3600_Ch10_probs

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online