PS4_Supp_Prob_Solutions - CHAPTER 7 ACIDS AND BASES H20 The...

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Unformatted text preview: CHAPTER 7 ACIDS AND BASES H20; The conjugate bases of strongecids are terrible bases (Ki, < I04“). 0C1“; ’t‘he conjugate bases of weak acids are weak bases (20“ < Kb < i); even though they are designated as weak bases, the conjugate bases of weak acids are as better bases than H20. N113; For a conjugate acid~base pair, K, X K, = Kw. From this reiationship, the stronger the acid the weaker the conjugate base (Kt, decreases as Ka increases). Since iiCzHZCIO2 is a stronger acid than NIL” (K.a for HC‘ZJEZRCKL)2 > K, for NHJ), then NH3 wilt be a stronger base than (121-126102‘ . 320 and CH3C02“ b. An acid-base reaction can be thought of as a competition bemeezx two opposing bases. Since this equilibritzm lies far to the iefi (K, < I), then CHSCO; is a stronger base than H20. c. The acetate ion is a better base than water and produces basic soiutions in water. When we put acetate ion into soletion as the only major basic species, the reaction is: c3300; + H20 e cmcozn + OH‘ Now the competition is between OHSCOE“ and OH“ for the proton. Hydroxide ion is the strongest base possibie in water. The above equilibn'um ties far to the Refit resulting in a Kb value iess than one. Those species we specificaiiy call weak. bases (10“1‘i < Kb < 1) tie between H20 and OH‘ in base strength. Weak bases are stronger bases than water but are weaker bases than 02?. 26. Zhe NH: ion is a weak acid because it Iies between Hg) and H30" (93") in terms of acid strength. Weak acids are better acids than water, thcs their aqua-cits soiations are acidic. They are weak acids because they are not as strong as 1330“ (I?) Weak acids only pmtially dissociate in water and have - K1, vetoes between 10”“ and 1. _ 27. In deciding whether a'substanc‘e is an acid or a base, strong or week, you should keep in mind a couple ideas: _ ' ' ' - 1. There are oniy a few common strong acids and strong bases alt of which shouid be memorized. Common strong acids 3 EC}, H213 Hi, W03, HCIO4 and E12304. Common strong bases = LiOH, NaOH, XOR, RbOH, CSOH, Ca(OH)2, SrfOH)2 and 32401;)? 2. Ail other acids and bases are weak and wiil have K3 and K1, vaiues less than one but greater than K.w (I914). Reference Tabie 7.2 for K“ values for some weak acids and ’Z‘able 7.3 for Ki, vaiues for some weak bases. There are too many weak acids and weak bases to memorize them ail. -'t‘herefore, use the tables of K3 and Kb values to hetp you identify weak acids and weak bases. Appendix 5 contains more complete tables of K13 and Kg voices. a. weak acid (K, m 4.8 X I84) b. stnong acid c. week base (Ka W 4.38 X 10“) - . d. strong base ,e. weak base (Kgm 1.8 X 10'5) f. weak acid (K11 m ”:12 >< 10") g, weak acid (K, m 1.8 >< 10“) h. stroog base i. strong acid CHAPTER 7 ACIDS AND BASES :69 .Sointions of Acids 31. Strong acids are assumed to compietely dissociate in water, c.g., HCKaq) + H300) ~+ H;O"(aq) 4* =} CNN) 0? H‘Cltiati) “* WM) + Citaq). . _- a. A 0.10 MEG! sointion gives 0. i0 Mil“+ and {3.10 MCI“ since HCi completeiy dissociates. The amount of I"? from H30 wiii be insignificant. 25-- pH m «log [If] = Jog (0.10) = 1.00 b. 5.0 MB? is produced when 5.0 MHCK).$ compietely dissociates. The amount of I? from H20 wiil be insignificant. pH = -iog {5.8) m 43.70 (Negative pH voices just inéicate vety oonoentrateé acid sointions.) o. 1.0 X 10‘“ M if is produced when 1.0 X 10“” M H1 compieteiy dissociates. Ifyon take the negative tog of 1.8 X 10*“ this gives pH = 11.00. This is impossibie! We dissolved an acid in water anti got a basic pH, What we must consider in this probiem is that water by itself donates 1.0 X 10‘? M if. We can normaliy ignore the smali amount of I? item 320 except when we have a very dilute solution of an acid (as in the case here). Thercfore, the pH is that ofneutrai ‘ water (pH w 7.08) since the amount of Hi present is insignificant. 1.19g x 38gHCI x ZmolHCi 100 g con. HCE soln 36.5 g 3.42 g x 10. g BN03 X imoi HNO3 m1. 200 g 5012'} 63.0 g ENG3 32. 50.0 in}; con. H62 5011: X = 0.62 moi HCI 20.0w}. con. HNOg 50111 x m 9.32 moi I~ZNO3 HCKaq) “t ifiaq) ~i~ Ci‘(aq)l and HNOJaq) “3 Hing) + NOflaq) (Both are strong acids.) So we wili have 0.62 ~1- _0.32 W 0.94 mol of H“ in the fins} solution. [‘H’} m ”£930“? e 0.94 M; pi: = ~iog {Ht} n —log (0194) m 9.92”: mess K M- OH =$=MM11><1014M {H 9] 0.94 33. a. Major species = ifiaq), Cl’(aq) and H206) (HCi is a strong aciti.) [W] m 8.258 M pH = Jog [If] m ~iog(0.250) W 0.602- b. I-flaq), Bt‘(eq) snot-£206) (381' is a strong acid.) pH = 0.602 c. H’Taq), (320.1313) and 1120(1) (RICK). is a strong acid.) pH = 0.602 :1. Meg), N03'(aq) and Egon) (W03 is a strong aeiti.) pH = 0.502 1 _________________________ __ ______________________ _____________________________________ 2-9: CHAPTER ? ACIDS AND BASES ' 173 The assumption x << 0.020 is not good (it is more that} 5% of 0.020). We must solve xzx’(0.020 - . x) m 7.2 X It)“‘ exactly by using either the quadratic formats or by the method of successive approximations (see Appendix 1.4 of text). Using scecessive approximations, we iet 0.016 M be a new approximation for {HF}. That is, in the denominator try 2cm 0. 0038 (the value ofx We ealcuiated making the normal assumption), so 0. 020 0. 0038* 0. 016, then some foraeew valee of x in the numerator x2 x2 - m =7.2x10~4,xr~3.4><10‘3 0.920 —x 0.016 We use this new value ofx to timber refine our estimate of {HF}, i.e., 0.020 -« x 2 0.020 ~ 0.0034 = 0.0266 (carrying an extra significant figure). x2 H x2 9.020 -x H 0.0366 We repeat, untii we get a seif-consisteot answer. This wouid he the same aostver we woeié get solving exactly using the quadratic equation. In this case it is: x = 3.5 X It?3 So: {11*} m {s} =3: e 3.5 x103 M; {03:} m Kw/[i—r] m2.9 x 2042M {HF} = 9.929 ~ 3: m 9.020 - 9.9035 x 9.917 M; pH = 2.49 . = 7.2 x 10*, x e 3.5 x 193 Note: When the 5% assemption faiis, use whichever method you are most comfortable with to soive exactiy. The method (if successive approximations is probably fastest when the percent em):- is Iess than ~25% (unless you have a graphing calcuiator). 36. Major species: HCZHZCIOZ (K.at a 1.35 x 103) 3116 H20; Major source of If“: HCIHZCIO2 HCIKQCiOZ we 193+ (slogan; . Iaitiai 0.10 M ~0 0 .1: mol/L HCZHZCZC); dissociates to reach equilibrium Change ~x‘ _ w} +34 or Equil. 0.10 - x .. . x I x 2 2 K,mi.35x104m x ~fimflx i.-2><19M - 0.10 w x 0 10 Checking the assumptions fines that x is 12% of 0.10 which fails the 5% mic. We must soive 1.35 X 10'3 = x2!(0.I0 » x) exactly using either the method of successive approximations or the ' quadratic equation. Using either method gives x = {HF} = 1.1 x 10‘1 M. pH = «log [11*] = ~iog (1.1 >< 10*)“;196. 3?. Major species: 310;, H20; Major source of H‘: Hi03 (a weak acid, Ka W 0.17) Hit}; W Ht + to; Ioitiai 0.010 M ~0 ' 0 x moi/L H103 dissociates to reach eqniiibrium Change 4x ms +3; +3: Equil. 0.010 - x x x W4 " CHAPTER 7 ACIDS AND BASES H+ ION I 2 '2 i N 3}“ x x .yxwg‘gé}; Chockassumption. K, m 0.17 = . s m {3103; 0.010 s x ooze Assumption is horrible, (x is more than 400% of 0.010). When tho assmnption is this poor, it is generaily quickest to soive exactly using the quadratic fonnuia (see Appendix 1.4 in text). Using the quadmtic formula anti canying extra significant figures: ' 2 9.17 w w, xzmmmomd), xz+9.17x— 1.7 x who _ 0.010 ~ 3: - - __-. "~0.17i[(0.17)2 « 4(2)(-1.7 x 10-51”; ~0.17¢0.189 2(1) _ . 2 x“ ,x*9.5X10”3M (3: must be positive) ' x = 9.5 X 10'3 M = [11"]; pHm ~iog(9‘5 X I0“) m 2132 38. HCalbiflO2 (K. m .1 .3 X 10") and Hg) (K, =“~ Kw = 2.0 x 10”“) are the major species present. HCgi-Ist wiil be the dominant producer of I?” since HCJISO2 is a stronger acid than H20. Soiving the weak aoié probiem; £63350? m ‘ if + (33171502. Initial} 0.100 M NO 0 1: 1:2qu 30311502 dissociates to reach equiiihrium Change “x -~> +3: +3: 39133. 0.109 ~ 3; x _ x Kn” 1.3 X I9‘3 3-; [3 +3 {C3H502w} W .x 2 I x2 [HC3H5023 ”0.200 A x x 0.200 x = [H‘] = 1.1 X103 M; pH n Jog (1.1 .9104): 2.96 Assumption {oiiow's the 5% mic (x is I} . }% of 0.100). [IT] = {CaflsOfl m 1.1 x303 M; [02?] w lefifl = 9.]. X 10'i2 M [HC3Hst1= 0.I00 ~ 1.3 X 10'3 = 0.099 M . + ,3 . Percozzt dissociation“ [H ] x100 ~ 1.1xi0 >< 100:1.1‘Vo {HC3H50230 moo 39. This is' a weak acié in water. We must soive a weak acid problem. Let 2:332: w Cal-LCOIH. 2 11101332 122.; g 0.56 g HBz x _m 4.6 X £03 moi; [8sz = 4.6‘ K IG'3 M CHAPTER 7 ACIDS AND BASES - T119, moza: soiubiiity of (351150021; is 4.2 x 1112 moUL. . 4.2 x 20'? mo] Céflscozs 122.1 g {26.1-15(30211 x - >< 8.100 E. m 0.5} g per 108.1131. soiution ' L moi CEHSCOZH . _ . . . 3.05 g 1 11101 M . 20.0 mi, glacial scene acid X X ~_ — 0.350 moi HCZH302 - mL 60.05 g Initial concentration of 11621130. - 5.1%.?” mol— = 1.40 M - O. 2500 L HCQH302 sw-‘w 11* 1» {221130; -K,, m 2.3 x 10-? _ Initial 1.40 M ~O 0 x mom; 1113221302 dissociates to reach equilibrium , Change «J: _> +x +x Bqnii. 1.49 - x x x {H f] [{TZHGOZM} m x 2 .. x 2 Kyisx 10.5.. [HC2H302] 1.40 — x '” 1.:10 x a {PP} 3. 5.8 K 30”3 .M; pH m 2.30 Assumptioas good (36 is 0.36% of 1.40). HF'and HOCGHL. are both weak acids with IQ vaiues of 7.2 x 10“ and 1. .6 x 10"“, respectiveiy. Since the K, value for HF is much greater than the Ka vaiuc for HOcfifls, then HF wiii be the dominant [producer of H“ (we can ignore the amount of 3* produced from E0C6H5 since it wili be imigaificant). HF # Hf + '14“ .44 1:111:21 8.10M ~11 _ 0 ' , x moifL HF dissociates to reach equilibrium Change —x _ w; +9: - +x Equii. 6.10 —x , _ x ‘ x K.m7.2><10~‘— {H‘EiF‘}: ”‘2 z "’62 {THE} 1.0 w): 1.0 = = .7 X 10' M; pH: ~ 0g {2.7 X 10“ ) == .57 Assumptions goo . 2 _ 2 1 2 1 d ' Sofving for {0061153 using Hogs. .1. 1-? + 00.11; equilibrium: ‘ - 11" OCT 11” 2.7 x '10“2 06 11* Ka=I.6XI9'm“[ 3i 6 SEQ N 6 5}.{ocsfls‘}w5.9x1o~9M 45. $005.53} 10 Note that this answer indicates that oniy S .9 X I8 9 M HOCsEIS dissociates which indicates that HZ? 15 truiy the only s1gnif cant producer of H‘ In this soiution. a. HCI' :3 a strong acid. it will produce 0.10 M If. HOCE is a weak acid. Let's consider the eqtzilihriilm: CENTER 7 ACIDS AND BASES . 177 H00} "05“» 21* + 001* K, m 3.5 x 10-8 Initiai 0.10M 0.10M . 0 x moi/L HOOK dissociates to reach equiiibrium _ Change ~x w ..... 3. +3: +3: Equil. 0.10~x 0.I0+x x Kw 3.5 >< 10* w {H '7 {0C} -1 — (3'10 ”7) (x) x 96,): m 3.5 x1000»: {H082} 0.30 ~ x Assumptions are great (x is 3.5 >< 105% of 0.10). We are seaily assuming that HC£ is {he only important source of H: which it is. The {5.11 contribution from HOCI, x, is negligible. Therefore, [H13 0.10 M'; pH = 1.00. 2). HNO; is a strong acid, giving an initial concemration of H" equal to 0.050 M Cénsiéer the equiiibrium: 510201302 «vs 21* _ + 0.330; {(31.8 x105 Initiai 0.50M 9.05001 0 3: may}; HCZH302 dissociates to reach eguilihriuzzz Change «x __> +x +x Equil. 0.50 ~ x 0.050 + x x (H? {Cg-H3027? .. (0.050 +505: 5 00505; {11620002} (0.50 — x.) 0.50 x ’5: M? X 10“; Assumptions are good (W811 witifizx the 5% mile). ' [Hi = 0.050 +x = 0.050 Mazzd pH = 1.30 Kamifix 10% 444 Both are strong acids. 0.0500 L x 0.050 2250125., = 2.5 Sc 103 mo! HCI m 2.5 x 103 mol 51* + 2.5 >< 103 incl at 0.1500 1;. x 0.30 000%” 1.5 x 10-2 100132003 m 1.5 x 10-2 231018.“le x 102 memo; W3 -2 K . EH?"- (2.5X10 + ZSXIO )mOI 30038.0(; {OH-I” w m11x10-2304 02000 L {H s; . “3 . -2 [02.1” 2.5 x 20 m0! “0.9131”; {NOJW 1.5 x10 mol new”; 0.2000 L ' . 0.2000 L H? W H‘ + F Znitiai _ 0300 M «0 0 x moUL E? dissociates to reach eqniiibfium Change J: ' ms . +x ' ‘ +x 0:12:31. 0100 «x' x 'x I78 . -- ' CHAP’I‘ER '7 ACIDS AND BASES {H*][F'] 3 x2 2 mm = -=a.03: x 0.109 $8.1X10‘3M K, [HF] 0.109»); [H‘] {F} ( M) ---3 2 [HF] m 0.190 — 8.1x :04 m 0.092 M; Kan W m ?.1><10“ - 0.092 45. ' m: was - E? + X‘ Initiai 3: ~0 o where I = {RX}, )1. mom, HX dissociates to reach equiiibrium Change —x w; +0: +x Equik. I «x 'x 3: From file problem, as m {125(1) and _I ~ 3: m 0.30 M. 2« 0.253) = 9.30 M,I= 9.40 Mantis = 025(049114): 0.18M K!“ {H‘fiX'} : x2 _ (0.10)2 {BX} {us 030 = {3.033 47. In ail parts of this probiem, acetic acid (HC2H302) is the has: weak acid present. We must solve a weak acid problem. ' a. 11023.02 . ”w“ 3* + 0.2130; Iaitiai 0.50 M ' ~‘0 0 x mom. H0234); dissociates to reach cquiiibrinm Change »x m) +x +x Equil. 0.59 ~ A: x x Kmlsxio’sw [H Reg-302} s 1‘2 - “j: [chHgog 0.50% .50 -x = [IF] = {C2H302‘} *—“ 3.9 x '10‘3 M Assfiznptions good. . -3 [H ] x100: 2.9.3519... x I00 =8.60% Percent dissociation = WWW [chHgozgo 0.58 b. The set-21f} for soiutions b and c are similar to soiuticn 3. except the 53231 equation is siightly éifi'erent, refiecfing the new concoct-aim: of HCEHSOZ. - 2 2 mLSXIO’smmi—mmm x K, 0.050 m x 0.050 x = {1?} m [C2H303‘] W 9.5 X 10" M Assumptions good. 9.5 x IOT‘1 0.950 % dissociation m x 100 wise/.5 £80 " CHAPTER 7 ACIDS AND BASES 49. pr 2.??, {PF} m 10‘“? m 2.? >< 20'3 M _ HOCN “Va—”3“- 1-? + OCN‘ initial 0,9100 w!) 0 Equii. (18100-3: x x x m {IPYJ m [OCN‘] m 1.7 X 10“ .M; [I-IOCN] — 0.0100 ~ 3: == 0.0108 - 0.801? = 0.8083 M gas or; was; g3 x10"3)2 {Hoes} 3.3 x 19-3 23.5 X 10" 50. I-ICIO,g is a stroug acid mm [H‘] = (L040 M. This squads the {H} in the {richlomacetic acié solution. . Set-up the problem using the K, equilibrium reaction for CCQCOZH. (361366281 ' # rs ' + ccuco; 1333an 8.058 M ~0 0 Equiic 8.050 ~ x ' x . x {H *1 [CCIECOQ'] 1c2 "- . . ‘ -- “ ; From the probiem, x m [21*] m 4.0 X 10‘2 M. iCCECQzfl] 0.050 w x (4.0 x 39:)? 0.050 w (4.0_>< EUR) #0116 S 1. Major species: HCOOH and H20; Meier source of if: KCOOH HCOOH # H’“ 4- HCOO‘ Initial C _ _ . -_ _ ~0 - 0 where C “—“* [HCOOHL x rrrolfL HCOOHdissociatcs to reach eqzr'iibrium ‘ Change «x ‘ w; +3 +3; Equiir C .~ :5 ' x x ' + w — 2 = 1.3 x104: ml}! 1 [H9091 : x whcrc'x= K“ {HGOOH} C » x [IF] . +- 2 1.8 x 10* m “331““; Since primzxm, then: [IF]=10'“° m 2.0 x :03 M C e {H *1 ' ---3 2 4; 1.8 x m“ m m, (142.0 x room 4'0 x 33136 =24 X10‘3M - C ~(2.0x 10—3) 2.8x10" - A 0.924 M {conic acid solution Win have 13H m 2.70. 52. When on acid dissociates, ions are proéuc'edr The conductivity of the solution is a measure of the number ofious. In audition the coiiigative properties, which will be discussed in Chapter 17, depend \ on the number of particles present. So measuremenis of osmotic pressure, vapor pressure lowering, 53. CI-iAP’i‘ER 7 ACIDS-AND BASES I 181 fi'eezing point éepression or inciting point eievation Wiii aiso attow as to detennine the extent of ionization. ' The reactions are: le'aAsO4 e H" + HZASO; K3; = 5 X 18‘3 HzAsO; He H" + HAsOf‘ 32 == 8 X 18'3 HMO? we I? + AsOf' Kan m 6 X 20“" We wit! doai with the reactions in order of importance, beginning with the iatgost Kn, Kay H+ ' A50" H3As04 w H’” + HEASO; K =5 x 10-h: W _ “1 {H3ASO4} Initiai 0.20M ' ”0. 8 Equii. 0.20 »x x _ x 2 2 5 >< 10'3‘m x - ~ “35”, x-- -3 x IO‘MM Assumption fails the 5% rnie 0.20 x 0 20 Solving by the method of successive approximations: 5 X 10‘3 m :c2 r’ (0.20 - 8.03), x m 3 X 10‘2 (consistent answer) [H’] m {HzAsO 3 m 3 x 10»2 M; {11325504} = 0.29 ~ 0.03 = 0.17 M {H nHAso: E ”3. .. . Since K32 “ WW “* 8 X 10 ts much smailer than the K value, then very itttie of [H A504} HZASO4 (and HASQE )dissociatcs as compared to H3A304. Therefore, [H ] and [HzAsQ1 } witl not change significantly by the K12 reaction Using the previonsly caioniated concentrations of H” and H1550; to calculate the concentration of HAsOf’: " 3 xit)”?'}-1As{}2" ' ‘ I 3 x1041: mellmmii, {HASOB‘}; s x WM 3 x 3072 Assumption that the Km2 reaction does not cbange {IF} and {HEASOII is good. We repeat the process using K33 to get [AsOf']. «6x101°_[H HMO: } _(3xzo”3)[Asoj‘] 1%. [nnso‘1 1 ' (snot!) {A5043 1*“ 1. 6 X 10 5 ~2 X 1015 M Assumption good So in 8.20 M anatyticai concentration of HgAsfifi: {HgASOJ m 0.17 M on} m {meson} _m 3 ><110*2 M; [HAsogj m s x 10* M; [A5043] = 2 x In” M; {OH-1m tome] x: 3 >< 10*” M CHAPTER 7' ACIDS AND BASES .. 187 MWmm—mmmmmmwmmwm 66. Major species: HZZNEIHEIHI2 (K,’ m 3.8 X 10"“) and H20 (Kb : KW : 1.0 X 20'”); ’Z‘he weak base Hfiw'il‘flfi2 . _ wili dominate 0H“ production. We must perform a weak base equilibrium caicuiation. £12m: + H20 W 2NNH; + OH- Kfiwxme Initz'ai ' 2.9 M ' 0 ~0 x mom. HZNNH2 reacts with H20 to reach equiiibrinm. Change —x fl.) +1: +3: Equii. 2.0 - x x x H NW1"?P 0H“ 3 2 - xb=3.0><1o6= { 2 i 3“ Jw x ~ x- (assomingx.<,<2.0) {HENNHZ} “’ 2.0 — x " 2.0 x = [OH] m 2.4 X 10'3 M; pOH-x 2.62; pH 2 11.38 Assumptions good (x is 0.12% of 2.0). 5.0x I0'3g X £11101 0.0100 1.. 299.4 g 67. = 1.? X {0‘3 M= {codeino]0; Lot cod 3 codeine, (SHERIZ‘K)3 Solving the weak base eqoiiibrium probiem: cod 4» H10 W cod}? + on K, m 10405 = 3.9 x 107? Initial 1.7 X 10'3 M ' . 0 4} x moHL codeine reacts with H20 to reach equilibrium Change -x _; +1: +3: Equil. If? X I94 ~-x x x 2 :2. ' K, m 3.9 x w” m x .. x , x w 3.9 x 10-5 Assumptions good. m x or3 or ' 2.? x10“3 {03“} m 3.9 X 16‘5 M; [H’] =“~ Kw/[OH'3 = 2.6 X 10"“3 M; pH at «iog [if] m 9.59 68. Codeine: CISHZINOS; .Coéeine suifate = C36H44NZOWS The formuia for codeine sulfate works out to (codeinc E5804} where codeine I"? *3 HClaflnNOf. Two coéeine moiccnles are protonated by H2804, forming the conjugate acié of codeine. The $0.5” then acts as the conflict ion to give a mantra! compound. Codeine sulfate is an ionic compound which is more solnhie in water than codeine, aiiowmg more of the (2ng into the bioodstream. 69. To soivc for percent ionization, just soivc the weak base equiiibrium problem. a. NH3 + 3820 W NH: + OH“ Kb W. 3.3 X 10‘5 Initial (MGM 0 «4) Equii. 0_.10~x x x 2 2 mum 30-5: m“ xi“, xm o3» =13 >< 106M; Assam tions ood. K” 0.10»; (no _[ 1 p fg v 'w- I —3 Percent ionizaiion = {05} x :00 - Mg X 100 m 1.3% {NHBJG {3.10M _ ______ § _______________________ g _________________________________________ ‘52? 188 ’ - - . CHAPTER 7 ACIDS AND BASES 6. NH. + 6130 W _. NH; + OH“ . Initiai 0.020 M 0 ~0 Equii. 0.010 ~ x _ x x 2 ’2 1.8 x 105 = x w x_ , xm {OI-1‘} = 4.2 X 10“ M; Assumptions good. 0.010 — x 0.010 _ - ~4 Percent ionization m M x 108 m 4.2% - 0.010 Note: For the same base, the percent ionization increases as the initini concentration of base decreases. ' c. _ " CH3N}i2 + 320 r—E CHENH; + OK“ Kym 4.38 x 19" Initial 8.10 M 0 ~0 Equil. 8.10 ~ 3: x x 2 2 4.36 x 10* = x 6 “3557...“, x m 6.6 x we; Assumption fails the 5% we (at is 6.6% Using successive approximations and carrying extra significant figures: 2 2 W m x = 4.38 X 20“, x = 6.4 X 10'3 (consistent answer) 0.18 - 0.0066 8.893 Percent ionization = 53%“ X I80 3 6.4% Iquninine 111101 quinine m . _3 . . _ u.“ . . _. 78. x _. £6 X I8 M umme, Let _ nznxne - C N O . 1.90001. 324.4 gqninine q Q q ”H“ 2 2 “3.6.. Q + H20 .6— Q}? _+ OH‘ Kfimiefimsx 10-5 Inifiai 1.6 x 10-3 M. ' ' 0- ~o x moi/Z. quinine reacts with H20 to reach eqniiibrinm Change ~x _ ' my +x +3: Eqnii. ' 1.6 X 10‘3 wt . x x Kmsx 10,6: [QHjioyl-g : x2 . x xi [Q1 1.6 x 29‘3 ~ :6 1.6 x 10—3 x = I X 10“; Assumption fails 5% mic (xis 6% of 0.0816). Using successive approximations: x2 W m 8 x 10”“, x = 1 X' 18“s M (consistent answer) 1.6x 10-3 ~1XZO’4 x: {03*} ——»~ 1 x 10‘4M; pt)H~'--4.0; pnm'z'oc CWTER 7 ACES AND BASES I89 WWMMWWWM 7 I. Usingthe K2, reaction to solve where ?T =..p—toiuidine, CH3C6H4NHz: ' PT+H20 W?7}F+OK' Initia} 0.016 M 0 ~0 - ‘ r x moHL ofPT reacts with H20 to reach equiijbrium ' ' ' Change «x - ' ms- +x +x Equii. 0.016 - x - x x Kb w my; {011“} m x 2 - {PT} 0.016 - x _ _ - _ Since pi: e 815th 130?! m 14.09 - 8.60 m 5.40 and {em =x m to“: 4.0 x 10-6 M (4.0 we"?2 «WWW-minnow l: 0.0I6 - 4.0 x 10"6 i- 72. Using the K, reactien'tc soive where ?Y-:'py1miidhze, C4H3NH: ' , FY + H20 W FYI"? + OH’ initial 1.00 x we M e 42 Equfi 2.00 X 10‘3 -—x x ' 3: x2 “MW?" LOOK IO wx ; Since 1311:1032: pOHm3.18and{OH‘]mic=I8‘3‘18W615x 10*‘M _ - ~42 '...
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