Dr. Steward s CHM152 Exam #2 Review Spring 2014 (Ch ) KEY. 2. For the acid/base reactions determine the conjugate acid/base pairs for each.

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1 Dr. Stewrd s CHM15 Em # Review Spring 014 (Ch ) 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 = , K HF = ): 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: M HBr ph = log(0.0051) =.9, poh = = [H + ] = 10.9 = M, [OH ] = = M 0.010M NOH poh = log(0.010) =.00, ph = = [OH ] = = 0.10M, [H + ] = 10 = M 0.010M Sr(OH) poh = log(0.010 ) = 1.70, ph = = 1.0 [OH ] = = 0.00M, [H + ] = = 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 M formic cid solution with K = ?

2 HCOOH + H O H O + + HCOO 0.100M 0M 0M M ph = log[h O + ] = log(4.410 ) =.7 7. Wht is the percent dissocition? 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 molNF.56g NF g NF molnf F + H O OH HF M 0M 0M M molnf MNF 4.50L 14 K Kb Kw ph = 14log[OH + ] = 14log( ) = 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 HA + H O H O + + A? (y) 0M 0M + + y [H O ph.59 ] ( ) y.5710 y 0.104M 0.104M.5710 M = 0.101M or 0.10M

3 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 + + ()() K % 1. Clculte the percent ioniztion if there is lso 0.50M concentrtion of NF in solution. HF + H O H O + + F I M M C + + E ()(0.500) % 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 [HO ] M (0.500)() % 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 L HCl 1 vol of NOH@ eq. molhcl 1LHCl HCl + NOH H O + NCl B mol mol Δ mol mol A mol 0 ph log[h ] log 1molNOH 1molHCl L NOH L NOH mol HCl totl 1 1L NOH.00molNOH mol of mol of L NOH 1.00mol NOH@ eq. (0.0050L) mol NOH 1L mol HCl. ( L) molNOH 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 mol log log(0.40) 0.40 eq L L

4 Wht would be the ph t the equivlence point? mol HCl 1mol NOH 1L NOH vol of NOH@ eq L HCl L NOH 1L HCl 1mol HCl.00mol NOH.00mol mol of NOH@ eq. ( L) mol NOH 1L H + + OH H O B mol mol Δ mol mol 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 ( L) mol NOH 1L H + + OH H O B mol mol Δ mol mol A mol After the equivlence point, you re just dding etr OH ions to the solution mol ph14 log L L L 16. You re titrting 10.00mL of M of benzoic cid (K is ) with.00m NOH. Wht would be the ph fter.50 ml hs been dded? HA + OH H O + A B mol mol 0 Δ mol mol mol A mol mol ph pk log ph 4.19 log mol mol 4.19

5 Wht would be the ph fter dding totl of 5.00 ml NOH? mol of mol of conjugtebse(a )teq. mol mol of HA. ( L) mol HAtstrt molof A 1L mol HA 1molNOH 1L NOH vol of NOH@eq L HA 1LHAl 1mol HA.00molNOH molof A molA [ A ] M totlvolume@eq L L A + H O OH + HA I M 0 0 C + + E Kw Kb K L NOH ()() ph = 14log[OH ] = 14log ( ) = 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 )(0.6667) Wht would be the ph fter dding totl of 5.75 ml NOH? HA + OH H O + A B mol mol 0 Δ mol mol mol A mol mol Totl mol.00mol of NOH ( L) mol 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 mol ph14 log L L L 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 )? 0.0M ph pk log 9.55 log M b 5 log pk 14 pk 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 mol 0.00mol 0.00mol Δ mol mol mol A mol 0.0mol 0.0mol molsof NH ndnh L 0.00molsof ech 1L.00mol molsof NOH dded L molsof NOH 1L ph pk log 9.5 log 0.0mol 0.08mol 9.8

6 Wht is the ph fter dding.00ml of M HCl? + HCl + NH NH 4 B mol 0.00mol 0.00mol Δ mol mol mol A mol 0.0mol molsof HCl mol dded L molsof HCl 1L ph pk log 9.5 log 0.08mol 0.0mol You hve 60.00mL of buffer solution consisting of 0.900M CH COOH nd 1.10M NCH COO. K b = Determine the ph of the solution fter dding 10.11mL of.m HBr mol CH COOH.06000L CH COOH 1L CH COOH mol CH COOH.mol HBr L HBr 0.05mol HBr K L HBr mol CH COO L CH COO mol CH COO 1L CH COO HBr + CH COO CH COOH + Br B 0.05mol mol mol ph pk Δ 0.05mol 0.05mol +0.05mol A mol mol log 4.74 log 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 L Wht would be the ph when t the point (tht you clculted bove?) HBr + CH COO CH COOH + Br B mol mol mol Δ mol mol mol A 0 0mol 0.100mol 0.100mol CH COOH 1.4M ( L 0.096L) CH COOH + H O H O + + CH COO I 1.4M 0 0 C + + E 1.4M ()() ( ph log(4.910 )(1.4) ).1

7 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 L NOH 1L NOG mol NOH mol CH COOH 1.10mol CH COO L CH COO mol CH COO 1L CH COO NOH + CH COOH CH COO + N + B mol mol mol ph pk Δ mol mol mol A mol mol log 4.74 log After dding 0.00mL of. KOH?.mol NOH L NOH 1L NOG mol NOH NOH + CH COOH CH COO + N + B mol mol mol Δ mol mol mol A mol 0 mol 0.100mol mol NOH 0.01M NOH ( L L) poh log[oh ] log( 0.01).0 ph 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 log (.0 ) (.0 ) log (.0 ) log (.0) M.00M 0.5M 1.75M

8 1. *If moles of AgNO is dded to L of 0.000M NCl, will AgCl precipitte form? (K sp AgCl = ). Eplin. AgCl(s) Ag + (q) + Cl (q) K sp = [Ag][Cl] Q sp = ( )(0.000) = 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] = ( )[Cl ] = [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 = ). 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 ] = () () =(7 )(4 ) I 0 0 C = = = molr solubility = E M [Ni + ] = ()( ) = M [PO 4 ] = ()( ) = 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 ] = () (0.100) = (7 )(0.0100) I C = 0.70 = E = molr solubility = M [Ni + ] = ()( ) = M [PO 4 ] = ()( ) 0.100M

9 5. How much (in g) brium sulfte (BSO 4, MW=.9) will dissolve in 500. ml of wter. For BSO 4 : K sp = B SO 4 + H O B + + SO 4 K sp = [B + ][SO 4 ] = I 0 0 C + = = molr solubility = M E molbso4.9g BSO L HO 1.10 gbso4 1L H O 1molBSO 4