3600_Ch11_probs - 11-4. 11-5. 2 x— — m 2 I = 3.11::10-3...

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Unformatted text preview: 11-4. 11-5. 2 x— — m 2 I = 3.11::10-3 = [14+] = rim-1:1}H = 2.51 EDIE] =1-[}|Z] [[FHM IH+1 “ [H33] = fllflfl—x = U.{I|':'IEI'.-H'-'I [AR-j = [bjl [H+] fi * f LUUIE lfl'fi' 4» 11H 2 {LEE} [HIV]. = E]- “If M [Hm] = Elfin—“5‘31 ; mm |'3 M [AI-J = $3155] = Imx m4 M H] [H'] [m —"'E — fi _: — cur — — -4 Ulmflq _ 2 .1' _[ ]. [um—3.1mm M -_;~ 11H: IfljE] [113']=U.1flfl x =99'Fx1fI-3M [Hm]: [FEM] _ Hm,c mm“ H H'u’t] [TWP] [311'] [twain-1113;“; 3.51 new In?- _y.11 22-: m3 1m I 10-3 Ill-JULIMMIHA em mm 117-3 mm lurl 1mm m-fi fillilflMHfigA H150 J.|:|{]M ml” ilfizs: NH 119?: 111-2 M}. HEM = n+ 4. HM' H. 2 L42}: LII-3 1"".1' .1' .'|.' [H2M] = 0.100—x = 0.089M [HM-1 = x = 1,12x10—2M [M21 =[—H[MH—fl]—K—2 = 2.01x10'6M _ K1K2(0.100) + Kle _ -5 _ (b) [H+] _ \f—~———-———K1 + 0.100 _ 5.30x 10 :> pH _ 4.28 " + [HM'] z 0.100M [H2M] = $947191; = 3-7><10'3M _ K2[HM_] 2 =~= .8 1-3M [M ] [H+] 3 x 0 The method of Box 11—2 would give more accurate answers, since [HM'] is not that much greater than [H2M] or [MZ'] in this case. (0) M2' + H20 “—’ HM' + OH' Kbl = Kw/Ka2 = 4.98x10'9 F — x x x mac—3:; = Kbl => x = 2.23x 10—5 => pH = —1og£xvl = 9.35 [MZ‘] = 0.100—x = 0.100M [HM’] = x = 223 x10—5M m = 7.04x10'12M [H2M] = K1 11-”. 11-14. lI-IT. 11-22. H'th 1|E[—|t-.Fn';,+ic11'- -- men;- [.fi' 1.13:1; |I215.U‘} ghluilj ' - ' '-I_ I: 1'. “"‘m ‘ m" 2] +1“'-L*tw.tm y;-.-'Ij:-‘.-1.t}| mm]: -- :- 1' _ 1.91%}; At pH 2.80, we have a mixture of 80%: and HSOA, since pKa for H8041 is 1.99. [80%;] [H8041] The reaction between H2304 and 80%" produces 2 moles of H304]: 2.80 = 1.99 + log => Hso;1 = 0.154950%] H2804 + 30% —> 2HSO;1 Initial mmol: x y — Final mmol: y — x 2x The Henderson-Hasselbalch equation told us that [H3011] = 0.154 9 [8031'] :9 2x = 0.154 9 (y — x). Since the total sulfur is 0.200 M, x + y = 0.200 mol. Substituting x = 0.200 — y into the equation 2x = 0.154 9 (y — x) gives NagSO4 = y = 0.186 6 mol = 26.50 g and H2804 = x = 0.013 4 mol 2 1.32 g. _ .I: 1:: {liufifln + N1fif_ _ _ 1:11 J-‘.u-Ll.nfiuu M Kligl’fh. |E[+_-— ' 31H- 4 “Sm-[j- "“ = |.‘H-: a In?I } 31H — '| 3'11 II-F'H' 31 ,. 4,1“- 1141-; + qu_| '- '1' --: “W Jill — L3! m" ~ |H.1Pn-II |H;:;]"|:.'|_1| K- H {n.immn + nit—- _ _ in] 1!.lli_l;:,-I::Ifi”['l h-‘l Eaiffifilzi l]lJ-]_ .J 1 ENE. F III-[Hun '-'r : iii-{M “1 IE] - :1 = ‘11:: |HQPHJI “1113:3141 a. a .‘_-. _, .- - -_._4|u- LII: - ...|—'1-3.“i . II'IL Iii-fluihl 1' H mm =:.|:m II-'~'£-' 7 “J” . “‘55.! "1'73 (a) HA (b) A" _ [A'] (C) pH " pKa + [HA] [A‘] _ 7.00 = 7.00+ log [HA] => [A ]/[HA] = 1.0 6.00 = 7.00 + log [A] => [A']/[HA] = 0.10 [HA] ...
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3600_Ch11_probs - 11-4. 11-5. 2 x— — m 2 I = 3.11::10-3...

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