH 4 + (q) + F (q) NH (q) + HF (q) More Products More rectnts OH (q) + HNO (q) H O (l) + NO (q) More Products More rectnts HSO 4 (q) + NH (q) NH 4 + (q) + SO 4 (q) More Products More rectnts H O (l) H O + (q) + OH (q) More Products More rectnts. For the cid/bse rections determine the conjugte cid/bse pirs for ech. NH 4 + / NH (cid / bse), F / HF (bse / cid) OH / H O (bse / cid), HNO / NO (cid / bse) HSO 4 / SO 4 (cid / bse), NH / NH 4 + (bse / cid) H O / H O + (bse / cid), H O / OH (cid / bse) The equilibrium of Bronsted/Lowry cid/bse rection will fvor the side of the eqution with the weker cid nd weker bse. We cn use K nd K b vlues to determine the weker cid/bse.. Eplin how NH is both Bronsted/Lowry bse nd Lewis bse. It is Bronsted/Lowry bse becuse it is ble to ccept protons. It is Lewis bse becuse it is ble to donte electrons. 4. Determine the ph, poh, [H O + ], nd [OH ] for the following solutions: 0.0051M HBr ph = log(0.0051) =.9, poh = 14.00.9 = 11.71 [H + ] = 10.9 = 0.0051M, [OH ] = 10 11.71 = 1.910 1 M 0.010M NOH poh = log(0.010) =.00, ph = 14.00.00 = [OH ] = 10.00 = 0.10M, [H + ] = 10 = 1.010 1 M 0.010M Sr(OH) poh = log(0.010 ) = 1.70, ph = 14.00 1.70 = 1.0 [OH ] = 10 1.7 = 0.00M, [H + ] = 10 1.0 = 5.010 1 M 5. Determine whether the following slts re cidic, bsic, or neutrl: LiCl Al(NO ) BF NH 4 I nuetrl cidic bsic cidic 6. Wht is the ph of 0.100 M formic cid solution with K = 1.8 10 4?
HCOOH + H O H O + + HCOO 0.100M 0M 0M + 0.100M + 1.810 0.100 1.810 4 1.810 0.100 5 5 1.810 4 ph = log[h O + ] = log(4.410 ) =.7 7. Wht is the percent dissocition? 4.% 4.410 4.410 0.100 100 4.% OK ssumption 8. If the concentrtion of the formic cid ws 4.0M insted, eplin how the ph, [OH ], nd % dissocition would be different (higher/lower/sme). The ph would be lower (more cidic/higher [H O + ]), [OH ] would be lower, % dissocition would be lower (high initil concentrtion = lover % dissocition for sme cid or bse). 9. Clculte the ph if.56 grms of sodium fluoride is dissolved in 4.50 liters of wter. K for hydrofluoric cid is 7.1 10 4 1molNF.56g NF 6.09710 41.99g NF molnf F + H O OH HF 1.510 M 0M 0M + + 1.510 M 6.09710 molnf 1.510 MNF 4.50L 14 K Kb Kw 1.010 1.010 14 1.4110 4 7.110 1.510 11 1.4110 11 1.510 1.4110 11 ph = 14log[OH + ] = 14log(4.710 7 ) = 7.64 1 1.9110 1 1.9110 4.710 7 10. Wht is the initil concentrtion of benzoic cid (C 6 H 5 COOH) in solution with ph =.59 (ssume the benzoic cid ws dded to the wter with no other ions initilly)? The cid dissocition constnt for this monoprotic cid is 6.5 10 5. HA + H O H O + + A? (y) 0M 0M + + y.5710.5710.5710 [H O ph.59 ] 10 10.5710 (. 5710 ) 5 6.510 y.5710 y 0.104M 0.104M.5710 M = 0.101M or 0.10M
11. Eplin the commonion effect. Wek cid or wek bse s % dissocition will decrese if they re plced in solutions with one of the products of dissocition. For emple, The % dissocition of HCN will be decrese if dditionl CN is dded to the solution (s NCN, for emple). HCN(q) + H O(l) H O + (q) + CN (q) 1. Clculte the percent ioniztion of M solution of HF (pk =.17). HF + H O H O + + F I M 0 0 C + + E + + ()() 6.7610 4 K 10.17 6.7610 4.610 4 6.7610.610 100.6% 1. Clculte the percent ioniztion if there is lso 0.50M concentrtion of NF in solution. HF + H O H O + + F I M 0 0.500M C + + E + 0.500+ ()(0.500) 6.7610 4 1.510 1.510 100 0.14% 14. Clculte the percent ioniztion if the ph of the solution before the HF ws dded ws 0.0. HF + H O H O + + F I M 0.500M 0 C + + E 0.500+ + 0.0 [HO ] 10 0.50M (0.500)() 6.7610 4 1.510 1.510 100 0.14% 15. You re titrting 10.00mL of M of HCl with.00m NOH. Wht would be the ph t ½ the equivlence point? NOH + HCl H O + NCl vol of NOH@eq. 0.01000L HCl 1 vol of NOH@ eq. molhcl 1LHCl HCl + NOH H O + NCl B 0.0100mol 0.00500mol Δ 0.00500mol 0.00500mol A 0.0050mol 0 ph log[h ] log 1molNOH 1molHCl 0.00500L NOH 0.0050L NOH mol HCl totl volume @ 1 1L NOH.00molNOH mol of mol of 0.00500L NOH 1.00mol NOH@ eq. (0.0050L) 0.00500mol NOH 1L mol HCl. (0.01000L) 0.0100molNOH 1L This mkes sense since only hlf of the cid should be left t ½ eq. If you relize this, you do not hve to use chnge tble. 0.0050mol log log(0.40) 0.40 eq 0.00500L 0.01000L
Wht would be the ph t the equivlence point? mol HCl 1mol NOH 1L NOH vol of NOH@ eq. 0.01000L HCl 0.00500L NOH 1L HCl 1mol HCl.00mol NOH.00mol mol of NOH@ eq. (0.00500L) 0.0100mol NOH 1L H + + OH H O B 0.0100mol 0.0100mol Δ 0.0100mol 0.0100mol A 0 0 There will only be neutrl slt t the equivlence point of Strong Acid/Strong Bde titrtion. ph = 7 Wht would be the ph fter dding 0.75mL of NOH fter the equivlence point? Totl mol.00mol of NOH (0.00575L) 0.0115mol NOH 1L H + + OH H O B 0.0100mol 0.0115mol Δ 0.0100mol 0.0100mol A 0 0.0015mol After the equivlence point, you re just dding etr OH ions to the solution 0.0015mol ph14 log 1.98 0.01000L 0.00500L 0.00075L 16. You re titrting 10.00mL of M of benzoic cid (K is 6.410 5 ) with.00m NOH. Wht would be the ph fter.50 ml hs been dded? HA + OH H O + A B 0.0100mol 0.00500mol 0 Δ 0.00500mol 0.00500mol +0.00500mol A 0.00500mol 0 +0.00500mol ph pk log ph 4.19 log 0.00500mol 0.00500mol 4.19
Wht would be the ph fter dding totl of 5.00 ml NOH? mol of cid(ha)@strt mol of conjugtebse(a )teq. mol mol of HA. (0.01000L) 0.0100mol HAtstrt molof A 1L mol HA 1molNOH 1L NOH vol of NOH@eq. 0.01000L HA 1LHAl 1mol HA.00molNOH molof A 0.0100molA [ A ] 0.6667M totlvolume@eq. 0.01000L 0.00500L A + H O OH + HA I 0.6667M 0 0 C + + E 0.6667 Kw Kb K 1.5610 10 (1.5610 @eq 0.00500L NOH ()() 0.6667 10 ph = 14log[OH ] = 14log (1.010 5 ) = 9.01 Think bout wht is left t the equivlence point of wek cid/strong bse titrtion. You could use chnge tble if you re unsure. But it s fster if you don t hve to. 1.5610 10 )(0.6667) 1.010 5 Wht would be the ph fter dding totl of 5.75 ml NOH? HA + OH H O + A B 0.0100mol 0.0115mol 0 Δ 0.0100mol 0.0100mol +0.0100mol A 0 0.0015mol +0.0100mol Totl mol.00mol of NOH (0.00575L) 0.0115mol NOH 1L Although you hve two bses fter the equivlence point, ssume tht the concentrtion is going to hve much greter effect on ph thn the wek bse. 0.0015mol ph14 log 1.98 0.01000L 0.00500L 0.00075L pk logk b 17. Wht is the ph of 100.0mL solution contining 0.0 M NH nd 0.0 M NH 4 NO (K b of NH is 1.810 5 )? 0.0M ph pk log 9.55 log 9.6 0.0M b 5 log1.810 4.745 pk 14 pk 14 4.745 9.55 Wht is the ph fter dding ml of.00m NOH Since this hs the components of buffer solution, we cn use HendersonHssleblch + NOH + NH 4 NH B 0.0000mol 0.00mol 0.00mol Δ 0.0000mol 0.0000mol +0.0000mol A 0 0.08mol 0.0mol 0.0mol molsof NH ndnh4 0.1000L 0.00molsof ech 1L.00mol molsof NOH dded 0.00100L 0.0000molsof NOH 1L ph pk log 9.5 log 0.0mol 0.08mol 9.8
Wht is the ph fter dding.00ml of M HCl? + HCl + NH NH 4 B 0.0000mol 0.00mol 0.00mol Δ 0.0000mol 0.0000mol +0.0000mol A 0 0.08mol 0.0mol molsof HCl mol dded 0.0000L 0.0000molsof HCl 1L ph pk log 9.5 log 0.08mol 0.0mol 9.17 18. You hve 60.00mL of buffer solution consisting of 0.900M CH COOH nd 1.10M NCH COO. K b = 5.610 10 Determine the ph of the solution fter dding 10.11mL of.m HBr. 0.900mol CH COOH.06000L CH COOH 1L CH COOH 0 0.0540mol CH COOH.mol HBr 14 0.01011L HBr 0.05mol HBr 1.010 5 K 1.8 10 1L HBr 10 5.610 1.10mol CH COO 0.06000L CH COO 0.0660mol CH COO 1L CH COO HBr + CH COO CH COOH + Br B 0.05mol 0.0660mol 0.0540mol ph pk Δ 0.05mol 0.05mol +0.05mol A 0 0.045mol 0.0860mol log 4.74 log 0.045 0.0860 4.44 How much totl HBr would you hve to dd until the buffer ws used up? 1.10mol CH COO 1mol HBr.06000L CH COO 1L CH COO 1mol CH COO 1L HBr. mol HBr 0 0.096L Wht would be the ph when t the point (tht you clculted bove?) HBr + CH COO CH COOH + Br B 0.0660mol 0.0660mol 0.0540mol Δ 0.0660mol 0.0660mol +0.0660mol A 0 0mol 0.100mol 0.100mol CH COOH 1.4M (0.06000L 0.096L) CH COOH + H O H O + + CH COO I 1.4M 0 0 C + + E 1.4M ()() 1.4 1.810 5 (1.810 5 ph log(4.910 )(1.4) 4.910 ).1
19. If you strt over with fresh buffer (sme conditions/volumes s originl), wht would be the ph of the solution fter dding 0.00mL of.m KOH? 0.900mol CH COOH.06000L CH COOH 1L CH COOH 0.mol NOH 0.0000L NOH 1L NOG 0.0446mol NOH 0.0540mol CH COOH 1.10mol CH COO 0.06000L CH COO 0.0660mol CH COO 1L CH COO NOH + CH COOH CH COO + N + B 0.0446mol 0.0660mol 0.0540mol ph pk Δ 0.0446mol 0.0446mol +0.0446mol A 0 0.014mol 0.0986mol log 4.74 log 0.0986 0.014 5.40 After dding 0.00mL of. KOH?.mol NOH 0.0000L NOH 1L NOG 0.0669mol NOH NOH + CH COOH CH COO + N + B 0.0669mol 0.0660mol 0.0540mol Δ 0.0660mol 0.0660mol +0.0660mol A 0.0009 mol 0 mol 0.100mol 0.0009mol NOH 0.01M NOH (0.06000L 0.0000L) poh log[oh ] log( 0.01).0 ph14.0.0 1.0 0. If you wnted to mke buffer with ph of.90 using the sme buffer s before with totl concentrtion of.00m, wht would the concentrtion of ech prt of the buffer need to be? ph pk log.90 4.74 log (.0 ) 0.145 (.0 ) 0.9 0.15 0.84 log (.0 ) 10 0.84 10 log (.0) 0.9 1.15 0.5M.00M 0.5M 1.75M 1.15 1.15
1. *If 0.00056 moles of AgNO is dded to L of 0.000M NCl, will AgCl precipitte form? (K sp AgCl = 1.810 10 ). Eplin. AgCl(s) Ag + (q) + Cl (q) K sp = [Ag][Cl] Q sp = (0.00056)(0.000) = 1.110 6 since Q > K will shift to the left (solid AgCl), so it will precipitte! Wht would the minimum concentrtion of NCl need to be to form precipitte? K sp = [Ag][Cl] = (0.00056)[Cl ] = 1.810 10 [Cl ] =.10 7 M. Ni(OH) is n insoluble compound in wter. How would the molr solubility be ffected in ech if the following situtions? NOH is dded Decrese solubility (common ion) NCl is dded No effect on solubility. Does not rect with Ni + or OH HCl is dded Increse solubility. H + rects with OH.. Assume you hve sturted solution of nickel (II) phosphte with some solid in the continer (K sp = 4.74 10 ). Wht is it s molr solubility? Determine the concentrtion of the Ni + nd PO 4 ions in the solution. Ni (PO 4 ) Ni + + PO 4 K sp = [Ni + ] [PO 4 ] 4.7410 = () () =(7 )(4 ) I 0 0 C + + 4.7410 = 108 5 =.110 7 = molr solubility = E.110 7 M [Ni + ] = ()(.110 7 ) = 6.910 7 M [PO 4 ] = ()(.110 7 ) = 4.610 7 M 4. Wht would be it s molr solubility be if nickel (II) phosphte if 0.100moles of N PO 4 ws dded to L of the solution bove? Wht would be the concentrtions of the Ni + nd PO 4 ions? Ni (PO 4 ) + H O Ni + + PO 4 K sp = [Ni + ] [PO 4 ] 4.7410 = () (0.100) = (7 )(0.0100) I 0 0.100 C + + 4.7410 = 0.70 = 5.6010 11 E 0.100+ = molr solubility = 5.6010 11 M [Ni + ] = ()(5.6010 11 ) = 1.6810 10 M [PO 4 ] = 0.100+()( 5.6010 11 ) 0.100M
5. How much (in g) brium sulfte (BSO 4, MW=.9) will dissolve in 500. ml of wter. For BSO 4 : K sp = 1.1 10 10 B SO 4 + H O B + + SO 4 K sp = [B + ][SO 4 ] 1.110 10 = I 0 0 C + = 1.0410 5 = molr solubility = 1.0410 5 M E 5 1.0410 molbso4.9g BSO4 0.500L HO 1.10 gbso4 1L H O 1molBSO 4