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Unformatted text preview: Chapter 16: Acids and Bases 16. First we determine the amount of KOH. and then the volume of the concentrated solution
required. pOH =14.00— pH =14.00—11.55 = 2.45 [OH‘]=10“m 210“5 a 00035 M 0.0035 M mol OH' X 1 moi KOH amount KOH = 25.0 LX = 0.088 mol KOH l L soln 1 1110] OH"
. . l
V50,"an = 0083 m0. Koﬂxmxmxﬂ= 29 ml, 501..
1 mol KOH 15.0 g KOH 1.14 g soln 1’? The volume of HCl(aq) needed is determined by first ﬁnding the amount of NH3(aq)
present. and then realizing that acid and base react in a 1:1 molar ratio. 0.265 mol NH3 X1 111011130+ x 1 mol HCl X l Lacid
1 L base 1 mol NH3 1 mo! H3O+ 6.15 mol HCl = 0.0539 L acid or 53.9 ml. acid. Vac. = 1.25 L basex 18. NH3 and HCl react in a 1:1 molar ratio. According to Avogadro’s hypothesis, equal
volumes of gases at the same temperature and pressure contain equal numbers of moles.
Thus, the volume of H2(g) at 762 mmHg and 21 .0° C that is equivalent to 28.2 L of H2(g)
at 742 mmHg and 25.0‘C will be equal to the volume of NII3(g) needed for stoichiometric neutralization.
mm 3.2 .0 K L NH
Vm ) = 28.2 L HCl(g)xﬂ—g—xwxl—ﬁﬂ = 27.1 L N113 (g)
=‘ 762 mmHg (273.2+25.0) K 1 LHCl(g) Alternatively, we can solve the ideal gas equation for the amount of each gas, and then by
equating these two expressions, we can find the volume of NH3 needed. MHCI}='.J'42 rnmIIg28.2L “{Nﬁa}=762 mmHgoV{NH3}
122982 K R2942 K 74 mmH  _ mmH  '
w=w This yields WNHJ} = 27,1L Isamg)
R 293.2 K R 294.2 K 1_9. Here we determine the amounts of 1130* and OH’ and then the amount of the one that is
in excess. We express molar concentration in millimoleslmilliliter, equivalent to moUL. 0.0155 mmol HI x 1 mo! H30+ = 0.775 mmol H30+
1 mL soln 1 mmol HI 50.00 mLx 0.0106 rnmol KOIl 1 mmol OH— x = 0.795 mmol 0H"
1 ml. soln 1 mo! KOH 75.00 mLx 550 Chapter 16: Acids and Bases The net reaction is H30+ (aq )+ OH‘ (aq ) —> ZHZOU). There is an excess of OH‘ of (0.795— 0.??‘5 =) 0.020 mmol 011'.
Thus, this is a basic solution. The total solution volume is (50.00 + 75.00:) 125.00 ml... [OH_] = 0.020 mmol OH" =1.6x10"‘ M, pOH 2 —log(l.6><10'4): 3.30. pH 210.20
125.00 mL 20. In each case, we need to determine the [11301 or [011'] of each solution being mixed, and then the amount of 1130* or OH" , so that we can determine the amount in excess.
We express molar concentratiOn in millimolesa’milliliter, equivalent to moUL. :I130*:=10'm = 7.6x10“3 M moles 1130+ = 25.00 mLx'l.6><10'3 M = 0.19 mmol H30” pOH =14.00—12.65 = 1.35 [OH']=10"'35 = 45x10"2 M
amount OH" = 25.00 i'ir1L><4.5X10'2 M = 1.13 mmol OH'
There is excess OH‘ in the amount of 0.94 mmol (= 1.13 mmol OH” 0.19 mmol 1130*) 0.94 mmol OH" [OH1: 25.00 1111.. + 25.00 1111.. =1.s§><10‘2 M pOH :—log(1.8§><10'2)=1.73 pH=12.27 Weak Acids, Weak Bases, and pH A1; We organize the solution around the balanced chemical equation.
Equatrom HNOJaq) + H200) T—_\ H30+(aq) + N02'(aq) 21;“: _ 0.143M =0 M 0M
use we ‘ M “M
' (0.143—x)M — xM xM
H 0+ N0 ' 2 2
K, :LELLLX— = 7.2x10" 2» x (itx«0.143)
[HNOZ] 0.143 ~ 1: 0.143 x = J0.143x7.2x10*‘ = 0.010 M We have assumed that x a: 0.143 M , an almost acceptable assumption. Another cycle of
approximations yields: 1: = \l(0.143 — 0.010)X7.2X10“‘ = 00098 M = [H30+] pH = —log(0.0098) = 2.01 This is the same result as is determined with the quadratic formula roots equation. 551 Chapter 16: Acids and Bases 21 Here we need to find the molarity S of the acid needed that yields[H30‘ ] = 10135 =1.4><10"3M Equation: HC7H502(aq) + H200) : H30*(aq) + C1H502‘(aq) Initial: s — 0 M 0 M
Changes: —0.0014 M _ +0.0014M +0.0014M
Equil: S —o.0014M * 00014 M 0.0014 M
H0*CHO‘ 0.0042 0.00142
Ka=[3_l_7_53_1=(___.1;:6.3x10'5 3—0.0014=(__e.2?=0_031
[HC.,H502] 3 —0.0014 6.3 x10
3 = 0.031 +0.0014 = 0.032 M = [110711502]
3500 mm 1 L x0.032 mol HC7H502 X122.1g11c,11502 = 1.4 g HQHSOZ
1000 mL 1 L soln 1 mol HC.,f1502 28. First we find [OH‘] and then use the Kb expression to ﬁnd [(CH3)3N]
pOH =14.00*p}1 :14.00—11.12 = 2.88 [0H ] = 1010“ = 101“ = 0.0013 M [(CH,),NH* ][OH‘] _ (0.0013)1 [(CH3),N]_F“ [(CHQSNLN
[(CH1‘ )3 NLM = 0.027 M(equil. concentration) + 0.0013 M(coneentration ionized)
[(0113 )3 101m 2 0.028 M Kb 2 6.3x10’5 = [(13113 )3NLIﬂ = 0.027 M (CI1,)3N a We use the balanced chemical equation, then solve using the quadratic formula. Equation : HCIO2 (aq) + H100) 1223— H30*(aq) + (3102' (aq) Initial: 0.55 M  == 0M 0M
Changes: —xM  +xM I ' I +xM
Equil: (0.55 —x) M — x M‘ x M :_.s____:____:1.lx10'2 =0.011 x2 =0.006140.011x
[HClOz] 0.55—x x2+0.011x0.0061=0
a + .1 +F——
ﬂ=wzﬂzmn=moﬂ 2a 2
The method of successive approximations converges to the same answer in four cycles. pH = —log[lIJO+] = —log(0.073) = 1.14 pOH 214.00— pH = 14.00~ 1.14 = 12.86
[0H‘1210‘W =10‘”‘“ =1.4:.<10"3 M I: 554 Chapter 16: Acids and Bases Percent Ionization :19; Let us first compute the [H30] in this solution. Equation: HC3H502(aq) + H200) ‘—“ H30*(aq} + C3H502'(aq)
Initial: M h _ 2 0 M x 0 M
C 3335' x M — +x M +x M
55““ ‘ (0.45%) M — x M x M
H C+ C H O ' 2 2 Ka :LLLLJ: x =10‘39=1_3x1{}5: x [HC3H502] 0.45 — x 0.45
x = 2.4 x10'3M ; We have assumed that; ca: 0.45 M , an assumption that clearly is
correct. I H 0+ I 3 (a) a = A : = 00053 = degree of ionization [HCJimeEI 0.45 M (b) % ionization = a x 100% = 0.0053x 100% = 053% For C2H5NH2 (ethylamine), Kb = 4.3 x 10“
C2H5NH2(aq) + H200) ‘—_' C2H5NH3+(3C]) + Initial 0.85 M "' 0 M 2 0 M
Change “J: M ‘" +x M +x M
Equil. (0.85 —x) M _ 1 M x M
2 2
4.3 x 10“1 = “Em— : x x = 0.019 M (x << 0.85, thus a valid assumption}
(0.851) M 0.85 M
Degree of ionizatiort o. = 0019M =0.022 Percent ionization : 0t X 100% z .3. 2’. ‘K:
0.85 M
Let x be the initial concentratioa of NH;” hence. the amount dissociated is 0.042 x
NH3(aq) + H200) K = 13 X 10" NILf'(aq) + OH'(aq)
Initial x M " 0 M = 0 M
Change 0.042x M “ 40.0423: M +0.042x M
Equil. (ix—0.042x) M “ 0.042): M 0.0421: M = 0.958 x 559 Chapter [6: Acids and Bases _ [m : [0.0421]2 Kb =1.3><10'S  = 0.0018gx
[N'Hanuu [0.958x1
—5
[NH31 .. = FM 2 0.0097354 = 0.00930:
“"31 0.001%
42.
HC3H502(21Q) + H206) M CsHsoz(aQ) + 1130130)
Initial 0.100 M ' 0 M = 0 M
Change —x M # +x M +x M
Equil (0.100 —x) M ” x M x M
2 2
1.3 X 10—5 = —L— = x x = 1.1 x 10'3 (x << 0.100, thus a valid assumption)
0.100 M —x 0.100
Percent ionization = 0001 1 x 100% = 1.1ﬁ %
0.100
Next we need to ﬁnd molarity of acetic acid that is 1.1% ionized
HC2H302(3C1) + H200) HIE—s C2H302F(aq) + H30+(aq)
‘__,_____
Initial X M _ 0 M = 0 M
Change Ji‘i XM _ +ﬁ XM +1—13— XM
100 100 100
Equﬂ Xzﬂx ﬂxM ﬂXM
100 F 100 100
=0.98§§XM =00111M =00111M
1.1g 2
1.8x10'5= ‘00 :1.3x10“(x); x=0.131M
09886 X Consequently, approximately 0.14 moles of acetic acid must be dissolved in one liter of
water in order to have the same percent ionization as 0.100 M propionic acid. 43. We would not expect these ionizations to be correct because the calculated degree of
ionization is based on the assumption that the [HCgJ'IgOglinmal =[HC2H302];,1M —
[HC2H3ozlawn , which is invalid at the 13 and 42 percent levels of ionization seen here. 44. chcno2 (trichloroacetic acid) pK. = 0.52 K; = 104152 = 0.30
For a 0.035 M solution. the assumption will not work (the concentration is too small and
K3 is too large) Thus, we must solve the problem using the quadratic equation. __ HC2C1301{aq) +1000) Viv czcnoﬁaq) + H30+(aq] Initial 0.035 M _ 0 M = 0 M Change ~x M _ +x M +x M Equil. (0.035  x) M "' x M x M
560 Chapter 16: Acids and Bases  2 0.30 =—I— or 0.010; a 0300:) = x2 or x2 + 0.30x ~ 0.010§ = 0 ; solve quadratic: 0.035— x
_ 0.30s «0.30)? —4(1)(0.010§)
x — 2(1) Therefore it = 0.032 M = {1130*}
Degree of ionization a = mz091 Percent ionization 2 (1 x 100% = 91. %
0.035 M
Polyprotic Acids ﬂ Because H3P04 is a weak acid, there is little [Ii‘04:" (produced in the 2“d ionization)
compared to the 1130* (produced in the 1st ionization). In turn, there is little P04} (produced in the 3Id ionization) compared to the HPO42', and very little compared to the
H3O+. 46. The main estimate involves assuming that the mass percents can be expressed as 0.057 g
of 75% H3P04 per 100. mL of solution and 0.034 g of 75% H3F’O‘i per 100. mL of solution. That is, that the density of the aqueous solution is essentially 1.00 gin1L, Based
on this assumption, we can compute the initial minimum and maximum concentrations of H3P04 .
0.057 gimpure H3PO4 x 100 7? g H3PE4PO x 93331 H1336
[Hde = mmeﬂﬁfea—szm : 00044101
100mLsoln><
10001111.
0.084 g impure II_,,PO4 X100 7:5 g H3PE‘PO x 9:33] [ngoA] : MW = 0.0064101
100 mLsoln x
1000mL
Equation: H3PO4 (aq) + H200) : H2P04' (aq) + H30+ (aq)
Initial; 0.0044 M _ 0 M 2 0 M
Changes: —xM ‘ +x M +x M
Equil: (0.0044x)M * x M x M
H PO  H 0* 2
in, 1—2734}?ng =W= 7.1x10‘3 x2 +0.0071x—3ix10“ = 0
3 4  _x
wbinz—tt  . :J. ‘5 . 4
x_ 2 at. =~0007l 50:10 +12X10 =30X10_3M:[H30+]
a 561 Chapter 16: Acids and Bases [H,0*][HOOC(CH2)4COO‘]=39X10_5= H x: K = :
[HOOC[CH2]4COOH] 0.10x 0.10 at x : 0.10x 3.9 x10" =2.0 x10'3M = [H301 We see that our simplifying assumption, that x << 0.10 M, is indeed valid. Now we
consider the second ionization. We shall see that very little Hp" is produced in this
ionization because of the small size of two numbers: Kat and[HOOC(CH2)4COO']. Again we base our calculation on the balanced chemical equation. Equation: IIC}(}‘C(CH2 )‘ COO' (aq)+ H20(l),T‘—' 00C (CH2 )‘ C00" (sq) + no (aq) Initial: 2_0x10'3M  0 M 2.0x10'3 M
Changes: —yM _ +y M +y M
Equil: (0.0020— y] M ' y M (0.0020+ y) M K _ [H30*I‘OOC(CH2)‘COO‘] 0% _ y(0.0020 + y) M 0.0020y
“1 [HOOC(CH2]4COO'] = 3.9x1 ._
0.0020— y 0.0020 y : 3.9x10" M = [‘00C(CH,),C00‘} Again, we see that our assumption, that y << 0.0020 M, is valid. In addition, we also note
that virtually no H3O+ is created in this second ionization. The concentrations of all species have been calculated above, with the exception of [OH’]. I lI.
[on'] = = $5373 = sumo—“M; [HOOC(CH2 1, coon] = 0.10M [1130*] = [Hooc (CH1 )‘ coo1: 2.0x10’3M; [ooc (CH, )‘ coo] = 3.9x 104M (3) Recall that a base is a proton acceptor, in this case, accepting II+ from H20.
First ionization: CmHMOIN2 +1120,_~~—_‘CmH2,,OZI~l,H+ +OH‘ pl‘r’b1 = 6.0 Second ionization: CNHM02N2H:+Hzo:cmnuozn,nf+on prz :93 b . .
( ) 1 .00 g quinine x ———1m01qmmpg— 324.4gquinine
1L lOOOI‘nL [CmIIMOZNZ] = =1.62x10‘3M 1900 mLX Kb1=10""° 21x106 Once again, we set up the ICE. table and solve for the {OH '] in this case: 565 “ Chapter 16: Acids and Bases Equation: cmHRozNgaq) + H200) : _CmHuozN2H+(aq) + OH'(aq) Initial: 0.00162 M _ 0 M = 0 M
Changes: —x M ’ +x M +1 M
W“ Equil: (000162 — x) M — x M x M
C H 0 N H" OH" 2 2
Kb :w:1x10'6=_ﬁf__z I x=4x10—5M
' [CmHnozNz] 0.00162 _ x 0.00162 The assumption x 4: 000162, is valid. 1301] = glog (4»)(10's )= 4.4 pH = 9.6 I“ 52. Hydrazine is made up of two NH; units held together by a nitrogennitrogen single bond:
H H
'“i \ /
1 N — N
.  / \
3} H H
in“ (a) N2H4(aq) + H200) —; N2H5*‘(aq) + OH‘(aq) . Kb. = 106‘" = 3.5 x10”
mm (b) N2H5+(aq) + H200) 2 N2H62+(aq) + OH'(aq) Io,2 = 104595 = 8.9 x1016
Because the OH' produced in (a) suppresses the reaction in (b) and since KM << sz,
"""= the solution’s pH is determined almost entirely by the ﬁrst base hydrolysis reaction
(a). Thus we need only solve the ICE. table for the hydrolysis of N2H4 in order to
ﬁnd the [OIThug, which will ultimately provide us with the pH via the relationship
“I. {0H']x[I130*] = 1.00 x 10'“.
I. Reaction: N2H4(aq) + H200) g N2H5+(aq) + OH‘(aq)
"" Initial: 0.245 M  0 M = 0 M
Changes: x M " H: M +x M
""h Equil: (0.245 —,r) M _ . x M x M
mm
N H + OH_ 2 2
b =Lil=s5x104=~iﬁ+= x x:0.0018§M={0H'] i,
' [NEIL] 0.245 —x 0.245 a“
14 Thus, [H30+ 1 = = 5.3’_}><10"2 M pH = log(5.31><10'12) = 11.26
5 0.00130 5—3. The species that hydrolyze are the cations of weak bases, namely, NH“+ and CWJISNHS+ ,
and the anions of weak acids, namely. N02" and CjHSOZF. (a) NH!+ (aq )+ N0; (aq )+ 11200): NH3 (aq) + NO,’ (aq )+ H30+ (aq) 566 Chapter 16: Acids and Bases 58. NH 4C1 dissociates completely in aqueous solution into NHf(aq) , which hydrolyzes,
and Cl'(aq) , which does not. We determine [H301 in a 0.123 M solution of NH“+ , ﬁnding the value of the hydrolysis constant from the ionization constant of NH3,
Kb =1.8x10'5. Equation: NHAWaq) + H200) :: NlI3(aq) + H3O*(aq) Initial: 0.123M — 0 M =0M
Changes: —x M — +x M +x M
Equil: (0.123—I)M — x M xM
14 + 2 2
Kb=£w_=l.0)<10_5 =5ﬁxwdn={NIIs][lrl+30 }: x z x
Ka 1.8X10 . [NH‘ ] 0123—): 0.123 8.3x104’M << 0.123, the assumption is valid x = 8.3x10*M = [H3O*]; pH = —Iog(3.3x10‘) = 5.03 gs; KC6H.J,02 dissociates completely in aqueous solution into K‘(aq). which does not hydrolyze, and the ion C6H702_ (aq) , which undergoes base hydrolysis. We determine
[err] in 0.37 M Kcénp2 solution using an 1.0.13. table. Note: Ka =10“ =10“ surmos Equation: CGH?02"(aq) + H200) r: HC6H702(aq) + onxaq) Initial: 0.37 M H OM = 0M
Changes: *xM — +xM +xM
Equil: (0.37 —x)M — xM xM
K =§=1.0x10"‘ : 59X10_w=_[_I£36H101][0H‘1 : x2 z x2 b a; 1.7 x10” ‘ [C6H,0;1 0.37 — x 0.37 x =1.5x10"M = [014'], 1101] = elog(l.5X10's ) = 4.32, pH = 14.00 —4.32 = 9.18 60. Equation: CSHSNH+(aq)+ H100) :CSHSqu) + H30*(aq) Initial: 0.0482M '  0M 30M Changes: —xM — +xM +xM Equil: (0.0482—x)M — xM xM
569 Chapter 16: Acids and Bases 5—_1.0><10'” __ 6 7xl0_6 _ [CSHSN][830+] 12 ~ 12
h 1.5x10'g ' [C5H5NH‘] 0.0482ex 0.0482 5.7 x10"M << 0.0482 M, the assumption is valid I = 5.7x10"M = [H301 . pH = —log(5.?x10“)= 3.24 61. (a) H30; +H20:.—_‘H30* +sof Ita = KWW = 6.2x10‘3 Kw _1.0x10'” 1130; + H20:OH' + st03 Kb =  d2 = 17x10“
mm; 1.3x10
Since it:al > Kb , solutions of HSO3' are acidic.
{b} HS' +H20:H30* +sz K, = Kawmm =1x10'”
‘14
HS' +H20:0H +IIZS Kb = Kw — LQZ‘L=1_0xio—1 Kalﬂmmc _ 1.0x10‘T Since K, < Kb , solutions of HS" an: alkaline, or basic. (c) IIPO‘2'+H20.:—‘_H30++P043_ 1g:15.1,!immm.“=4.2><10‘13 14
HPOf— +1120 :03 +H2PO4' Kb : “iv—— = 939)— =1.s><1oT Kazmwm 6.3x10'“ Since K3 4. Kb , solutions of HPOf‘ are alkaline, or basic. 62. pH = 8.65 (basic). Need a salt made up of a weakly polarizing cation and an anion of a weak acid, which will hydrolyze to produce a basic solution. The salt (c); KNOg. satisﬁes
these requirements. NH 401 is the salt of the cation of a weak base and the anion of a strong acid, and should form an acidic solution. (b) KHSQl and (d) NaNO3 both of these
salts have weakly polarizing cations and anions of strong acids; they form pH—neutral solutions. pOH =14.00—8.65 = 5.35 [014‘]: 19535 = 4.5x10*M Equation: NO; (aq) + H200) ;‘ HNOJaq) + 0H" (aq)
Initial: S — 0 M = 0 M
Changes: —4.Sx10"’M w +4.5x10'6M +4.5x10"‘M
Equil: (s —4.Sx10"M) — 4.5x104M 4.5x10'ﬁM —14 HNo. OH' 4.5)(1045 2
Kh=§=ﬂ§l~07=1.4xio'”=[ K, 7.2x10 [NOz ] S—4.5x10
4 (4.5x10‘t)2 S—4.5><10 maﬁa—am“ 3:1.4M+4.5x10“=1.4M=[KN01] . X 5'30 Chapter 16: Acids and Bases H
 ..  .
ﬂ. Structure of H3P03 shown on the rt ght. ‘ Hf0_?_O_H
The two ionizable protons are bound to oxygen. " :Cl) : "
FEATURE PROBLEMS 97. (a) From the combustion analysis we can determine the empirical formula. Note that the mass of oxygen is determined by difference. ﬂxﬂ‘l‘c— = 0.03633 mol C
44.01 g (302 1 “‘01 C02 12.011gC
1 mol C amount C = 1.599 3 CO2 x mass of C = 0.03633 mol Cx = 0.4364 g C ﬂ). 2 “‘“l H = 0.03629 moi H amount H = 0.327 g HEOX
18.02 g H20 1 mol H20 mass 01H : 0.03629 mol Hxl'l—r— = 0.03658 g H 1molO amount 0 = (1.054 g sample — 0.4364 g C  0.03568 g H)X ————— = 0.0364 mol 0
16.00g 0 There are equal moles of the three elements. The empirical formula is CHO.
The freezingpoint depresSion data are used to determine the molar mass. Arr =_Kfm mzﬂzﬂguzlm
—Kr —3.90 Cr’m
1 kg 0.21 mol solute = 00053 mm x
000 g 1 kg solvent amount of solute = 25.10 g solventx 1 = 0'6” g some = 1.2 x lOZgImol
0.0053 mol solute
The formula mass of the empirical formula is: 12.0g C +1.0g H + 16.0g 0 = 29.0g! mol Thus, there are four empirical units in "a molecule, the molecular formula
is C4IL,O4 . and the molar mass is 116.1 glmol. (1)) Here we determine the mass of maleic acid that reacts with one mol OH" , mass 0.4250 gmaleic acrd = 5803 glmol OH_ molOII' 34'03measex 1L >(0.2152molKOII
IOOOmL lLbase This means that one mole of maleic acid (116.1 gfmol) reacts with two moles of
hydroxide ion. Maleic acid is a diprotic acid: 112C4HzO4 . (c) Maleic acid has two—COOH groups joined together by a bridging C2112 group. A
plausible Lewis structure is sh0wn below: [(IZ‘H H H IO! llllﬂ H—giC—C=C—C——g—H 536 Chapter 16: Acids and Bases (d) We first determine [no ] = 10'0" = 10W = 0.016 M and then the initial coricentration of acid. [(CHCOOH)2] =3ﬂx1000 "‘Lx 1m“ 'm'ia' 50.00 mi. 1L 116.1 g
We use the first ionization to determine the value of K”:I Equation: (CHCOOH)2 (aq) + 11200) :4“ H(CHCOO); (aq)+ H3O*(aq) = 0.0370 M Initial: 0.0370M — OM :OM
Changes: —0.016M  +0.016M +0.016M
Equil: 0.02m _ 0.016M 0.01am
K _ [H(CHC00)2' ][H30‘] _ (0.016)(0.016) _1 M 04 ‘_ [(CHCOOH)2] ' . 0.021 " ' Ka2 could be determined if we had some way to measure the total concentration of all ions in solution, or if we could determine [(CHCOO)? ] = K, ‘1 (e) Practically all the [H301 arises from the first ionization. Equation: (CHCOOH)2(aq)+ H200) : H(CHCOO)2_ (aq)+ H30"(aq) Initial: 0.0500M — 0 M zOM Changes: — x M — +x M +xM Equil: (0.0500— x) M — xM xM
 +  2 K. =—..._—[H(CHCOO)2 HHSO 1— x =1.2><10'2 [(CHCOOPDz] ' 0.0500w x
x2 = 000060—0012 x x2 +0012 x —0.00060 = 0 _ b_‘t\/b2 —4ac _ —0.012:J0.00014+0.0024
2a 2 I = 0.019 M = [H30]
pH = —log(0.019)= 1.72 58? —_——_——————w Chapter 16: Acids and Bases 2 2 93. (a) __"__ s X s 4.2x107‘ x, e 0.00102 0.00250 m .i 0.00250 1 2
_.__X__.__ e I : 4.2): me in :0.000?88
0.00250—0.00102 0.00148 ' 2 2 _#x_#_= x :4.2x10" x3 20.000848
0.00250 —0.000783 0.00171 x2 x2 M : = 4.2x10" ini : 0.000833 ;_
0.00250—0.000848 0.00165 i 1 2 é. = x z 4.2x10“‘ x5 = 0.000837
0.00250—0.000833 0.00167 i
x2 .l:2 5 : 4.2x107‘ x6 =0.000836 0.00250 — 0.000337 = 0.00166 .
x6 2: 0.000836 or = 8.4XI04, i
which is the same as the value obtained using the quadratic equation. 03) We organize the solution around the balanced chemical equation, as we have done before.
Equation: HC102(aq) + H100) # C10 2' (aq) + H3O+ (aq) _
Initial: 0.500M — 0M =0 M Changes: — x M — +xM +11: M '
Equil: 1M — xM xM CEO ‘ H 0+ 2 2
KI=L;][—3——]=—x—=l.1x10'zz 1 Assuming x<<0.500M, [H001] 0.500 x 0.500 x = if? = 0.500><1.1><10‘2 = 0.074M Not signiﬁcantly smaller than 0.500 M. Assume x = 0.074 x = A0500—0.0774)><1.l><10‘2 = 0.063M Try once more.
Assume x = 0.063 x = J(0.500 4 0.063)><1.1x10‘2 = 0.069M One more time.
Assume x = 0.069 x = J10.500 — 0.069)><1.1><10‘2 = 0.069 M Final result! [H301 = 0.069 M. pH = —log[H10+]= —16g (0.069): 1.16 538 ...
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This note was uploaded on 05/05/2008 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell.
 Spring '07
 DAVIS,F
 Chemistry

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