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Unformatted text preview: Chapter 1'7: Additional Aspects of AcidBase Equilibria Buffer Solutions 7. [H301 = 10““ = 8.7 >< 105M. We let S = [cnoz'lm Equation: HCH02(aq) +H20(l): CHO;(aq) + H30*(aq) Initial: 0.366M — SM 3 0M
Changes: —8.7x10'5M — +8.7x10'5M +8.?x10‘5M
Equil: 0.366M ~ (5 + 3.7 x10"s)M 8.7 x1075 M K zmﬁgxwa =(S+37X10'5)37><10'5 w 8.7x10'sS
. [H3102] 0.366 0.366 To determine S , we assumedS 3» 8.7 x10‘5M , which is clearly a valid assumption.
Or, we could have used the HendersonHasselbalch equation (see below). pic, = —log(1.s x104)= 3.24 ;S=0.76M [01410;] _ [CHOZ']
[HCHOJ [HCHOZ] The difference in the two answers is due simply to rounding. 4.06 = 3.74+log = 2.1; [CHOZ‘] = 2.1x 0.366 = 0.77M 8. We use the HendersonHasselbalch equation to ﬁnd the required [NHg].
pr = —log(1.s x10“)= 4.74 pK, =14.00_pr =14.00—4.74 = 9.26 pH = 9.12 = 926+ log——[W3+] (NH. 1
[L141] =10"°'” = 072 [NH3]= 0.72x[NH;] = 0.72x 0.73204 = 0.53M
[Na '] .a (8) Equation: HC7H502 (3(1) + H200) : 0:11:0sz + H3019“) Initial: 0.012 M ~ 0.033 M e 0 M
Changes: xM — 41: M +x M
Equil: (0.012 —x) M ~ (0.033+x)M x M K 2 [H3O*][C?HSO;] = 6.3x“? = x(0.033 +x) 3 0.033;: x= 2.3XIU"SM [HC,H502] 0012—): 0.012 To determine the value of x , we assumed x <: 0.012 M, which is an assumption that
clearly is correct. [H301 = 2.3 x ID‘SM pH = 403(2'3 x10"‘)= 454 ﬂ Chapter 17: Additional Aspects of AcidBase Equilibria (0) Equation: NH,(aq) +H,0(1) a: NH,+(aq) +0H'(aq) Initial: 0.408 M  0.153 M a: 0 M
Changes: —x M — +x M + x M
Equil: (0.408  x) M — (0.153 + x) M x M [NH. ][0H‘] = 1.8x 10,, = x(0.153+x) m 0.153;:
[NH3] 0.408—x 0.403 To determine the value of x , we assumed x <§ 0.153 , which clearly is a valid assumption. [014’] = 4.8x10‘5M; pOH = —log(4.8x10") = 4.32; pH =14.00—4.32 = 9.68 x=4.3x10‘5M Kb= 10. Since the mixture is a buffer, we can use the HendersonHasselbalch equation to determine
Ka of lactic acid. [c H o _] = 1.00 gNac,H,o, x 1000 mL x 1 tool 1430,1133, x ImolCJHSO; = 00892 M .
3 ‘ ’ 100.0 mi. soln 1 Lsoln 112.1 gNac,H,o, 1molNaC,H,0, 
[CaﬂsOJ] 0 0892 M i = _ = + __..: + _——’ = K + .251 2
pH 4 11 pKI log [Hcaﬂsoa] pK, log 0.0500 M p , 0 1 pK, = 4.11—0.251 = 3.86; K, =10"ms =1.4><10“ l_l. (a) 0.100 M NaCl is not a buffer solution. Neither ion reacts with water to a detectable
extent. (1)) 0.100 M NaCl—0.100 M NH4Cl is not a buffer solution. Although a weak acid, NH: ,
is present, its conjugate base, N113, is not. (c) 0.100 M CH31\lIl2 and 0.150 M CH,NH;C1' is a buffer solution. Both the weak
base, CH aNH2 , and its conjugate acid, CH3NH; , are present in approximately
equal concentrations. (d) 0.100 M HCl—0.050 M NaNO2 is not a buffer solution. All the NO; has converted
to HNO2 and thus the solution is a mixture of a strong acid and a weak acid. (0) 0.100 M HCl—0.200 M NaC2H302 is a buffer solution. All of the HCl reacts with _
half of the C2H30; to form a solution with 0.100 M lIC,1l,02 , a weak acid, and
0.100 M C,H,0; , its conjugate base. ' (f) 0.100 M llCzlI302 and 0.125 M NaC3H502 is not a buffer in the strict sense because it does not contain a weak acid and its conjugate base, but rather the
conjugate base of another weak acid. These two weak acids (acetic, K, = 1.8 x 10" and propionic, K, = 1.35 x 105) have approximately the same strength, however, this solution would resist changes in its pH on the addition of strong acid or strong base,
consequently, it could be argued that this system should also be called a buffer. 608 Chapter 17: Additional Aspects of AcidBase Equilibria (b) The capacity of the buffer is reached when all of the weak base or all of the conjugate
acid has been neutralized by the added strong acid or strong base. Because their
concentrations are the same, the number of moles of base are equal to the number of moles of conjugate acid in the same volume of solution. amount ofweak base = 125 mLxW= 6.25 mmol (:11.an or CH3NH; lmL Thus, the buffer capacity is 6.25 millimoles of acid or base per 125 mL buffer solution. [anon1  — log(1‘3x104)+log 3.5mm01/7S.0m]_, H= K + o —————
9 p " 15.5mmows.0mL ‘ g m
= 3.74 — 0.26 = 3.48
[Note: the solution is not a good buffer, as [CHOz']=1.1>< 10" , which is only ~ 600
times K3]
2 mmol OH'
1 rnmol 133:1(0H)1 The OH‘ added reacts with the formic acid and produces formate ion.
Equation: chozmq) + OH’(aq) v—i CHO;(aq) + 141.0(1) 03) Amount of added OH' = 0.25 mmol 35:1(011)2 x = 0.50 mmol 0H" Buffer: 15.5 mmol z 0M 8.5mmol —
Add base: +0.50 mmol
React: —0.50 mmol —0.50 mmol +0.50 mmol —
Final: 15.0 mol 0 mmoi 9.0 mmol —
CHO F
pH = PK; +logBl—Cﬁ—ET3 = —log(l.8x10“)+log%%
= 3.74 — 0.22 = 3.52 12 mol HCl x 1 mmol Hp"
1 mL acid 1 mmol HCl
The Hp" added reacts with the formate ion and produces formic acid. Equation: cno;(aq) + 1130*(aq) .=~ £10110:qu + 1120(1) (c) Amount of added 1130‘ =1.os m1. acidx =13 mmol 11,0+ Buffer: 8.5 mmol as 0 mmol 15.5 mmol —
Add acid : + 13 mmol React: “8.5 mmol ”8.5 mmol I +3.5 mmoi —
Final ' 0mmol 4.5 mmol 24.0mrnol — The buffer's capacity has been exceeded. The pH of the solution is determined by the
excess strong acid present. + 4.5mmol
H0 ='—“_"‘"_—=0.05 ; =4 _ = ‘
[3 ] 75.0n1L+l.05mL 9“ F“ 0303059) 123 614 23. (a) The pH of this buffer solution is determined with the HendersonHasselbaleh equation. ——_—_—l—__—————V— Chapter l7: Additional Aspects of AcidBase Equilibria 2_S._ (a) We use the HendersonHasselbalch equation to determine the pH of the solution.
The total solution volume is 36.00 mL + 64.00 mL = 100.00 ml... pK_ = 14.00 — ox, = 14.00 + 1og(1.s x10") = 9.26 36.0011th 0.200MNH3 : M = 0.0720 M
100.00mL 100.0 mL [W3] = [m4.]=ﬂ.00mx0.2wMW, =12.8mmo1NH, =01st 100.00mL 100.0mL 1
[NH,] 0.0720M '
H= K +1 =9.25+1 __=9_01 9.00
p p ' {going} 03 0.1st x (o) The solution has [0H']=10“'” = 1.0 x 10'5 M
The HendersonHasselbaleh equation depends on the assumption that: [1~111,]:e1.0x10'5 M e: [hing]
Ifthe solution is diluted to 1.00 L, [N11,] = "1.20 x 10'3 M, and [NI14*] = 1.28 x 10‘2 M. These concentrations are consistent with the assumption. However, if the solution is diluted to 1000. L, [NH 3] = 7.2 x 10" M , and
[NIl3] = 1.28 x 102’ M , and these two concentrations are not consistent with the
assumption. Thus, in 1000. L of solution, the given quantities of NH3 and NH: will not produce a solution with pH = 9.00. With sufﬁcient dilution, the solution
will become indistinguishable ﬁ‘om pure water (i.e. its pH will equal 7.00). i (e) The 0.20 mL of added 1.00 M HCl does not signiﬁcantly affect the volume of the
solution, but it does add 0.20mL x 1.00 M HCi = 0.20 mm] H 30+ . This added H J0+ '
reacts with NH3, decreasing its amount from 7.20 mmol NH3 to 7.00 mmol NH“ and increasing the amount of NH,+ from 12.8 mmol NH,+ to 13.0 mmol NH,+ , as
the reaction: NH3 + H30+ —> NH,+ +H20 10011111101 NH3 {100.20mL
13.0 mmol NH,* 1100201111. (11) We see in the calculation of part (c) that the total volume of the solution does not
affect the pOH of the solution, at least as long as the Henderson—Hasselbaich ‘ equation is obeyed. We let x represent the number of millimoles of H30+ added,
through 1.00 M HCl. This increases the amount of NH f and decreases the amount
of NH“ through the reaction NH3 + H30+ —> NH," + H20 pH = 9.26 + log = 8.99 616 Chapter 1?: Additional Aspects of AcidBase Equilibria (d) We determine the [011'] due to the added 13:1(011)2 .
1 113 GB 
[OH H_]_5.00gBa (OII) x mo a[ )2 x 2molOH =0.0778M
0750L 171.34g Ba(OH)2 1molBa(0H)2 This is sufficient [OI1‘] to react with the existing [HNOZ] and leave an excess [OH’] = 0.0778 M ~ 0.025? M = 0.0021 M. pOIl = —log (0.0021) = 2.68.
pH = 14.00 — 2.68 = 1 1.32 The indicator is blue in this solution. 3; Moles ofHCl = C >< V= 0.04050 M x 0.01000 L = 4.050 x 10'4 moles 34. Moles ofBa(OH)2 at endpoint = C x V= 0.01120 M x 0.01790 L = 2.005 x 10“ moles.
Moles of HCl that react with Ba(0H)2 = 2 x moles B31011); Moles off1C1 in excess 4.050 x 10" moles — 4.010 x 10" moles = 4.05 x 1006 moles
Total volume at the equivalence point = (10.00 mL + 17.90 mL) = 27.90 mL 4.04x10* mole HO] .4 .4
c1 =————————=l.45 10 M; H=—l 1.45 10 =3.s4
[H 1““ 0.02790 L x p °g( x ) (a) The approximate pKH... = 3.84 (generally : 1 pH unit) (1)) This is a relatively good indicator (with as 1 % of the equivalence point volume),
however, pKHin is not very close to the theoretical pH at the equivalence point
(pH = 7.000) For very accurate work, a better indicator is needed (Le.
bromothymol blue (13th11 = 7.1) Note: 2.4dinitrophenol works relatively well here
because the pH near the equivalence point of a strong acidfsuong base titration rises
very sharply (as 6 pH units for an addition of only 2 drops (0.10 mL)) solution (a): 100.0 mL of 0.100 M HC], [1130*] = 0.100 M and pH ='1.000 (yellow)
Solution (b): 150 mL of 0.100 M NaC2H302 K. «140211303 = 1.8 x 10‘5 Kb oszHsoz' = 5.6 x 10'”
C2H302‘(aq) + H200) K = 5.5.. Iv“ HCzH302(aq) + 011*(aq)
initial 0.100 M — 0 M ~ 0 M
change —x — +3: +3:
equil. 0.100 —x — x x Assumex is small: 5.6 x 10'”= x2; x= 3’. 48 x 10“6 M (assmnption valid by inspection) {0111— x= 7 43 x 10‘6 M pOH= 5.13 and pH= s. s? (greenblue)
Mixture of solution (a) and (b). Total volume= 250.0 mi. an;= CxV= 0..=1000Lx0100M 0.0100molHCl mzmof= C x V: 0.1500 L x 0.100 M: 0.0150 mol CZH30f
HCl is the limiting reagent. Assume 100% reaction.
Therefore, 0.0050 mole C2H302’ left umeacted, and 0.0100 moles of HC2H302 form. 0.0050 moi 0.0100 mol [C2H302'] = = ————~—~= 0.020 M [HCgHgOz] = = —————~—= 0.0400 M 1 1
V 0.250L V 0.2501. _—'————_————* Chapter 17: Additional Aspects of AcidBase Equilibria HC2H30201CD + H200) K 18xw’ C2H302_(aq) + H30*(aq) initial ' 0.0400 M — 0.020 M ~ 0 M change —x — +x +x equil. 0.0400 —x — 0.020 + x x
1.8 x 10.5 = x(0.020+x) s: x(0.020) x = 3.6 x 10.5 0.0400 — x 0.0400 (proof 0. 18 % < 5%, the assumption was valid) 5
[1130*] = 3.6 x 105 pH = 4.44 Color of thyme] blue at various pIIs: pH pH pH PH
1.2 2.8 8.0 9.8
4— Red Orang Yellow 4—G Blue—I i
Solution (a) Solution (0} Solution (b)
RED YELLOW GREEN Neutralization Reactions 35. The reaction (to 2"d equiv. pt.) is: H3P04 (aq)+2 KOH (aq) ——> KllIPO4 (aq) + zuzoa).
The molarity of the HJPO4 solution is determined in the following manner. 0.2420n1molKOH x lrnmolH3PO‘ 31.15mL KOIIsoln x
H3PO4 molarity = w u 0.1508M 25 .00mL HSPO‘ soln 36. The reaction (1“ to 2'“I equiv. pt.) is:
NaHlPQ, (aq) + NaOlI (aq) —) Nail[1’04 (aq) + H200) . The molarity of the H 3P0. solution is determined in the following manner. 0.1885 mmol NaOH X lmmol H3P04
lrnL NaOH soln lrmnolNaOlI
20.00 mL HJPO‘ soln 18.67_mL NaOIi soin x HJPO‘ molarity = = 0.1760M 37. Here we must determine amount of Hp" or OH' in each solution, and the amount of
excess reagent.
0.01501nmol H2304 x 2 mmol H30+ A.’ amount H30+ =50.00mL><
lmLsoln ImmoleSO. =1.50m:m1H30* (assuming complete ionization of H280; and HSOK in the presence of OH") 622 Chapter 17: Additional Aspects of AcidBase Equilibria 46. (a) (b) (0) (d) This part simply involves calculating the pH of a 0.275 M NH3 solution.
Equation: NH3 (aq) + H200) :2." NH; (aq) + OH‘ (aq) Initial: 0.275 M — 0M 2: 0 M
Changes: va — +xM +xM
Equil: (0.275 —x)M — xM xM
NH + OH‘ 2 2
Kb=w=l.8x10's=—x——z x x=2.2x10'3M={OII']
[N11,] (ms—x 0.275 (x << 0.225, thus the approximation is valid)
pOH = —mg(2.2x10'3 ) = 2.66 pH =11.34M This is the volume of titrant needed to reach the equivalence point.
The relevant titIation reaction is NH 3(aq) + HI(aq)—> NH 41(aq) 0.275 mmol NH lmmol HI 1 mL HI 50111
= . mL N —3 —
V” 20 00 H3 (3‘0" lmL NH3 50111 x lmmol NH3 x 0.325 mmol H1 V... =16.9mL HI 50111 The pOH at the half—equivalence point of the titration of a weak base with a strong
acid is equal to the pr of the weak base. pOH == pKh = 4.74; pH =14.00—4.74 = 9.26 NH: is fanned during the titration, and its hydrolysis determines the pH of the
solution. Total volume of solution = 20.001111. + 16.9111L = 36.91111. ' 0.275 mmol NH3 x lmmol NH: = 5.50 mmol NH:
1 mL NH. soln I mmoi NH3 mmol NH.+ = 20.00mL NHll (sq) x + 5.50mmol NH +
NH = —‘ = 0.149M
[ ‘ ] 36.91111, soln Equation: NI—L+ (aq) + H100) : NH3(aq) + H30+ (3‘31) Initial: 0.149M — 0M mom
Changes: — x M — +x M +x M
Equil: (0.149—x)M — xM xM
= g = 1.0x10‘“ = [NH3][H30*] _ x2 .. :c2
‘ Kb 1.8x10" [mg] 0.149“): ” 0.149 (3: << 0.149, thus the approximation is valid)
x = 9.1x10'6 M = [1130*] pH = —log(9.lXI0")= 5.04 629 Chapter 1?: Additional Aspects of AcidBase Equilibria 78. We shall represent piprazine as Pip in what follows. The cation resulting from the ﬁrst
ionization is HPiP”, and that resulting from the second ionization is HgPip2+.
1.00g CdeNIﬁ H20x lmol Cal110N145 H20 =0 0515M (”0 [PIP] = 0.1001. 194.22g €4H10N26H20
Equation: Pt'p(aq) + H200) .=" l[Pip+ (aq) + 0H'(aq)
Initial: 0.0515 M  0 M a: 0 M
Changes: —x M — +x M + x M
Equil: (0.0515 — x) M — x M x M C's/Ks = 858; thus, the approximation is n_ot valid. The full quadratic equation must be solved Ks =10433=6_0x105 = W=$
‘ [Pip] 0.0515 — x From the roots ofthe equation, x = [OI1'] = 1.?)(10‘3 M pOII = 2.?6 pH =11.24 (b) At the half—equivalence point of the 1" step in the titration, pOH = PM: = 4.22
pH = 14.00 —pOH = 14.00 — 4.22 = 9.78 0.0515mmolPip x lnunolHCIX lmLtitrant v 1 fHCl=100.mL —'
(c) 0 time 0 x imL lmmol Pip 0.500 mmolHCl = 10.3mL (d) At the ﬁrst equivalence point we have a solution of HPip‘”. This ion can react as a base
with H20 to form HgPip * or it can react as an acid with water forming Pip (i.e. HPip+ is
amphoteric). The solution’s pH is determined as follows (base hydrolysis predominates):
pOH = 1A (prt + 131(13): Vs (4.22 + 8.67) = 6.44, hence pH = 14.00 — 6.44 = 7.56 (e) The pOII at the half—equivalence point in the second step of the titration equals pth.
pOH=prz=8.67 pH= l4.00—8.67=5.33 . (f) The volume needed to reach the second equivalence point is twice the volume needed to
reach the ﬁrst equivalence point, that is 2 X 10.3 mL = 20.6 mL (g) The pH at the second equivalence point is determined by the hydrolysis of the Hemp”
cation, of which there is 5.15 mmol in solution, resulting from the reaction of HPipJr
with HCl. The total solution volume is 100. mL + 20.6 mL = 120.6 mL [HsPip’j = %%= 0.042? M K53 =10—“’ =2.1x10‘9 Equation: HzPip"*(aq) + 1120(1) .—_—‘ HPip*(aq) + H30*(aq) Initial: 0.042? M — 0 M e o M Changes: — x M — +x M + x M Equil: (0.0427 —x) M — x M x M K, = Kw =1.00x10: = [HPip*]'[H,O*] :43“? = 33): a x1
Kb: 2.1 x10 [H2P1p2*] 0.0427—x 0.0427 x=[H30*]=Jo.o427x4.sxlo* =4.5x10“M pH =3.34g
(x << 0.0427; thus, the approximation is valid). 65? ...
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 Spring '07
 DAVIS,F
 Chemistry

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