3600_Ch10_probs - 10-1. HBr (or any other acid or base)...

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Unformatted text preview: 10-1. HBr (or any other acid or base) drives the reaction H20 ‘——‘ H+ + OH' to the left, according to Le Chatelier's principle. If, for example, the solution contains 10'4 M HBr, the concentration of OH” from H20 is Kw/[H+] = 10'10 M. The concentration of H+ from H20 must also be 10‘10 M, since H+ and OH“ are created in equimolar quantities. 10-2. (a) pH = —log [H+] = —log (1.0x 103) = 3.00 (b) [H+] = Kw /[OH'] = (1.0x10'14)/(1.Ox10'2) = 1.0x 10'12 M. pH = ~log [H+] = 12.00 10-3. [H+] = [OH'] +[C104'L] => [OH‘] = [H+]—5.0><10'8 [H+] [OH'] = Kw [H+] ([H+] —5.0>< 103) = 1.0x 10'14 => [H+] = 1.28 x 10'7M pH = —log [H+] = 6.89 [OH'] = Kw/[H+] = 7.8x lO‘8M :> [H+] from H20 = 7.8x 10‘8M 7.8 ><10‘8 M 1.28 x10'7 M 10-5. (a) @CO2H .—_= @COZ‘ + H+ Ka (b) @{05 + H20 .—= @COZH + OH' Kb + (c) @NH2 + H20 .—= NH3 + OH' Kb + (d) NH3 \-—‘ ©NH2 + H+ Ka Fraction of total [H+] from H20 = = 0.61 K 10-7. BH+ F? B + H+ Ka = Kw/Kb = 1.OO><10'10 0.100—x x x x2 m = 1-00X10'10 => x = [B]=[H+]= 3.16x10-6M => pH=5.50 10-8. (CH3)3NH+ ‘——‘ (CH3)3N + H+ K, = 1.58x10'10 F—x x x 2 x m = Ka=> x = 3.08x10'6 2) pH = 5,51 [(CH3)3N] = x = 3.1 x 10'6M , [(CH3)3NH+] =F——x = 0.060M 10-11. @COZH :2 ©405+m F _ 10—2.78 10-2.78 102.78 (lo—2.78)2 Ka = 0.045 0 — 10-2-78 10-17. 10-18. 10-19. 10-20. 10-21. 10-26. 10-27. The "fishy" smell comes from volatile amines (RNHZ). Lemon juice protonates the amines, giving much less volatile ammonium ions (RNHgL). x2 m = 1.00x 10'5 Letx = [OH'] = [BH+] and 0.100—x = [B]. Kw =>x= 9.95 x10'4M => [H+] = T =1_005 X1041 ::>pH = 11.00 ' [BH+l 9.95 x 104 a = = = -3 [B] + [BH+] 0.100 9'95 X 10 (CH3)3N + H20 "-‘ (CH3)3NH+ + OH" Kb = Kw/Ka = 6.33 x 10-5 F—x x x ———x2 -K=> —1910—3 H-IKW 0.060—x" b 95- ~2>< =91) —~0g7=11.28 [(CH3)3NH+] = x = 1.92x 10-3M, [(CH3)3N] =F_x = 00ng CN' + H20 = HCN + OH' Kb = KW/Ka = 1.6X10—5 F—x x x L—K: —8910-4 H—1_K_w__ 0.050—x— b x— ‘X =>P —-0gx —10.95 CH3CO§ + H20 :3 CH3C02H + OH” Kb = Kw/Ka -_- 5_71X10-10 F—x x x ~-“—-)£2~#*—-K=> —75610-6 —x— (1.00x10'1)—x ‘ b x - ~ >< => a — F — 0.00756% ————x2~——K=> —23910-6 —£—0 (1.00x10'2)—x ‘ b x- - >< =>oc—F— .0239% For 1.00 x 10'12 M sodium acetate, pH = 7.00 and we can say [HA] = $13; = 5.71x10-3[A-] [0H] -3 — 0P [HA] _ 5.71x10 [A] =0.568% _ [HA] + [A-] _ (5.71 x 10—3 +1)[A'] The more dilute the solution, the greater is (X. The pH of a buffer depends on the ratio of the concentrations of HA and A‘ (pH = pKa + log [A‘]/[HA]). When the volume of solution is changed, both concentrations are affected equally and their ratio does not change. Buffer capacity measures the ability to maintain the original [A‘]/[HA] ratio when acid or base is added. A more concentrated buffer has more molecules of A' and HA, so a smaller fraction of A” or HA is consumed by the added acid or base and there is a smaller change in the ratio [A']/ [HA]. 10-29. When pH = pKa, the ratio of concentrations [A‘]/ [HA] is unity. A given increment of added acid or base has the least effect on the ratio [A']/ [HA] when the concentrations of A" and HA are initially equal. , _.-i'| _ . tin-m .. III-3E. I111 .- _.1h_, -- |::-!_'. EH...“ '- fillll + up “In”; 4_.-[J 111—113. {'Hfi .|_..-.!E[_. - tl' —- 1E[-.-{'][3NH': liaitialmmnl: !.-1|-: .a |:_§]:||I1I|nn|: I. PH 1 PH; 4 lug J: H_';'l_-1_|;INH_.| 'f'||_~.{'|':_:‘~lll'!| .. . I J.- Ill ' ||.|.".I_J “LEW-El”. + It‘j'. 1 ” '.‘* .I.' ='-'- '31-.H'IIL1:II||'_L:I'_ _ I _ [ll-Hi”. Il'mnl __ . .H II 1.11 “IL. [1.24rlll_l1|:J-I._lll- .1._-.-|-1 . 10-37. (a) N +H _ <\ + H20 = / \ + OH Kb H H + H 1/ ) .-——\ \ + H+ Ka H H (b) FM of imidazole = 68.077. PM of imidazole hydrochloride = 104.538. LOO/68.077 pH = 6.993 + log = 7.18 (c) B + H+ —> BH+ Initialmmol: 14.69 2.46 9.57 Final mmol: 12.23 —— 12.03 12.23 pH = 6.993 + log 1203 = 7.00 (d) The imidazole must be half neutralized to obtain pH = pKa. Since there are 14.69 mmol of imidazole, this will require 514.69) = 7.34 mmol of HClO4 = 6.86 mL. ...
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3600_Ch10_probs - 10-1. HBr (or any other acid or base)...

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