Question

46. What is the percent ionization of a 0.491 M aqueous solution of carbonic acid? Ka1= 4.2x10^-7; Ka2= 4.8x10^-11

47. What is the percent ionization of a 0.356 M aqueous solution of hydrosulfuric acid? Ka1= 9.5x10^-8; Ka2= 1x10^-19

48. What is the pH of a 0.387 M aqueous solution of acetic acid? Ka (CH3COOH) = 1.8x10^-5

49. What is the pH of a 0.0398 M aqueous solution of benzoic acid? Ka (C6H5COOH) = 6.5x10^-5

50. What is the pH of a 0.415 M aqueous solution of hydrofluoric acid? Ka (HF) = 7.1x10^-4

Please Help & Show Work! Thanks!

Answer 1

(46) Formula to find percent ionization is : Percent ionization = [17] X 100 - THAIgnitial (C) change -n H2CO3 ( carbonic acid) b a dibrotic acid. 9t's gonization occur as: Ho comuns Hont H603; ka, = 42x107 → Teorema nyt first coa pah kq = 4.Rajo" → using a Hz Coz + Hсоз Let us suppose at equilibaum GlInitial 0.491 (I) concentration 0 0 y amount of it and HCO3 are foomed from tu tu 0.491 M H₂CO3 (E) Equilibrum 0491-x x x in Remaining conce. of Hacoa at equilibrium = 0.491- Ka – (Ht] [HC0g) [H2CO3] = 4.2x70= x,x 01491_n Assume a iş very small can compared to 0.491 sox can be neglected je; 0.491-2 ♡ 0.491 [because Rais very very small] = 42x1o? = ? 0.491 x²= 4,2x0.491 XTOT - x2= 2:06x107 ar x²= 20.68108 as se = √ 20.68108 - 4.53X104 ✓ 4.

, and (HCO3] = x .. (16934.53110m At equilibrium (H+) = x 2 :[ht] = 4.53X1ÖLMt using - HCO H+ + cor- Initial 4.537104 4.538109 o Chamae -x + +z - Equilibrium 4,53x10 - 4.53xłote se Ką = [ft] [co2] [46oo) m § 4.8x1őtt 14953X70*4). x º = 4,53x1o?_ u<<< 4.530104, so x can be neglected i 4.52x106 x Ŭ 4.52 x 104 and 4.537]otze 4 4.52x10" 4.8x10"- 4.53770 - -U b . = x= 4.8x10" At equilibrium , [ht] - 4.53X16"+ x is [ht] = 4.52x16"+ 4.8x16" 4,53x16" M - very very 4+2 4. 53 xjölmt small

0.491M r Initial conce [HA] = Put these values in equo Percent impatiens 4053xjoy Percentionnation = -X100 = 0.092 % ✓ 0.497 ; Percent ionizatim - 0.092.04| > pas (99) (29) Hys(tyehostulphurish acid dissociates mi hs, 2 H+ + HS ; Ka = 9.6x1080 HS (98) 2 4 + + 52 (as) ; Kq = ixlola 6 using ♡ H₂S 2 4+ + HS' a gnitial 0.356 o o Change - +7 + Equilibrium 0.356-2 n n ICE Table - Ka = [ht] [n s'] - [hs - .3 XD - 2012 0-356-2a can be neglected Assume usaa 0.356, so x 9.5x108 = 2 0.356

is 9.50.356x168 x = 338 xtö8 ar r= 13:38 xjö8 = 1.84x164 At equilibrium (H+) = (45) = i [H +] = [H Ś] = 1.89 X O'M ✓ being @ - H5 ~ 4+ + ² gmitial 1.84x164 1:84X10 0 change - te te equilibrium 1.39x09_n 1.87xjote e - Ka = [ht] [52] (45) => 7810!9 (184x154+x)= x Thisy x104_41) A rcca 1.84xlou ; 1.84xłotne tor 189 xloy f 1.84 X10-a u 1.89 xloy > Ixio" =_******* J. 20 x = 1x1019 At equitiböym, [H+) = 1.84x76_ " : [H+] = 1 x 4 x16+ + x10' (+) = l. Byxlöya v - 1 X X16 M - very very Small 77 of 80

Sritied Conce, [HA] = 0.356 ; Percent ionigetim = 1.348109 xloo 01356 = 0.052016 Percent ionizatime 0.0520 | 43 Chz cool = Chycooked + as) 194) Initial 0.387 0 - te Changen tu Equilibrium 0.387-2 - Ka = [CH₂ coo] [at] (сн сон) = 1.8x105 = 2.0 0.387-X Assume *<<< 0.387 i 0.387-4 0.387 = 18 x 105 = x 0.387 u= 1.8X0-387 X10S x = N 0-696x465 0646x460-_2.63X163 At equilibrum, [ht] = x [+) = 2163x16SMV

phu give by При = -lba Cot] Bu =-lag (2163Xčo?) [iky axbelog at lg 6] PH=-{log 2063 + log [03] [log ab = bloga] pu= -(0.419 - 3 loglo ] [log 2.63 = 049 by=-[0.419-3 ] toglo=1 bly - - (-2.58] = +2.58 [bu = 2:58] Z C6 Hg coo tHt Cols CoolH 0.0398 Initial Change - +и и Equilibrium 0.0398- n x [Conscoo] [et] [Chg cook] 6.5X105 - 11.2 0.0398-1 → Assume u<<<< 0.0398 0.0398-1 ñ 0.0398 = 6:5X165 = x? - 00998 - x²= er x At equilibrium 2.58x106 22.58x100 = 1.61x103 ✓ [ht] =n :(H+) = 1161X103 MV 1.61 X 10

PH=-log Cut] s-log (1.61% 103] =-(loytet thoros] =-[0.207 -3 ] = 2.79 [P4 = 2.79 | V [long log1= 0.207] 50 Initial change Equilibrium HF = 0.415 an 0.415-1 4 o n n + F o a r Ra= [ht] [F] (HF] 2 . n * 7.1 x6 = x 0:415-X Assume n <<< 0.415 0.415-4 " 0.415 7 7.18164 = u? 0415 ne 7.) xjö x 0.415 is n = 1.7 x 102 √2.95*254 = At equilibrium [ht] = x .: (ht] = 1.7 x10 m ✓ pH = -log (1.7810² ) E-Llog 1.72 X log do J [hung 17 = 6.23 = -(0.23 -2 ] = 1: 77 bu = 1,77|r