CHAPTER 11. Practice Questions (a) OH (b) I. (h) NH 3 CH 3 CO 2 (j) C 6 H 5 O - (k) (CH 3 ) 3 N conjugate pair

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1 CAPTER 11 Practice Questios 11.1 (a) O (b) I (c) NO 2 (d) 2 PO 4 (e) 2 PO 4 (f) 3 PO 4 (g) SO 4 (h) N 3 (i) C 3 CO 2 (j) C 6 5 O - (k) (C 3 ) 3 N 11.3 cojugate pair PO 3 4 (aq) C 3 COO(aq) PO 2 4 (aq) C 3 COO (aq) base acid acid base cojugate pair 11.5 p = log[ 3 O ] = log[ ] = 3.44 po = p = = The solutio is acidic, sice the p is less tha (a) [ 3 O ] = = M (b) (c) (d) [O ] = / M = M The solutio is acidic. [ 3 O ] = = M [O ] = / M = M The solutio is acidic. [ 3 O ] = = M [O ] = / M = M The solutio is basic. [ 3 O ] = = M [O ] = / M = M The solutio is acidic. (e) [ 3 O ] = = M [O ] = / M = M The solutio is basic [ 3 O ] = = M

2 11.11 The stroger acid has the lower pk a, therefore X is the stroger acid. The K a values are: p a For X: K a 10 K = = For Y: K a 10 K p a = = x = M = [ 3 O ] p = log[ 3 O ] = log( ) = x = M = [ 3 O ] p = log[ 3 O ] = log( ) = x = M = [O ] po = log[o ] = log( ) = 5.93 p = po = Sice K b (CN ) is larger tha K a (N 4 ), the N 4 CN solutio is basic (a) Br is the stroger acid sice biary acid stregth icreases across a period. (b) 2 Te is the stroger acid sice biary acid stregth icreases dow a group. (c) C 3 S sice acid stregth icreases dow a group I all cases, the acid with the greater umber of O atoms is the stroger. (a) IO 4 (b) 2 TeO 4 (c) 3 AsO Use the edersoasselbalch equatio: p = pk a log [A ] [A] = 9.26 log = p chage is ( ) = 0.09 p uits O, N 3 ad P 3 all have at least oe loe pair o the cetral atom ad are therefore Lewis bases. ardess icreases across a period ad decreases dow a group so the order of icreasig hardess is P 3 < N 3 < 2 O. Review Questios 11.1 A Brøsted-Lowry acid is a proto door ad a Brøsted-Lowry base is a proto acceptor SO 4 is ot the cojugate acid of SO 4 2 because 2 SO 4 has two more hydroge ios tha SO 4 2.

3 11.5 (a) C 3 C 3 C 2 N (b) N (c) O N 11.7 I pure water, the cocetratio of 3 O ad O are equal ad their product is K w. As K w = at 25 o C, the maximum possible cocetratio of 3 O at this temperature is (the square root of K w ) A strog acid is oe that reacts completely with water to give quatitative formatio of 3 O N 2 2 O N 3 O As N 2 is such a strog base, this reactio proceeds to completio. The N 2 io accepts a proto from water ad gives quatitative formatio of N 3. As N 3 is a very weak acid, the reverse reactio does ot occur (a) NO 2 2 O 3 O NO 2 K = a [3O ][NO 2 ] [NO ] 2 (b) 3 PO 4 2 O 3 O 2 PO 4 K = a [3O ][ 2PO 4 ] [ PO ] 3 4 (c) AsO O 3 O AsO 4 3

4 [ O ][AsO ] a 2- [AsO 4 ] K = (d) (C 3 ) 3 N 2 O 3 O (C 3 ) 3 N [ O ][(C ) N] K = [(C ) N ] a (a) (C 3 ) 3 N 2 O (C 3 ) 3 N O K = b [(C 3) 3N ][O ] [(C ) N] 3 3 (b) AsO O AsO 4 2 O [AsO ][O ] K = [AsO ] 2-4 b 3-4 (c) NO 2 2 O NO 2 O K = b [NO 2][O ] [NO ] 2 (d) (C 3 ) 2 N O (C 3 ) 2 N 2 3 O K = b [(C 3) 2N2 3 ][O ] [(C ) N ] The sodium salt of acetylsalicylic acid cotais the cojugate base of this acid, the acetylsalicylate io. Beig the cojugate base of a weak acid, this io has appreciable basic properties ad a solutio of the sodium salt is therefore basic Sice ammoium io is the cojugate acid of a weak base (N 3 ), it is appreciably acidic. The aio i the fertiliser, NO 3, is the cojugate base of a strog acid ad is therefore, eutral. Cosequetly, addig ammoium itrate lowers the p of soil If [A] iitial 400 K a, the iitial cocetratio of the acid is equal to the equilibrium cocetratio. The same criterio applies to bases Acid stregth icreases across a period ad dow a group I itric acid, there are more oxyge atoms boud to the itroge atom tha i itrous acid. As the umber of oxyge atoms icreases, the pull o the electros i the O bod icreases, withdrawig electros away from the hydroge atom ad icreasig the polarity of the O bod which makes it easier to lose a proto.

5 11.27 NO 2 is a stroger base tha SO 3 2 because its cojugate acid, NO 2, is a weaker acid tha the cojugate acid of SO 3 2, SO 3. SO 3 is a stroger acid tha NO 2 sice it cotais a greater umber of loe oxyge atoms The equilibrium lies to the left, i the directio which favours formatio of the weaker acid ad base; so the strogest base reacts with the strogest acid This equilibrium lies almost completely to the left, sice NO 3 is a strog acid ad NO 3 is a weak base I water, formic acid ad acetic acid have slightly differet dissociatio costats. ece, it is possible to differetiate their acid stregths. I liquid N 3 however, formic acid ad acetic acid are fully dissociated ad appear to be the same stregth because N 3 is a much stroger base tha 2 O ad ca more readily accept a proto Buffer 1 is made with a excess of N 3 ad is therefore, better able to resist chages i p o additio of acid, as this reacts with the N 3 i the buffer. Buffer 2 is made with a excess of N 4 ad is therefore, better able to resist chages i p o additio of base. Cosequetly, buffer 1 maitais a steady p o additio of a strog acid At the equivalece poit, the titrated hydrazie solutio is acidic ad is a solutio of hydraziium chloride. The hydraziium io, beig the cojugate acid of a weak base, has appreciable acidic properties A acid-base idicator is a weak acid that chages colour whe it coverts from its acidic form to its basic form. Because idicators are weak acids, they react with the titrat ad cosequetly, small amouts of idicator are used so the results are ot affected excessively by iteractio of the titrat with the idicator Potassium hydroxide is a strog base ad hydrobromic acid is a strog acid. Cosequetly, the p at the equivalece poit is 7. Ay idicator that chages colour aroud p 7 is appropriate. Pheolphthalei is also a good idicator sice the p chages rapidly ear the equivalece poit i titratios of a strog acid with a strog base. Oly a small volume of added acid, frequetly less tha oe drop, causes a large chage i p. I additio, sice the colour chage for pheolphthalei is from colourless to pik, it is easily observed A Lewis acid is a electro pair acceptor. A Lewis base is a electro pair door C N F B F C N F B F F F

6 11.47 (a) (b) (c) 11.49

7 11.51 (a) N 3 < N 3 < NF 3 (NF 3 is hardest) (b) O - 2 < O - 3 < O - 4 (O - 4 is hardest) (c) Pb 2 < Pb 4 < Z 2 (Z 2 is hardest) (d) Sb 3 < P 3, PF 3 (PF 3 is hardest) Review Problems (a) F (b) N 2 5 (c) C 5 5 N (d) O 2 (e) 2 CrO (a) cojugate pair NO 3 N 2 4 N 2 5 NO 3 acid base acid base cojugate pair (b) cojugate pair N 2 5 N 3 N 4 N 2 4 acid base acid base cojugate pair (c) cojugate pair 2 PO 4 2 CO 3 CO 3 2 PO 4 acid base acid base cojugate pair

8 (d) cojugate pair IO 3 C 2 O 4 2 C 2 O 4 IO 3 acid base acid base cojugate pair At 25 C, K w = = [ 3 O ] [O ] (a) [ 3 O ] = ( ) / (0.0024) = M (b) [ 3 O ] = ( ) / ( ) = M (c) [ 3 O ] = ( ) / ( ) = M (d) [ 3 O ] = ( ) / ( ) = M p = log[ 3 O ] [ 3 O ] = M p = [ 3 O ] = M p = 9.15 [ 3 O ] = M p = 5.74 [ 3 O ] = M p = (a) [O ] = 10 po = = M p = po = = 5.74 [ 3 O ] = 10 p = = M (b) (c) (d) (e) [O ] = 10 po = = M p = po = = 3.75 [ 3 O ] = 10 p = = M [O ] = 10 po = = M p = po = = 9.35 [ 3 O ] = 10 p = = M [O ] = 10 po = = M p = po = = 7.82 [ 3 O ] = 10 p = = M [O ] = 10 po = = M p = po = = 4.30 [ 3 O ] = 10 p = = M m 6.0 g mol M g mol NaO mol c M V 1 L As NaO is a strog base, it dissociates completely. Thus [O ] = M.

9 po = log[o ] = log(0.15) = 0.82 p = po = = [ 3 O ] = 10 p = = M V mol L = 16.9 ml -1 c mol L At 25 C, K a K b = K a = K w /K b = / = K a K b = K a = K w /K b = = (a) The cojugate base is IO 3 pk b = pk a = = K b = 10 pk b = = (b) IO 3 is a weaker base tha acetate aio, because its K b is smaller tha that of acetate aio. Coversely, IO 3 is a stroger acid tha C 3 COO, so its cojugate base is weaker tha that of C 3 COO (0.038)(0.038) K a = = pk a = log(k a ) = log( ) = K b = = pk b = log(k b ) = log( ) = 6.96 [x][x] 5 3 K a = = x [3O ] 0.15 p p = po = = Iitial cocetratio of N 3 is 0.15 M. 500 ml = 0.5 L of solutio, therefore mol L L = mol. N 3 cv xx 11 K b = = x Sice (0.40 x) 0.40: x = M = [O ] = [NO 2 ]

10 po = log[o ] = log( ) = 5.63 p = po = = xx 9 5 K a = 0.10 = x [3O ] p K a K b = K w, K b = K w / K a = / = ` xx 7 4 K b = x [O ] 0.67 po 3.33 p p = log ( ) = p = log ( ) = (a) Br: Br larger tha therefore Br bod is loger ad weaker. (b) F: more electroegative F polarises ad weakes the F bod. (c) Br: Br larger tha S therefore Br bod is loger ad weaker (a) O 2, because it has more loe oxyge atoms. (b) 2 SeO 4, because it has more loe oxyge atoms (a) O 3, because is more electroegative, (b) O 3, because it has more loe oxyge atoms ad is more electroegative tha I. (c) BrO 4, because it has more loe oxyge atoms The equilibrium reactio is: N 3 2 O N 4 O Usig the edersoasselbalch equatio: p = pk a log [A ] [A] = 9.26 log = [ C 3 COO] = [C 3 COO] fial [C 3 COO] iitial = M 0.15 M = 0.07 M [ C 3 COO ] = [C 3 COO ] fial [C 3 COO ] iitial = 0.21 M 0.25 M = 0.04 M

11 m M 82.0 g mol mol = 22 g NaOOCC m M g mol mol = 0.96 g N Iitial p p = pk a log A A = 4.74 log = 4.78 Fial p A p = pk a log A = 4.74 log = 4.76 If the same volume of was added to 125 ml of pure water, rememberig that dissociates completely: mol c mol L 1 = [ 3 O ] V L p = log(0.0167) = 1.78 ece, the p chage is ( ) = 5.22 p uits which shows how a buffer solutio maitais a costat p Lactic acid, C 3 C(O)COO, is a mooprotic acid which reacts with NaO i a 1:1 mole ratio accordig to the equatio: C 3 C(O)COO O C 3 C(O)COO 2 O NaO cv mol L 1 ( L) = mol From the reactio stoichiometry, this must also be the umber of moles of lactic acid preset The % Na by mass i the origial mixture is: (mass Na/total mass)x100 = (0.72 g / g) 100 = 58 % Na p = po = = p = (a) p = log [ 3 O ] = log ( ) = (b) p = pk a log [A ] [A] = 4.74 log

12 (c) log [A ] = 0 ad p = pk a = 4.74 [A] (d) p = po = = N 2 is the Lewis base ad 3 O is the Lewis acid. _ N O N O Each Al atom acts as a Lewis acid, acceptig a electro pair from. Al Al Al Al (a) Lewis acid is Al 3 : Lewis base is - (b) Lewis acid is Z 2 : Lewis base is CN - (c) Lewis acid is I 2 : Lewis base is I - (d) Lewis acid is CO 2 : Lewis base is O The order of icreasig polarisability is : V 3 < Fe 3 < Fe 2 < Pb 2. Polarisability icreases as ioic size icreases The order of icreasig softess is : (a) BF 3 < B 3 < Al 3 (b) Al 3 < Tl 3 < Tl (c) Al 3 < AlBr 3 < AlI 3 Softess icreases dow groups Both SO 2 ad SO 3 have a loe pair of electros o S but SO 3 is less polarisable tha SO 2 because this loe pair is closer to S i SO 3 tha i SO 2, due to the greater electroegativity effect of 3 oxyges compared to 2 oxyges.

13 Additioal Exercises Cojugate acid: (C 3 ) 2 N 2 ad cojugate base: (C 3 ) 2 N p = log( ) = p = pk a log A = 9.26 log A Sice the ammoium io is the salt of the weak base N 3, it is acidic. The cyaide io is the salt of the weak acid, CN, so it is basic. I order to determie if the solutio is acidic or basic, the relative stregths of these two compoets eed to be determied: K a (N 4 ) = K w / K b (N 3 ) = / = K b (CN ) = K w / K a (CN) = / = Sice K b (CN ) is larger tha K a (N 4 ), CN acts as a base to a larger extet tha N 4 acts as a acid. Therefore the N 4 CN solutio is basic.

14 This is a example of a titratio of a strog acid with a strog base. Prior to reachig the equivalece poit, there is a excess of acid preset. The umber of moles of acid preset is determied by subtractig the umber of moles of base added from the amout of acid iitially preset. The ew cocetratio of acid is determied takig the volume chage ito accout to calculate the p. After the equivalece poit, there is a excess of base. The excess is determied by subtractig the iitial umber of moles of acid preset from the umber of moles of base added. The resultig cocetratio of base is determied to calculate the p. The umber of moles of 3 O iitially preset are: cv mol L 1 ( L) = mol 3O (a) Iitially, there is oly M preset: [ 3 O ] = M, p = log[ 3 O ] = (b) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added 3O mol mol = mol mol [ 3 O ] = M L p = log[ 3 O ] = 1.37 (c) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added 3O mol mol = mol mol [ 3 O ] = M L p = log[ 3 O ] = 3.70 (d) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added 3O mol mol = mol mol [ 3 O ] = M L p = log[ 3 O ] = 4.70 (e) After ml of added base, the solutio volume is ml The solutio ow cotais equal umbers of moles of acid ad base, so it is eutral. [ 3 O ] = [O ] = M

15 p = log( ) = 7.00 (f) After ml of added base, the solutio volume is ml From here o, the solutios cotai excess base ad the ay excess over x 10 3 mol, the amout of base required to reach the equivalece poit, is the total hydroxide io cocetratio i the solutio. cv mol L 1 ( L) = mol O added mol x 10 3 mol = mol O excess [O ] = V mol L po = log[o ] = 4.70 p = po = M (g) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added mol x 10 3 mol = mol O excess [O ] = V mol L po = log[o ] = 3.70 p = po = M (h) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added mol x 10 3 mol = mol O excess [O ] = V mol L po = log[o ] = 2.71 p = po = M (i) After ml of added base, the solutio volume is ml cv mol L 1 ( L) = mol O added mol x 10 3 mol = mol O excess [O ] = V mol L M

16 po = log[o ] = 1.48 p = po = The titratio graph is o the ext page: p ml NaO added (a) Al 3 LiC 3 Li [AlC 3 3 ] - (b) SO 3 2 O 2 SO 4 (c) SbF 5 LiF Li [SbF 6 ] - (d) SF 4 As 5 As 3 SF 4 2

What Is Required? You need to find the ph at a certain point in the titration using hypobromous acid and potassium hydroxide solutions.

What Is Required? You need to find the ph at a certain point in the titration using hypobromous acid and potassium hydroxide solutions. 104. A chemist titrated 35.00 ml of a 0.150 mol/l solutio of hypobromous acid, HBrO(aq), K a = 2.8 10 9. Calculate the resultig ph after the additio of 15.00 ml of 0.1000 mol/l sodium hydroxide solutio,