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PHN NG TO KT TA
TS Vi Anh TunKhoa ha hc Trng i hc KHTN - HQG H Ni
Phn ng to kt ta l phn ng to thnh cht rn t cc cht tan trong dung
dch.
Th d:
Ag+ + Cl- AgCl (r)
Ca2+ + C2O42- CaC2O4 (r)
Trong ho phn tch, phn ng to kt ta c s dng :
Tch cht cn xc nh khi cc cht cn tr.
Phn tch khi lng.Phn tch gin tip.
Chun kt ta.
1. Tch s tan v tan
1.1 Tch s tan
Qu trnh ho tan l qu trnh thun nghch, do cng tun theo nh lut tc
dng
khi lng. Xt cn bng ha tan (Mn+ l ion kim loi, Xm- l gc axit hoc
OH-):
MmXn mMn+ + nXm- T = [M]m[X]n
(*)
T c gi l tch s tan (solubility product).
Tch s tan c s dng :
So snh tan ca cc cht t tan "ng dng".
Xem mt dung dch bo ho hay cha:
n
X
m
MCCQ = > T: dung dch qu bo ho => xut hin kt ta.
n
X
m
MCCQ = = T: dung dch bo ho.
n
X
m
MCCQ = < T: dung dch cha bo ho => khng xut hin kt ta.
Tnh tan ca cc cht t tan (mui, hidroxit).
Cu 1.1. So snh tan ca AgCl v AgBr trong nc ct. Bit TAgCl =
10-10, TAgBr = 10-
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Hng dn gii (AgCl > AgBr)
*Ch : Mc d TAgCl= 10-10 > TMg(OH)2= 1,2.10-11, nhng trong nc
ct, tan ca
Mg(OH)2 li ln hn tan ca AgCl.
Cu 1.2. (a) Trn 1 ml dung dch K2CrO4 0,12M vi 2 ml dung dch
Ba(OH)2 0,009M.
C kt ta BaCrO4 to thnh khng? Bit TBaCrO4= 1,2. 10-10
.(b) Tnh nng cn bng ca cc cu t sau khi trn.
Hng dn gii
(a. Q= 0,04 0,006 = 2,4.10-4 > T => c kt ta to thnh;
(b) TPGH: CrO42-: 0,034 M
BaCrO4 Ba2+ + CrO42-
Cb x 0,034 + xT = x (0,034 + x) = 1,2.10-10
x = 3,53. 10-9 M.
[CrO42-] = 0,034 M;
[Ba2+] = 3,53.10-9 M)
Cu 1.3. Metylamin, CH3NH2, l mt baz yu phn li trong dung dch nh
sau:
CH3NH2 + H2O CH3NH3+ + OH-
(a) 25C, phn trm ion ho ca dung dch CH3NH2 0,160M l 4,7%. Hy tnh
[OH-],
[CH3NH3+], [CH3NH2], [H3O+] v pH ca dung dch.
(b) Hy tnh Kb ca metylamin.
(c) Nu thm 0,05 mol La(NO3)3 vo 1,00 L dung dch cha 0,20 mol CH
3NH2 v 0,20
mol CH3NH3Cl. C kt ta La(OH)3 xut hin khng? Cho tch s tan ca
La(OH)3 l 1.10-
19.Hng dn gii
(a) [CH3NH2]= 0,152 M; [CH3NH3+]=[OH-]= 7,5.10-3; pH= 11,9
(b) 3,7.10-4
(c) Q = 2,56.10-12 > T, c kt ta)
Cu 1.4. MgF2(r) Mg2+(aq) + 2 F-(aq)
Trong dung dch bo ho MgF2 18 C, nng ca Mg2+ l 1,21.10-3 M.(a) Hy
vit biu thc tch s tan, T, v tnh gi tr ny 18 C.
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(b) Hy tnh nng cn bng ca Mg2+ trong 1,000 L dung dch MgF2 bo ho
18C
cha 0,100 mol KF.
(c) Hy d on kt ta MgF2 c to thnh khng khi trn 100,0 mL dung dch
Mg(NO3)2
3.10-3 M vi 200,0 mL dung dch NaF 2,00.10-3 M 18C.
(d) 27C nng ca Mg2+
trong dung dch bo ho MgF2 l 1,17.10-3
M. Hy cho bitqu trnh ho tan MgF2 l to nhit hay thu nhit? Gii
thch.
Hng dn gii
(a) 7,09.10-9
(b) 7,09.10-7M
(c) Q < T, khng c kt ta
(d) To nhit)Cu 1.5. Dung dch bo ha H2S c nng 0,100 M.
Hng s axit ca H2S: K1 = 1,0 10-7 v K2 = 1,3 10-13.
(a) Tnh nng ion sunfua trong dung dch H2S 0,100 M khi iu chnh pH
= 2,0.
(b) Mt dung dch A cha cc cation Mn2+, Co2+, v Ag+ vi nng ban u
ca mi ion
u bng 0,010 M. Ho tan H2S vo A n bo ho v iu chnh pH = 2,0 th ion
no to
kt ta?Cho: TMnS = 2,5 10-10 ; TCoS = 4,0 10-21 ; TAg2S = 6,3
10-50.
(c) Hy cho bit c bao nhiu gam kt ta ch(II) sunfua c tch ra t
1,00 lit dung dch
bo ha ch(II) sunfat? bit nng sunfua c iu chnh n 1,00 .10-17 M?
Cho cc
gi tr tch s tan: TPbSO4 = 1,6 10-8 v TPbS = 2,5 10-27.
Hng dn gii
a) 17211
2212 10.3,1
][][][ 2
++
=
++
= SH
aaa
aa CKKKHH
KKS
b) C: [Mn2+] [S2-] = 10-21,3 .10-17 = 1,3 .10-19 < TMnS = 2,5
.10-10 ;
khng c kt ta
[Co2+] [ S2-] = 10-21,3 .10-17 = 1,3 .10-19 > TCoS = 4,0
.10-21 ; c kt ta
CoS
[Ag+]2[S2-] = (10-2)21,3 .10-17 = 1,3 .1021 > TAg2S = 6,3
.10-50 ; c kt ta
Ag2S
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c) C: [Pb2+][SO42-] = 1,6.10-8.
[Pb2+] = [SO42-] = 1,265.10-4.
Khi nng sunfua t 1,00.10-17 M th nng Pb2+ cn li trong dung dch
l:
[Pb2+] = 2,5.10-27/ 1,00.10-17 = 2,5.10-10.
mggammPbS 3,3010.03,312,239)10.5,210.265,1( 2104 === )
1.2 Quan h gia tan v tch s tan
tan (S,solubility) ca mt cht l nng ca cht trong dung dch bo
ho.
tan thng c biu din theo nng mol/l.
tan v tch s tan l nhng i lng c trng cho dung dch bo ho ca cht
t
tan. Do , tch s tan v tan c mi quan h vi nhau, iu c ngha l ta c
th tnh
c tan ca mt cht t tan t tch s tan ca n v ngc li.
MmXn m Mn+ + n Xm-
mS nS
C: T = [M]m[X]n = [mS]m[nS]n nmnmnm
TS
+
=
1
*Nhn xt: Cng thc trn ch ng nu Mn+ v Xm- khng tham phn ng no
khc.
Cu 1.6. Cho tch s tan ca Ag2CrO4 25o
C l 2,6.10-12
.(a) Hy vit biu thc tch s tan ca Ag2CrO4.
(b) Hy tnh [Ag+] trong dung dch bo ha Ag2CrO4.
(c) Hy tnh khi lng Ag2CrO4 c th tan ti a trong 100 ml nc
25oC.
(d) Thm 0,1 mol AgNO3 vo 1,0 lit dung dch bo ha Ag2CrO4. Gi thit
th tch dung
dch khng thay i. Hy cho bit [CrO42-] tng, gim hay khng i? Gii
thch.
Trong dung dch bo ha Ag3PO4 25
o
C, nng Ag
+
l 5,3.10
-5
M.(e) Hy tnh tch s tan ca Ag3PO4 25oC.
(g) Lm bay hi 1,00 lit dung dch bo ha Ag3PO4 25oC n cn 500 ml.
Hy tnh
[Ag+] trong dung dch thu c.
p s
b. 8,66.10-5 M.
c. 2,88.10-3 gam;
d. gim;
e. 2,63.10-18.4
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g. khng i, 5,3.10-5 M)
2. Kt ta phn on
Nu trong dung dch c cha hai hay nhiu ion c kh nng to kt ta vi
cng mt
ion khc, nhng cc kt ta hnh thnh c tan khc nhau nhiu th khi thm
cht to kt
ta vo dung dch, cc kt ta s ln lt c to thnh. Hin tng to thnh ln
ltcc kt ta trong dung dch c gi l kt ta phn on.
*iu kin kt ta hon ton:
[X] < 10-6M, hoc
%X cn li trong dung dch < 0,1%
Cu 2.1. Thm AgNO3 rn vo dung dch NaCl 0,10 M v Na2CrO4 0,0010 M.
Cho tchs tan ca AgCl l 1,8.10-10 v ca Ag2CrO4 l 2,4.10-12.
(a) Hy tnh nng Ag+ cn thit bt u xut hin kt ta AgCl.
(b) Hy tnh nng Ag+ cn thit bt u xut hin kt ta Ag2CrO4.
(c) Kt ta no c to thnh trc khi cho AgNO3 vo dung dch trn?
(d) Hy tnh phn trm ion Cl- cn li trong dung dch khi Ag2CrO4 bt u
kt ta?
p s(a) 1,8.10-9M
(b) 4,9.10-5M
(c) AgCl
(d) 3,7.10-3%)
Cu 2.2. tan l mt yu t quan trng tan l mt yu t quan trng dng nh
gi mc gy nhim mi
trng ca mui. tan ca mui ph thuc nhiu vo bn cht ca mui, dung mi
vcc iu kin th nghim nh nhit , pH v s to phc.
Mt dung dch cha BaCl2 v SrCl2 c cng nng l 0,01 M. Cu hi t ra l
liu
c th tch hon ton hai mui ny ra khi nhau bng cch thm dung dch bo
ha natri
sunfat hay khng. Bit iu kin tch hon ton l t nht 99,9% Ba2+ b kt
ta
dng BaSO4 v SrSO4 chim khng qu 0,1 % khi lng kt ta. Bit cc gi tr
tch s
tan nh sau: TBaSO4 = 1 10
-10
v TSrSO4 = 3 10
-7
.
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(a) Hy tnh nng ca Ba2+ cn li trong dung dch khi 99,9% Ba2+ b kt
ta v cho
bit phng php ny c dng c tch hon ton hai mui ra khi nhau hay
khng?
S to phc c th lm tng ng k tan. Bit tch s tan ca AgCl l 1,7
10-10,
hng s bn tng cng ca phc Ag(NH3)2+ l 1,5 107.
(b) Hy chng minh (bng php tnh c th) tan ca AgCl trong dung dch
amoniac 1,0
M cao hn so vi tan trong nc ct.
Hng dn gii
a. MBa 52 10.0,101,0100
9,99100][ + =
=
Sau khi 99,9% Ba2+ b kt ta th nng SO42- trong dung dch l:
MBa
T
SOBaSO 5
5
10
2
2
4 1010.0,1
10.1
][][4
+
===
MMSO
TSr
SrSO01,010.3
10.0,1
10.3
][][ 2
5
7
2
4
2 4>===
+
Sr2+ cha kt ta. Vy c th s dng phng php ny tch hon ton hai mui
ra
khi nhau.
b. tan ca AgCl trong nc ct:
MTAgS AgCl5
1 10.30,1][+
===
Tnh tan ca AgCl trong dung dch amoniac 1,0 M.
AgCl + 2 NH3Ag(NH3)2+ + Cl3107
10.55,210.7,110.5,1
==K
b 1,0
cb 1,0 - 2x x x
3
2
2
10.55,2
)20,1(
=
=
x
xK x = 4,59.10-2 M
S2 = x = 4,59.10-2 M; lanSS 3
1
2 10.6,4= )
3. Cc yu t nh hng n tan
Trong thc t, ion kim loi ca kt ta c th to phc vi OH- v anion ca
kt ta
c th phn ng vi H+ trong dung dch. Ngoi ra, nhng cu t khc c trong
dung dch
cng c th tham gia phn ng vi cc ion ca kt ta hoc t nht cng lm bin
i h s
hot ca chng. Nhng yu t u nh hng n tan ca kt ta.
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3.1 nh hng ca pH
Cu 3.1. (a) Hy cho bit dung dch ca cc mui sau c tnh axit, baz
hay trung tnh?
Gii thch. Natri photphat, ng (II) nitrat v xesi clorua.
(b) Hy tnh khi lng bc photphat cn dng pha 10 lit dung dch bo ha.
Khi tnh
b qua s thy phn ca ion photphat.Bit bc photphat c T = 1,3
.1020.
(c) Hy cho bit trong thc t nu ha tan lng bc photphat tnh c phn
(b) vo 10
lit nc th dung dch thu c bo ha hay cha? Gii thch.
Hng dn gii
a. Na3PO4: baz; Cu(NO3)2: axit; CsCl: trung tnh;
b. Ag3PO4 3 Ag+
+ PO43-
3S S
SST 3)3(=
MT
S 6420
4 10.68,427
10.3,1
27
===
mAg3PO4 = 4,68.10-610419 = 1,96.10-2 gam
c. Cha, v PO43-
b thy phn lm tng tan ca mui)Cu 3.2. Tnh tan ca AgOCN trong dung
dch HNO3 0,001M.
Cho TAgOCN= 2,3.10-7; HOCN c Ka=3,3.10-4.
Hng dn gii
AgOCN Ag+ + OCN- T = [Ag+][OCN-] (1)
OCN- + H+ HOCN ][]][[
HOCN
OCNHKa
+
= (2)
Lp phng trnh
[Ag+] = [OCN-] + [HOCN] (3)
[H+] + [HOCN] = 10-3 (4)
Gii h:
(2, 4) ][
]])[[10(10.3,3
34
HOCN
OCNHOCN =
][10.3,3].[10][
4
3
+=
OCNOCNHOCN (5)
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(3, 5) ][10.3,3
][10][][
4
3
+
+
+=
OCN
OCNOCNAg (6)
t [OCN-]= x
(1,6) 743
10.3,2)10.3,3
10(
=+
+ xx
xx
x3 + 1,33.10-3x2 - 2,3.10-7 x - 7,59.10-11 = 0
x= 2,98.10-4 = [OCN-]
(5) [HOCN]= 4,75.10-4
(4) [H+]= 5,25.10-4
(1) => [Ag+]= 7,72.10-4 = S.
*Nhn xt: v nng ca ion cc ion v phn t gn bng nhau nn khng th gii
gnng c)
Cu 3.3. (a) 100 ml nc 25(a) 100 ml nc 25ooC ha tan c ti a 440 ml
kh HC ha tan c ti a 440 ml kh H22S ( ktc). Hy tnhS ( ktc). Hy
tnh
nng mol ca Hnng mol ca H22S trong dung dch bo ha. Gi thit rng qu
trnh ha tan HS trong dung dch bo ha. Gi thit rng qu trnh ha tan
H22S khngS khng
lm thay i th tch ca dung dch.lm thay i th tch ca dung dch.
(b) Dung dch FeCl(b) Dung dch FeCl22 0,010 M c bo ha H0,010 M c
bo ha H22S bng cch xc lin tc dng kh HS bng cch xc lin tc dng kh H
22S voS vo
dung dch. Cho Tdung dch. Cho TFeSFeS = 8,0 .10= 8,0 .10-19-19.
H. H22S c KS c Ka1a1 = 9,5 .10= 9,5 .10-8-8 v Kv Ka2a2 = 1,3 .10=
1,3 .10-14-14. Hng s ion ca. Hng s ion ca
nc Knc Kww = 1 .10= 1 .10-14-14.. Hy cho bit thu c nhiu kt ta
FeS hn th cn phi tng hayHy cho bit thu c nhiu kt ta FeS hn th cn
phi tng hay
gim pH ca dung dch?gim pH ca dung dch?
(c) Hy tnh pH cn thit lp nng Fe(c) Hy tnh pH cn thit lp nng
Fe2+2+ gim t 0,010 M xung cn 1,0 .10gim t 0,010 M xung cn 1,0
.10-8-8 M.M.
(d) Ngi ta thm axit axetic vo dung dch phn (b) nng u ca axit
axetic t(d) Ngi ta thm axit axetic vo dung dch phn (b) nng u ca
axit axetic t
0,10 M. Hy tnh nng u ca natri axetat cn thit lp nng Fe0,10 M. Hy
tnh nng u ca natri axetat cn thit lp nng Fe 2+2+ trong dungtrong
dung
dch thu c l 1,0.10dch thu c l 1,0.10-8-8 M. Khi tnh ch s to thnh
HM. Khi tnh ch s to thnh H ++ do phn ng: Fedo phn ng: Fe2+2+ + H+
H22SS
FeS (r) + 2HFeS (r) + 2H++. Bit axit axetic c K. Bit axit axetic
c Kaa = 1,8 .10= 1,8 .10-5-5. Gi s vic thm axit axetic v natri. Gi
s vic thm axit axetic v natri
axetat khng lm thay i th tch ca dung dch.axetat khng lm thay i
th tch ca dung dch.
(e) Hy tnh pH ca dung dch m trc khi xc kh H(e) Hy tnh pH ca dung
dch m trc khi xc kh H22S.S.
Hng dn gii
(a.(a. MCSH SH 196,01,0
4,22
44,0
][22
=== (H(H22S phn li khng ng k)S phn li khng ng k)
b. Tng pH.b. Tng pH.
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c. C:c. C: 11819
2
2 10.0,810.0,1
10.0,8
][][
+
===
Fe
TS FeS
Mt khc:Mt khc: 22122
][
][][
+
=
H
KKSHS aa
MS
KKSH
H
aa 6
11
148
2
212
10.77,110.8
10.3,110.5,9196,0
][
][
][
+=
==
pH = 5,75;pH = 5,75;
d.d. FeFe2+2+ + H+ H22SS FeS (r) + 2 HFeS (r) + 2 H++
0,01 0,020,01 0,02
CHCH33COOCOO-- + H+ H++ CHCH33COOHCOOH
b a 0,02 0,1b a 0,02 0,1
cb a-0,02 - 0,1 + 0,02cb a-0,02 - 0,1 + 0,02
C:C:][
][log
3
3
COOHCH
COOCHpKpH a
+=
12,002,0
log74,475,5
+=a
a = 1,25 Ma = 1,25 M
e.e. 84,51,025,1log74,4
][][log
3
3 =+=+=
COOHCHCOOCHpKpH a ))
Cu 3.4. (QG 2007) Mt dung dch c ba cht HCl, BaCl(QG 2007) Mt
dung dch c ba cht HCl, BaCl 22, FeCl, FeCl33 cng nng cng nng
0,0150M. Sc kh CO0,0150M. Sc kh CO22 vo dung dch ny cho n bo ho.
Sau thm t t NaOH vovo dung dch ny cho n bo ho. Sau thm t t NaOH
vo
dung dch n nng 0,120M. Cho bit: nng COdung dch n nng 0,120M. Cho
bit: nng CO22 trong dung dch bo ho l 3.10trong dung dch bo ho l
3.10--
22M; th tch ca dung dch khng thay i khi cho COM; th tch ca dung
dch khng thay i khi cho CO 22 v NaOH vo; cc hng s: pKv NaOH vo; cc
hng s: pKaa
ca Hca H22COCO33 l 6,35 v 10,33; pKl 6,35 v 10,33; pKss ca
Fe(OH)ca Fe(OH)33 l 37,5 v ca BaCOl 37,5 v ca BaCO33 l 8,30; pKl
8,30; pKaa ca Feca Fe3+3+
l 2,17. Hy tnh pH ca dung dch thu c.l 2,17. Hy tnh pH ca dung
dch thu c.
Hng dn gii
HH++ + OH+ OH-- HH22OO
0,015 0,0150,015 0,015
COCO22 + 2 OH+ 2 OH-- COCO332-2- + H+ H22OO
0,03 0,06 0,030,03 0,06 0,03
FeFe3+3+ + 3 OH+ 3 OH-- Fe(OH)Fe(OH)33
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0,015 0,0450,015 0,045
BaBa2+2+ + CO+ CO332-2- BaCOBaCO33
0,015 0,0150,015 0,015
TPGH: COTPGH: CO332-2-: 0,015 M;: 0,015 M;
COCO332-2- + H+ H22OO HCOHCO33-- + OH+ OH-- KKb1b1 = 10=
10-3,67-3,67
0,015-x x x0,015-x x x
67,3
2
1 10015,0
=
=x
xKb
x = 1,69.10x = 1,69.10-3-3 MM
pH = 14 + log (1,69.10pH = 14 + log (1,69.10-3-3) = 11,23)) =
11,23)
Cu 3.5. Du hiu cho thy mt ngi c nguy c mc bnh gout l nng axit
uricgout l nng axit uric(HUr) v urat (Ur(HUr) v urat (Ur--) trong
mu ca ngi qu cao. Bnh vim khp xut hin do s kt) trong mu ca ngi qu
cao. Bnh vim khp xut hin do s kt
ta ca natri urat trong cc khp ni. Cho cc cn bng:ta ca natri urat
trong cc khp ni. Cho cc cn bng:
HUr (aq) + HHUr (aq) + H22OO UrUr-- (aq) + H(aq) + H33OO++
(aq)(aq) pK = 5,4 37CpK = 5,4 37C
UrUr-- (aq) + Na(aq) + Na++ (aq)(aq) NaUr (r)NaUr (r)
37C, 1,0 lit nc ha tan c ti a 8,0 mmol natri urat. 37C, 1,0 lit
nc ha tan c ti a 8,0 mmol natri urat.
(a) Hy tnh tch s tan ca natri urat. B qua s thy phn ca ion
urat.(a) Hy tnh tch s tan ca natri urat. B qua s thy phn ca ion
urat.Trong mu (c pH = 7,4 v 37C) nng NaTrong mu (c pH = 7,4 v 37C)
nng Na++ l 130 mmol/L.l 130 mmol/L.
(b) Hy tnh nng urat ti a trong mu khng c kt ta natri urat xut
hin.(b) Hy tnh nng urat ti a trong mu khng c kt ta natri urat xut
hin.
Gi tr tch s tan ph thuc vo nhit . Bit thm rng bnh gout thng xut
hinGi tr tch s tan ph thuc vo nhit . Bit thm rng bnh gout thng xut
hin
u tin cc t ngn chn v ngn tay.u tin cc t ngn chn v ngn tay.
(c) Hy cho bit tch s tan ph thuc vo nhit nh th no?(c) Hy cho bit
tch s tan ph thuc vo nhit nh th no?
tan ca axit uric trong nc 37C l 0,5 mmol/L. tan ca axit uric
trong nc 37C l 0,5 mmol/L.(d) Chng minh rng nu khng c kt ta natri
urat xut hin th cng s khng c kt ta(d) Chng minh rng nu khng c kt ta
natri urat xut hin th cng s khng c kt ta
axit uric xut hin.axit uric xut hin.
Gi thit rng ch c HUr v UrGi thit rng ch c HUr v Ur-- l nh hng n
gi tr pH ca dung dch. Si thnl nh hng n gi tr pH ca dung dch. Si
thn
thng c axit uric. Nguyn nhn l nng qu cao ca axit uric v urat c
trong ncthng c axit uric. Nguyn nhn l nng qu cao ca axit uric v
urat c trong nc
tiu v pH thp ca nc tiu (pH = 5 - 6).tiu v pH thp ca nc tiu (pH =
5 - 6).
(e) Hy tnh gi tr pH ti si (cha axit uric khng tan) c hnh thnh t
nc tiu(e) Hy tnh gi tr pH ti si (cha axit uric khng tan) c hnh thnh
t nc tiu
ca bnh nhn. Gi thit rng nng tng cng ca axit uric v urat l 2,0
mmol/L.ca bnh nhn. Gi thit rng nng tng cng ca axit uric v urat l
2,0 mmol/L.
Hng dn gii10
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a. 6,4 10a. 6,4 10-5-5;;
b. 4,910b. 4,910-4-4 M;M;
c. Nhit gim th tch s tan gim.c. Nhit gim th tch s tan gim.
d. Cd. C][
][log
HUr
UrpKpH a
+=
24,54,7][
][log ===
apKpHHUr
Ur
10010][
][ 2==
HUr
Ur
V trong mu khng c kt ta NaUr nn [UrV trong mu khng c kt ta NaUr
nn [Ur--] < 4,910] < 4,910-4-4 (kt qu tnh c phn (b)).(kt qu
tnh c phn (b)).
MSUr
HUr HUr46
4
10.510.9,4100
10.9,4
100
][][
=
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a. CuBr Cu+ + Br-
C: 4-7,4 10.00,210][CuS + ===
Mt khc: 410.00,2143,35V
1S ==
=> V = 34,9 lit;b. CuBr Cu
+ + Br- pT = 7,4
Cu+ + NH3 [Cu(NH3)]+ lg1 = 6,18
[Cu(NH3)]+ + NH3 [Cu(NH3)2]+ lg2 = 4,69
C: [Br -] = [Cu+] + [Cu(NH3)+]+ [Cu(NH3)2+] (1)
[NH3] + [Cu(NH3)+]+ 2[Cu(NH3)2+] = 0,1 (2)
Gi s: [Cu(NH3)2+] >> [Cu+], [Cu(NH3)+]
(1) [Br-] = [Cu(NH3)2+]
(2) [NH3] + 2[Cu(NH3)2+] = 0,1
C:
])2[Br-(0,1]r[
10
]r[
])2[Br-](0,1[Cu
]r[]][NH[Cu
])[Cu(NH
-
-
7,4-
-
-
-
3
232,1
B
BB===
++
+
[Br
] = 0,05 ; [Cu+
] = 1,99.10-6
; [Cu(NH3)2+
] = [Br-
] = 0,057
87,106
2,1
233 10.39,3
1010.99,1
05,0
][Cu
])[Cu(NH][NH
+
+
=
==
[Cu(NH3)+] = 1[Cu+][NH3] = 106,181,99.10-63,39.10-7 =
1,02.10-6
KTGT: tha mn;
05,0][35,143
1
2
=== BrV
S V2 = 0,140 lit
c. T= ([Cu+]+[Cu(NH3)+] + [Cu(NH3)2+]) [Br]
= (1,99106 +3,39107 +0,05) 0,05 = 2,5103
Cu 3.7. Bit tch s tan ca Zn(OH)2 l 1,80 10-17.
(a) Hy tnh tan ca Zn(OH)2 trong nc.
(b) Hy tnh pH ca dung dch Zn(OH)2 bo ha.
Cho cc gi tr th kh chun:
[Zn(OH)4]2- + 2 e Zn (r) + 4 OH- E = -1,285 V
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Zn2+ + 2e Zn (r) E = - 0,762 V
(c) Hy tnh hng s bn tng cng ca phc tetrahidroxozincat(II).
(d) Hy tnh tan ca Zn(OH)2 trong dung dch m c pH = 9,58. B qua s
to phc
[Zn(OH)4]2-.
(e) Hy tnh tan ca Zn(OH)2 trong dung dch m c pH = 9,58 v c tnh n
s tothnh phc [Zn(OH)4]2-.
(g) Hy so snh kt qu tm c (d) v (e) v rt ra nhn xt.
Hng dn gii
a. b qua c s phn li ca nc; S = 1,65.10-6;
b. 8,52;
c. Cch 1: Thit lp cng thc tnh o ZnOHZnE /)( 24 theoo
ZnZnE /2+ .
C 42
4
/
2
// ][
])([lg
2
0592,0]lg[
2
0592,0222
+
+=+=+++
OH
OHZnEZnEE o
ZnZn
o
ZnZnZnZn
42
4
/ ][
])([lg
2
0592,0lg
2
0592,02
+=+
OH
OHZnEo
ZnZn
Khi [Zn(OH)42-] = [OH-] = 1 M th:
lg2
0592,0
//)(/ 2242==
++
o
ZnZn
o
ZnOHZnZnZn EEE
= 4,67.1017;
Cch 2:
Zn(r) + 4 OH- [Zn(OH)4]2- + 2 e E 1 = +1.285 V
G1 = -zFE1 = -247.97 kJ/mol
Zn2+ + 2e- Zn(r) E2 = -0.762 V
G2 = -zFE2 = 147.04 kJ/mol
Zn2+ + 4 OH- [Zn(OH)4]2- G = G1 + G2 = -100.92 kJ/mol
17298314,8
100920
10.90,4===
eeK RT
G
d. MOH
TZn 8
2
2 10.25,1][
][
+== .
e. S = [Zn2+] + [Zn(OH)42-] = [Zn2+] + [Zn2+][OH-]4
= MOHOH
T 842
10.56,2)][1(][
=+ .
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g. Kt qu khc nhau: (2,56- 1,25)/2,56 = 51%; rt ln; nh vy s to
phc nh hng
ng k n tan)
Cu 3.8. (IChO 43) PbO l mt oxit lng tnh. Khi ha tan vo nc xy ra
cc cn
bng:
PbO (r) + H2O Pb2+(aq) + 2 OH- (aq) T = 8,010-16
PbO (r) + 2 H2O Pb(OH)3- (aq) + H3O+ (aq) Ka = 1,010-15
(a) Hy tnh gi tr pH ca dung dch ti dung dch Pb2+ 1,0010-2 M bt u
c kt ta
PbO xut hin?
(b) T gi tr pH tnh c phn (a), ngi ta tng pH ca dung dch n mt gi
tr
nht nh th kt ta bt u tan hon ton. Hy tnh gi tr pH ny?
(c) Hy vit biu thc tnh tan ca PbO.
(d) tan ca PbO t gi tr cc tiu ti pH =9,40. Hy tnh nng ca cc cu t
v
tan ca PbO ti gi tr pH ny.
(e) Hy tnh khong pH ti tan ca PbO nh hn 1,010-3 M.
Hng dn gii
a. [Pb2+][OH-]2 = 8.10-16;
[OH-] = 2,83.10-7 pH = 7,45;
b. [Pb(OH)3-][H3O+] = 1.10-15
[H3O+]= 1.10-13 pH = 13;
c. S = [Pb2+] + [Pb(OH)3-];
d. [Pb2+]= 8.10-16/ [OH-]2 = 1,27.10-6M;
[Pb(OH)3-] = 10-15/ [H3O+]= 2,51.10-6 M;
S = 3,78.10-6M;
M rng: chng minh rng Smin ti gi tr pH = 9,40;
][
10][10.8
][
10
][OH
8.10][Pb(OH)][PbS
15212
15
2-
-16-
3
2
+
+
+
++=+=+=H
HH
0][
10][10.16'S
2
1512
==+
+
HH
[H+]= 3,97.10-10 (pH = 9,40);
e. 315
21210
][
10][10.8S
+
+=+=
HH
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010][10][10.8 153312 = ++ HH
[H+]1 = 1,12.10-8; pH1 = 7,95;
[H+]2 = 1,0.10-12; pH2 = 12,00;
7,95 pH12,00)
3.3. nh hng ng thi ca pH v phn ng to phc
Cu 3.9. Tnh tan ca AgI trong dung dch NH3 0,1M. Bit TAgI =
8,3.10-17; NH3 c Kb
= 1,75.10-5 v:
Ag+ + 2NH3 Ag(NH3)2+ ; 1,2 = 1,7.107
Hng dn gii
Cc cn bng xy ra:
AgI Ag+ + I-
Ag+ + 2 NH3 Ag(NH3)2+
NH3 + H2O NH4+ + OH-
Thit lp cc phng trnh:
T = [Ag+][I-] = 8,3.10-17 (1)
72
3
232,1 10.7,1
]][[])([ ==
+
+
NHAgNHAg (2)
5
3
4 10.75,1][
]][[
+
==
NH
OHNHK
b (3)
S = [I-] = [Ag+] + [Ag(NH3)2+] (4)
[NH3] + 2 [Ag(NH3)2+] + [NH4+] = 0,1 M (5)
[NH4+] = [OH-] (6)
Gi s [NH 4+]
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(2) 7
2
23
1723 10.7,1
1,0])([
10.3,8
])([=
+
+
NHAg
NHAg
[Ag(NH3)2+] = 3,76.10-6 M
11
6
17
23
17
10.21,210.76,310.3,8
])([10.3,8][
+
+
===
NHAgAg
KTGT: tha mn
S = [I-] = [Ag(NH3)2+] = 3,76.10-6M)
Cu 3.10. Tnh nng cn bng ca cc ion Ag+, Br-, Cl-, Ag(NH3)2+, NH4+
v OH-
trong dung dch bo ho AgCl v AgBr vi NH3 0,020M. Gi thit rng phc
Ag(NH3)+
to thnh khng ng k.Cho TAgCl = 10-10; TAgBr = 5.10-13; 1,2 = 108
v Kb= 1,8.10-5.
Hng dn gii
AgCl Ag+ + Cl TAgCl = 10-10
AgBr Ag+ + Br TAgBr= 5.10-13
Ag+ + 2 NH3 Ag(NH3)2+ 1,2 = 108
NH3 + H2O NH4+ + OH- Kb = 1,8.10-5
C: [AgL2] + [Ag+] = [Cl] + [Br] (1)
[NH3] + [NH4+] + 2[AgL2] = 0,02 (2)
Gi s: [Ag+]
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MAg
TBr
AgBr 610.33,8][
][ +
==
[OH-] = [NH4+] = M452 10.48,510.8,110.67,1 =
KTGT: tho mn)
Cu 3.11. Thm 0,1 ml Na2S 1M vo 10 ml dung dch Cu + 10-2M v CN-
1M pH= 12.
Tnh xem c kt ta mu en Cu2S xut hin khng?
Bit: TCu2S= 10-47,6. Phc Cu(CN)43- c 1,4= 1030,3,
HCN c pKa= 9, H2S c pK1= 7 v pK2= 12,9.
Hng dn gii
([S2-]= 1,11.10-3M; [Cu+]= 5,9.10-33 => Q= 3,86.10-68 < T
=> khng c kt ta)
Cu 3.12. Cho TCho TCu(OH)2Cu(OH)2 = 4,50 .10= 4,50 .102121; M;
MWW (Cu(OH)(Cu(OH)22) = 97,59 g.mol) = 97,59 g.mol11
v pKv pKbb (NH(NH33) = 4,76.) = 4,76.
(a) i. Hy tnh tan ca Cu(OH)(a) i. Hy tnh tan ca Cu(OH)22 trong
nc theo n v g/100 mL. B qua qu trnh ttrong nc theo n v g/100 mL. B
qua qu trnh t
phn li ca nc.phn li ca nc.
ii. Hy tnh pH ca dung dch bo ha Cu(OH)ii. Hy tnh pH ca dung dch
bo ha Cu(OH)22..
(b) tan ca nhiu hidroxit kim loi c tng ln nh qu trnh to phc ca
ion kim(b) tan ca nhiu hidroxit kim loi c tng ln nh qu trnh to phc
ca ion kim
loi vi phi t nh amoniac. Trong mt th nghim, ngi ta ha tan hon
ton 5,00 mgloi vi phi t nh amoniac. Trong mt th nghim, ngi ta ha
tan hon ton 5,00 mg
Cu(OH)Cu(OH)22 trong 25,00 mL dung dch NHtrong 25,00 mL dung dch
NH33. Bit nng cn bng ca NH. Bit nng cn bng ca NH33 trong dung
dchtrong dung dch
thu c l 1,00 .10thu c l 1,00 .1033 M, hng s bn tng cng ca phc
Cu(NHM, hng s bn tng cng ca phc Cu(NH33))442+2+ ll 1,41,4 = 10=
1011,7511,75..
i. Hy tnh nng mol tng cng ca ng trong dung dch thu c.i. Hy tnh
nng mol tng cng ca ng trong dung dch thu c.
ii. Hy tnh nng cn bng ca cc cu t cha ng trong dung dch.ii. Hy
tnh nng cn bng ca cc cu t cha ng trong dung dch.
iii. Hy tnh nng cn bng ca NHiii. Hy tnh nng cn bng ca
NH44++..
iv. Hy tnh pH ca dung dch.iv. Hy tnh pH ca dung dch.
v. Hy tnh nng ca dung dch NHv. Hy tnh nng ca dung dch NH33 ban
u.ban u.
Hng dn gii
a. i.a. i. Cu(OH)Cu(OH)22 CuCu2+2+ + 2 OH+ 2 OH--
S 2SS 2S
C:C: 21222 10.50,4)2(]][[ + === SSOHCuT
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MS 7321
10.04,14
10.50,4
==
mlgS 100/10.01,159,971,010.04,1'67
==
ii. C:ii. C: [OH[OH--]= 2S = 2]= 2S = 21,04 .101,04 .10-7-7 =
2,08 .10= 2,08 .10-7-7
pH = 14 + log[OHpH = 14 + log[OH--] = 14 + log (2,08 .10] = 14 +
log (2,08 .10-7-7) = 7,32;) = 7,32;
b. i.b. i. MCCu
3
3
10.05,2025,0
59,97
10.00,5
2
==+
ii.ii. Cu(OH)Cu(OH)22 CuCu2+2+ + 2 OH+ 2 OH-- T = 4,50.10T =
4,50.10-21-21
CuCu2+2+ + 4 NH+ 4 NH33 Cu(NHCu(NH33))442+2+ 1,41,4 = 10=
1011,7511,75
C:C: CCCu2+Cu2+ = [Cu= [Cu2+2+] + [Cu(NH] + [Cu(NH33))442+2+] =
2,05.10] = 2,05.10-3-3 (1)(1)
4
3
2
2
434,1
]][[
])([
NHCu
NHCu+
+
= (2)(2)
(2)(2) ])([778,1)10(10
])([
][
])([][ 2434375,11
2
43
4
34,1
2
432 +
++
+
=
== NHCuNHCu
NH
NHCuCu
(3)(3)
(1, 3)(1, 3) [Cu[Cu2+2+] = 1,31.10] = 1,31.10-3-3
[Cu(NH[Cu(NH33))442+2+] = 7,38.10] = 7,38.10-4-4
iii.iii. NHNH33 + H+ H22OO NHNH44++ + OH+ OH-- KKbb
=10=10-4,76-4,76
C:C:][
][
][
]][[
3
2
4
3
4
NH
NH
NH
OHNHKb
++
==
MNHKNH b4376,4
34 10.32,110.00,110][][+
===
iv. C:iv. C: [OH[OH--] = [NH] = [NH44++]= 1,32.10]=
1,32.10-4-4
pH = 14 + log[OHpH = 14 + log[OH--] = 14 + log(1,32.10] = 14 +
log(1,32.10-4-4) = 10,12) = 10,12
v.v. CCNH3NH3 = [NH= [NH33] + [NH] + [NH44++] + 4[Cu(NH] +
4[Cu(NH33))442+2+] = 1.10] = 1.10-3-3 + 1,32.10+ 1,32.10-4-4 + 4+
47,38.107,38.10-4-4 ==4,08.104,08.10-3-3 M)M)
4. Xc nh tch s tan
4.1. Da vo tan
Cu 4.1. Thm t t dung dch bari nitrat 0,0010 M vo 200 ml dung dch
NaF 0,040 M.
Khi 35 ml dung dch bari nitrat c thm vo th thy kt ta BaF2 bt u
xut hin.
Hy tnh tch s tan ca BaF2.p s: (1,72.10-7)
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Cu 4.2. Dung dch bo ha Cd(OH)2 c pH = 9,56. Hy tnh tch s tan ca
Cd(OH)2.
p s: (2,39.10-14)
Cu 4.3. Bit 1 lit dung dch NH3 1M ha tan c ti a 0,33 gam AgBr.
Hy tnh
TAgBr. Bit phc Ag(NH3)2+ c 1,2 = 5,88.106.
Hng dn gii
[Ag(NH3)2+] = [Br-] = 0,33/188 = 1,76.10-3 M.
[NH3] = 1 2[Ag(NH3)2+] = 0,996 M
10
2
32,1
23 10.02,3][
])([][
+
+
==
NH
NHAgAg
T = [Ag+][Br-]= 5,32.10-13)
Cu 4.4. Tnh tch s tan ca Ca(IO3)2
Th nghim 1. Chun ha dung dch natri thiosunfat.
Ly 10,0 ml dung dch KIO3 0,0120 M cho vo bnh nn. Thm 2 gam KI v
10 ml
dung dch HCl 1M. Dung dch c mu nu thm. Chun bng dung dch Na
2S2O3 n
mu vng rm. Thm 5 ml h tinh bt v tip tc chun n mt mu xanh ca
phc
tinh bt vi I3- thy ht 20,55 ml.
Th nghim 2. Tnh tan ca Ca(IO3)2 trong nc ct.Ly 10,0 ml dung dch
bo ha Ca(IO3)2 cho vo bnh nn. Thm 2 gam KI v 10 ml
HCl 1M. Tin hnh chun dung dch thu c bng dung dch Na2S2O3 trn thy
ht
19,20 ml. Hy:
(a) vit cc phn ng c m t trong th nghim.
(b) tnh nng dung dch Na2S2O3.
(c) tnh nng ca IO3-
.(d) tnh tan ca Ca(IO3)2 trong nc.
(e) tnh tch s tan ca Ca(IO3)2.
p s: a. IO3- + 5 I + 6 H+ 3 I2 + 3 H2O
I2 + 2 S2O32- 2 I + S4O62-
b. 0,0350M;
c. 0,0122 M.d. 5,6.10-3 M.
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e. 7,1.10-7)
4.2. Da vo gi tr th kh chun
Cu 4.5. Cho VEo
HgHg789,0
/22=
+ ; VEo HgClHg 268,0/22 = . Hy tnh tch s tan v tan ca
Hg2Cl2.
Hng dn gii
Hg22+ + 2 e 2 Hg
Hg22+ + 2 Cl- Hg2Cl2 ; T = [Hg22+][Cl-]2
Thit lp cng thc tnh o HgClHgE /22 theoo
HgHgE
/22+ .
C 2/22// ][
lg2
0592,0]lg[
2
0592,022
22
22
+
+=+=+++
Cl
TEHgEE o
HgHg
o
HgHgHgHg
]lg[0592,0lg2
0592,0/
22
+= + ClTE
o
HgHg
Khi [Cl] = 1 M th:
TEEE oHgHg
o
HgHgClHgHgCllg
2
0592,0/// 2222
+==+
T = 2,51.10-18
C S(2S)2 = T MT
S 73 10.56,8
4
== )
Cu 4.6. Cho 3 pin in ha vi cc sc in ng tng ng 298K:
(1) Hg/HgCl2, KCl (bo ha ) // Ag+ (0,0100 M)/Ag E1= 0,439 V
(2) Hg/HgCl2, KCl (bo ha ) // AgI (bo ha)/Ag E2= 0,089 V
(3) Ag/AgI (bo ha), PbI2 (bo ha ) // KCl (bo ha), HgCl2/Hg E3=
0,230 V
a) Hy tnh tch s tan ca bc idodua.
b) Hy tnh tch s tan ca ch (II) iodua.Cho VE
o
AgAg799,0
/=+ , R = 8,314 J/mol/K, F = 96487 C/mol.
p s: a) 1,37.10-16;
b) [Ag+] = 4,58.10-14; [I] = 2,99.10-3;
[Pb2+]= 0,5 ([I] [Ag+]) = 1,5.10-3; T = 1,34.10-8
Cu 4.7. (IChO 42) Cho cc gi tr th kh chun sau:(IChO 42) Cho cc
gi tr th kh chun sau:
Bn phn ngBn phn ng EE00
, V (298K), V (298K)SnSn2+2+ + 2e+ 2e SnSn -0,14-0,14
SnSn4+4+ + 2e+ 2e SnSn2+2+ +0,15+0,15
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HgHg222+2+ + 2e+ 2e 2 Hg2 Hg +0,79+0,79
HgHg22ClCl22 + 2e+ 2e 2 Hg2 Hg
+ 2 Cl+ 2 Cl
+0,27+0,27
(a) Hy tnh hng s cn bng ca phn ng sau 298 K:(a) Hy tnh hng s cn
bng ca phn ng sau 298 K:
Sn (r) + SnSn (r) + Sn4+4+ (aq)(aq) 2 Sn2 Sn2+2+ (aq)(aq)
(b) Hy tnh tan ca Hg(b) Hy tnh tan ca Hg22ClCl22 trong nc 298 K
(theo n v mol/l).trong nc 298 K (theo n v mol/l).
(c) Hy tnh sut in ng chun, E, ca pin nhin liu s dng phn ng
sau:(c) Hy tnh sut in ng chun, E, ca pin nhin liu s dng phn ng
sau:
HH22 (k) + 1/2 O(k) + 1/2 O22 (k)(k) HH22O (l)O (l) G = 237,1
kJ.molG = 237,1 kJ.mol11
Hng dn gii
a.a. 90592,0)14,015,0(2
10.27,610 ==
+
K
b. Xc nh tch s tan:b. Xc nh tch s tan:Cch 1:Cch 1:
-1-1 HgHg222+2+ + 2e+ 2e 2 Hg2 Hg
GG0011 = -nFE= -nFE0011 = -2= -2 96485964850,79 = -152,4.100,79
= -152,4.1033 JJ
11 HgHg22ClCl22 + 2e+ 2e 2 Hg + 2 Cl2 Hg + 2 Cl--
GG0022 = -nFE= -nFE0022 = -2= -2 96485964850,27 = -52,1.100,27 =
-52,1.1033 JJ
HgHg22ClCl22 HgHg222+2+ + 2 Cl+ 2 Cl--
GG0033 = -= -GG0011++ GG0022 = 100,3.10= 100,3.1033 J = -RTlnTJ
= -RTlnT
48,40298314,8
10.3,100ln
30
3=
=
=RT
GT
T = 2,62.10T = 2,62.10-18-18
Cch 2:Cch 2:
Hg22+ + 2 e 2 Hg
Hg22+ + 2 Cl- Hg2Cl2 ; T = [Hg22+][Cl-]2
Thit lp cng thc tnh o HgClHgE /22 theoo
HgHgE
/22+ .
C 2/22// ][
lg2
0592,0]lg[
2
0592,022
22
22
+
+=+=+++
Cl
TEHgEE o
HgHg
o
HgHgHgHg
]lg[0592,0lg2
0592,0/
22
+= + ClTE
o
HgHg
21
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Khi [Cl-] = 1 M th:
TEEE oHgHg
o
HgHgClHgHgCllg
2
0592,0/// 2222
+==+
T = 2,71.10-18
Tnh tan:
C HgHg22ClCl22 HgHg222+2+ + 2 Cl+ 2 Cl--
S 2SS 2S
S(2S)2 = T
MT
S 7318
3 10.68,84
10.62,2
4
===
c. C:c. C: GG00 = -nFE= -nFE00pinpin
VnF
GEpin 23,1
964852
10.1,237 30
0 =
=
=
Cu 4.8. K thut in ho thng c dng xc nh tan ca cc mui kh tan.
Do sc in ng l hm bc nht theo logarit ca nng nn c th xc nh c
cc
nng d rt nh.
Bi tp ny s dng mt pin in ho gm hai phn, c ni vi nhau bng cumui.
Phn bn tri ca s pin l mt thanh Zn nhng trong dung dch Zn(NO 3)2
0,200M; cn phn bn phi l mt thanh Ag nhng trong dung dch AgNO 3
0,100M. Mi
dung dch c th tch 1,00L 250C.
(a) V s pin v vit cc bn phn ng xy ra mi cc.
(b) Hy tnh sc in ng ca pin v vit phng trnh phn ng xy ra khi pin
phng
in.Gi s pin phng in hon ton v lng Zn c d.
(c) Hy tnh in lng c phng thch trong qu trnh phng in.
Trong mt th nghim khc, KCl c thm vo dung dch AgNO3 pha bn
phi
ca pin ban u. Xy ra phn ng to kt ta AgCl v lm thay i sc in ng.
Sau khi
thm xong. Sc in ng bng ca pin bng 1,04V v [K+] = 0,300M.
(d) Hy tnh [Ag+
] ti trng thi cn bng.(e) Hy tnh [Cl-] ti trng thi cn bng v
TAgCl.
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Cho: EoZn2+/Zn= -0,76V; EoAg+/Ag= 0,80V.
p s: a. Zn| Zn2+||Ag+ |Ag b. 1,52V
c. 9649C d. 7,3.10-10M
e. [Cl-] = 0,2M; T = 1,5.10-10)
Cu 4.9. Xem xt pin in ha sau:Xem xt pin in ha sau:
Pt |HPt |H22 (p = 1 atm)|H(p = 1 atm)|H22SOSO44 0,01 M|PbSO0,01
M|PbSO44(r)|Pb(r).(r)|Pb(r).
(a) Hy tnh nng cn bng ca SO(a) Hy tnh nng cn bng ca SO442-2- v
pH ca dung dch trong pin trn.v pH ca dung dch trong pin trn.
(b) Hy vit phn ng xy ra khi pin phng in.(b) Hy vit phn ng xy ra
khi pin phng in.
Sut in ng ca pin trn 298,15 K l 0,188 V. Gi thit rng trong phn
(c) vSut in ng ca pin trn 298,15 K l 0,188 V. Gi thit rng trong phn
(c) v
(d) nng cn bng ca SO(d) nng cn bng ca SO442-2- l 510l 510-3-3 M
v ca HM v ca H33OO++ l 1510l 1510-3-3 M (cc gi tr ny c thM (cc gi
tr ny c th
khc gi tr tnh c phn (a)).khc gi tr tnh c phn (a)).(c) Hy tnh tch
s tan ca PbSO(c) Hy tnh tch s tan ca PbSO44..
(d) Hy cho bit sut in ng ca pin trn tng hay gim bao nhiu V khi p
sut ca(d) Hy cho bit sut in ng ca pin trn tng hay gim bao nhiu V
khi p sut ca
hidro gim mt na?hidro gim mt na?
Vng kim loi khng tan trong dung dch axit nitric nhng tan c trong
ncVng kim loi khng tan trong dung dch axit nitric nhng tan c trong
nc
cng toan (l hn hp gm axit clohidric c v axit nitric c c t l th
tch tng ngcng toan (l hn hp gm axit clohidric c v axit nitric c c t
l th tch tng ng
l 3 : 1). Vng phn ng vi nc cng toan to thnh ion phc [AuCll 3 :
1). Vng phn ng vi nc cng toan to thnh ion phc [AuCl44]]--
..(e) S dng cc gi tr th kh chun cho di y, hy tnh hng s bn tng
cng ca(e) S dng cc gi tr th kh chun cho di y, hy tnh hng s bn tng
cng ca
phc [AuClphc [AuCl44]]--..
Cho:Cho: pKpKa2a2 (H(H22SOSO44) = 1,92;) = 1,92;
E(PbE(Pb2+2+/Pb) = - 0,126 V/Pb) = - 0,126 V
E(AuE(Au3+3+/Au) = + 1,50 V/Au) = + 1,50 V
E([AuClE([AuCl44]]--/Au + 4 Cl/Au + 4 Cl--) = + 1,00 V) = + 1,00
V
Hng dn gii
a.a. HH22SOSO44 HH++
+ HSO+ HSO44--
0,01 0,010,01 0,01
HSOHSO44-- HH++ + SO+ SO442-2-
cb 0,01 - x 0,01 + x xcb 0,01 - x 0,01 + x x
92,1
2 1001,0
)01,0( =
+=
x
xxKa
x = 4,53.10x = 4,53.10-3-3 MM
[SO[SO442-2-]= x = 4,53.10]= x = 4,53.10-3-3 MM
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[H+] = 0,01 + x = 0,0145 M[H+] = 0,01 + x = 0,0145 M
pH = 1,84;pH = 1,84;
b.b. Catot (+)Catot (+) PbSOPbSO44 + 2e+ 2e Pb + SOPb +
SO442-2-
Anot (-)Anot (-) HH22 2 H2 H++ + 2e+ 2e
PbSOPbSO44 + H+ H22 Pb + 2 HPb + 2 H++ + SO+ SO442-2-
c. C:c. C: ]lg[2
0592,0 20/
2
++= + PbEE PbPbcatot
Vp
HEE
HHHanot
108,01
)10.15(lg
2
0592,00
][lg
2
0592,0 2320/
2
2
=+=+=
+
+
188,0108,0]lg[20592,0
20/
2 =++==+
+ PbEEEE PbPbanotcatotpin
[Pb[Pb2+2+]= 1,81.10]= 1,81.10-6-6 MM
T = [PbT = [Pb2+2+][SO][SO442-2-] = 1,81.10] = 1,81.10-6-6
5.105.10 = 9,05.10= 9,05.10-9-9
d. Khi p sut ca hidro gim mt na:d. Khi p sut ca hidro gim mt
na:
Vp
HEE
HHHanot
099,05,0
)10.15(lg
2
0592,00
][lg
2
0592,0 2320/
2
2
=+=+=
+
+
EEcatotcatot khng i, Ekhng i, Eanotanot tng (-0,099 + 0,108) =
0,009V, vy Etng (-0,099 + 0,108) = 0,009V, vy Epinpin gim 0,009
V;gim 0,009 V;
e.e. +1+1 AuAu3+3+ + 3e+ 3e AuAu
GG0011 = -nFE= -nFE0011 = -3= -3 96485964851,50 = -434,2.101,50
= -434,2.1033 JJ
-1-1 AuClAuCl44-- + 3e+ 3e Au + 4 ClAu + 4 Cl--
GG0022 = -nFE= -nFE0022 = -3= -3 96485964851,00 = -289,5.101,00
= -289,5.1033 JJ
AuAu3+3+ + 4Cl+ 4Cl-- AuClAuCl44--
GG0033 == GG0011-- GG0022 = -144,7.10= -144,7.1033 J = -RTlnJ =
-RTln1,41,4
4,54298314,8
10.7,144ln
30
34,1 =
=
=RT
G
1,41,4 = 2,31.10= 2,31.102525))
