Dr. Stewrd s CHM152 Em #2 Review Spring 2014 (Ch. 16. 17) KEY 1. 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 2 O(l) H O + (q) + CN (q) 2. Clculte the percent ioniztion of M solution of HF (pk =.17). HF + H 2 O H O + + F I M 0 0 C + + E + + ()() 6.7610 4 K 10.17 6.7610 4 2.610 2 4 2 6.7610 2.610 100 2.6%. Clculte the percent ioniztion if there is lso 0.50M concentrtion of NF in solution. HF + H 2 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% 4. Clculte the percent ioniztion if the ph of the solution before the HF ws dded ws 0.0. HF + H 2 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% 5. You re titrting 10.00mL of M of HCl with 2.00M NOH. Wht would be the ph fter 2.50 ml of NOH re dded? NOH + HCl H 2 O + NCl mol 2.00mol of NOH (0.00250L) 0.00500mol NOH 1L HCl + NOH H 2 O + NCl B 0.0100mol 0.00500mol Δ 0.00500mol 0.00500mol A 0.0050mol 0 mol mol of HCl. (0.01000L) 0.0100molNOH 1L
ph log[h ] log mol HCl totl volume 0.0050mol log log(0.40) 0.40 0.002500L 0.01000L Wht would be the ph fter 5.00ml totl of NOH re dded? mol 2.00mol of NOH (0.00500L) 0.0100mol NOH 1L H + + OH H 2 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 totl of 5.75mL of NOH? Totl mol 2.00mol of NOH (0.00575L) 0.0115mol NOH 1L H + + OH H 2 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 12.98 0.01000L 0.00575L 6. You re titrting 10.00mL of M of benzoic cid (K is 6.410 5 ) with 2.00M NOH. Wht would be the ph fter 2.50 ml hs been dded? HA + OH H 2 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? HA + OH H 2 O + A B 0.0100mol 0.0100mol 0 Δ 0.0100mol 0.0100mol +0.0100mol A 0 0mol +0.0100mol mol of A 0.0100mol A [ A ] 0.6667M totl volume 0.01000L 0.00500L A + H 2 O OH + HA I 0.6667M 0 0 C + + E 0.6667 Kw Kb K 1.5610 10 (1.5610 ()() 0.6667 10 1.5610 10 )(0.6667) 1.010 ph = 14log[OH ] = 14log (1.010 5 ) = 9.01 5 Wht would be the ph fter dding totl of 5.75 ml NOH? HA + OH H 2 O + A B 0.0100mol 0.0115mol 0 Δ 0.0100mol 0.0100mol +0.0100mol A 0 0.0015mol +0.0100mol Totl mol 2.00mol 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 12.98 0.01000L 0.00575L pk logk b 7. 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 9.255 log 9.26 0.0M b 5 log1.810 4.745 pk 14 pk 14 4.745 9.255 Wht is the ph fter dding ml of 2.00M NOH Since this hs the components of buffer solution, we cn use HendersonHssleblch + NOH + NH 4 NH B 0.00200mol 0.00mol 0.00mol 0.0mol molsof NH ndnh4 0.1000L 0.00molsof ech 1L 2.00mol molsof NOH dded 0.00100L 0.00200molsof NOH 1L
Δ 0.00200mol 0.00200mol +0.00200mol A 0 0.028mol 0.02mol ph pk log 9.255 log 0.02mol 0.028mol 9.1 Wht is the ph fter dding 2.00mL of M HCl? + HCl + NH NH 4 B 0.00200mol 0.00mol 0.00mol Δ 0.00200mol 0.00200mol +0.00200mol A 0 0.028mol 0.02mol molsof HCl mol dded 0.00200L 0.00200molsof HCl 1L ph pk log 9.255 log 0.028mol 0.02mol 9.20 8. 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 2.2M HBr. 0.900mol CH COOH.06000L CH COOH 1L CH COOH 0 0.0540mol CH COOH 2.2mol HBr 14 0.01011L HBr 0.0225mol 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 0.0225mol 0.0660mol 0.0540mol ph pk log 4.74 log Δ 0.0225mol 0.0225mol +0.0225mol A 0 0.045mol 0.0765mol 0.045 0.0765 4.49 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 2.2 mol HBr 0 0.0296L 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 0.1200mol 0.1200mol CH COOH 1.4M (0.06000L 0.0296L) CH COOH + H 2 O H O + + CH COO I 1.4M 0 0 (1.810 5 )(1.4) 4.910
C + + E 1.4M ()() 1.4 5 1.810 ph log(4.910 ) 2.1 9. If you strt over with fresh buffer (sme conditions/volumes s originl), wht would be the ph of the solution fter dding 20.00mL of 2.2M KOH? 0.900mol CH COOH.06000L CH COOH 1L CH COOH 0 2.2mol NOH 0.02000L 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 0.0214mol 0.0986mol log 4.74 log 0.0986 0.0214 5.40 After dding 0.00mL of 2.2 KOH? 2.2mol 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 0.1200mol 0.0009mol NOH 0.01M NOH (0.06000L 0.0000L) poh log[oh ] log( 0.01) 2.0 ph14.0 2.0 12.0 10. If you wnted to mke buffer with ph of.90 using the sme buffer s before with totl concentrtion of 2.00M, wht would the concentrtion of ech prt of the buffer need to be? ph pk log.90 4.74 log (2.0 ) 0.145 (2.0 ) 0.29 0.15 0.84 log (2.0 ) 10 0.84 10 log (2.0) 0.29 1.15 0.25M 2.00M 0.25M 1.75M 1.15 1.15
11. *If 0.00056 moles of AgNO is dded to L of 0.0020M 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)(0.0020) = 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 ] =.210 7 M 12. Ni(OH) 2 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 2+ or OH HCl is dded Increse solubility. H + rects with OH. 1. Assume you hve sturted solution of nickel (II) phosphte with some solid in the continer (K sp = 4.74 10 2 ). Wht is it s molr solubility? Determine the concentrtion of the Ni 2+ nd PO 4 ions in the solution. Ni (PO 4 ) 2 Ni 2+ + 2PO 4 K sp = [Ni 2+ ] [PO 4 ] 2 4.7410 2 = () (2) 2 =(27 )(4 2 ) I 0 0 C + +2 4.7410 2 = 108 5 = 2.110 7 = molr solubility = E 2 2.110 7 M [Ni 2+ ] = ()(2.110 7 ) = 6.910 7 M [PO 4 ] = (2)(2.110 7 ) = 4.2610 7 M 14. Wht would be its molr solubility be of nickel (II) phosphte if 0.100moles of N 2 PO 4 ws dded to L of the solution bove? Wht would be the concentrtions of the Ni 2+ nd PO 4 ions? Ni (PO 4 ) 2 + H 2 O Ni 2+ + 2PO 4 (27 )(0.0100) I 0 0.100 K sp = [Ni 2+ ] [PO 4 ] 2 4.7410 2 = () (0.100) 2 =
C + +2 4.7410 2 = 0.270 = 5.6010 11 E 0.100+2 = molr solubility = 5.6010 11 M [Ni 2+ ] = ()(5.6010 11 ) = 1.6810 10 M [PO 4 ] = 0.100+(2)( 5.6010 11 ) 0.100M 15. How much (in g) brium sulfte (BSO 4, MW=2.9) will dissolve in 500. ml of wter. For BSO 4 : K sp = 1.1 10 10 B 2 SO 4 + H 2 O B 2+ 2 + SO 4 K sp = [B 2+ ][SO 2 4 ] 1.110 10 = 2 I 0 0 C + = 1.0410 5 = molr solubility = 1.0410 5 M E 5 1.0410 molbso4 2.9g BSO4 0.500L H2O 1.210 gbso4 1L H O 1molBSO 2 4