Chuyên đề Phương pháp giải bài tập hidroxit lưỡng tính.
27
Trường THPT Chơn Thành--------------------------- Phương pháp giải bài tập hiđroxit lưỡng tính CHUYÊN !" PHƯƠNG PHÁP GIẢI BÀI TẬP HIĐROXIT LƯỠNG TÍNH #$" Tr%n &' Phương------ -------------------------------- --------------------------------------------- 1
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Transcript of Chuyên đề Phương pháp giải bài tập hidroxit lưỡng tính.
A
Trng THPT Chn ThnhPhng php gii bi tp hiroxit lng tnh
CHUYN :PHNG PHP GII BI TP
HIROXIT LNG TNHA. C S L THUYT
I. Khi nim v hiroxit lng tnh
1. Theo thuyt A-re-ni-ut
Hiroxit lng tnh l hiroxit khi tan trong nc va c th phn li nh axit, va c th phn li nh baz.
V d: Zn(OH)2
- Phn li theo kiu baz: Zn(OH)2 Zn2+ + 2OH-
- Phn li theo kiu axit:Zn(OH)2 2H+ +
2. Theo thuyt Bron-stet
Hiroxit lng tnh l hiroxit va c th nhn H+, va c th nhng H+.
V d: Zn(OH)2
- Kh nng nhn H+: Zn(OH)2 + 2H3O+ Zn2+ + 4H2O- Kh nng nhng H+:Zn(OH)2 + 4H2O [Zn(OH)4]2- + 2H3O+Tm li: Hiroxit lng tnh l hiroxit va c kh nng phn ng vi axit, va c kh nng phn ng vi baz.
V d: Zn(OH)2.
Zn(OH)2 + 2HCl ( ZnCl2 + 2H2O
Zn(OH)2 + 2NaOH ( Na2ZnO2 + 2H2O hoc Zn(OH)2 + 2NaOH ( Na2[Zn(OH)4]
3. Mt s hiroxit lng tnh thng gpDng bazDng axit
Hiroxit kim loi ha tr 2M(OH)2H2MO2
Zn(OH)2H2ZnO2
Sn(OH)2H2SnO2
Pb(OH)2H2PbO2
Be(OH)2H2BeO2
Hiroxit kim loi ha tr 3M(OH)3HMO2.H2O
Al(OH)3HAlO2.H2O
Cr(OH)3HCrO2.H2O
II. Cc dng ton thng gp1. Dng 1: Thm dung dch baz (OH-) vo dung dch mui Al3+ hoc Zn2+a. Dung dch mui Al3+
Hin tng: u tin c kt ta trng Al(OH)3 xut hin, sau kt ta tan dn khi OH- d:
Al3+ + 3OH- Al(OH)3(
(1)
Al(OH)3 + OH- Al(OH)4-hoc Al(OH)3 + OH- AlO2- + 2H2O (2)
Al3+ + 4OH- Al(OH)4-
hoc Al3+ + 4OH- AlO2- + 2H2O(3)t
Al(OH)3Al(OH)4- hoc AlO2-
3
4Nhn xt:- T = 3 => = 3: Lng kt ta cc i tnh theo (1)
- T ( 4 => ( 4: Lng kt ta cc tiu tnh theo (3)
- T < 4 => < 4: iu kin c kt ta
Nu < : ng vi mi gi tr c th c tng ng 2 gi tr khc nhau.
+ Trng hp 1: Kt ta ng vi gi tr cc i, ch xy ra phn ng (1): = 3 (Lng OH- tiu tn t nht).
+ Trng hp 2: Kt ta cn li sau khi b ha tan mt phn, xy ra phn ng (1) v (2): (lng OH- tiu tn nhiu nht).CC CCH GII:
1. Cch gii thng thng (theo phng trnh ion).Al3+ + 3OH- Al(OH)3(
Al(OH)3 + OH- Al(OH)4- hoc Al(OH)3 + OH- AlO2- + 2H2O
2. S dng s v p dng bo ton nguyn t vi Al v OH-Al3+ + OH- ( Al(OH)3( + Al(OH)4- hoc Al3+ + OH- ( Al(OH)3( + AlO2- + H2O- Bo ton nguyn t Al:
=
- Bo ton nhm OH-:
=
3. Theo cng thc tnh nhanh:
- Lng OH- tiu tn t nht:
- Lng OH- tiu tn nhiu nht:
Ch : Nu cho NaOH vo hn hp gm (mui Al3+ v axit H+) th cng thm s mol H+ v 2 cng thc trn, tc l:
+
+
4. Phng php dng th
Rt t t dung dch kim n d vo dung dch cha a mol mui Al3+. Sau phn ng thu c b mol kt ta.
S mol Al(OH)3
S mol OH- x 3a y 4a S mol OH-(min) phn ng l: x = 3b (mol)
S mol OH-(max) phn ng l: y = 4a - b (mol).b. Dung dch mui Zn2+Hin tng: u tin c kt ta trng Zn(OH)2 xut hin, sau kt ta tan dn khi OH- d:
Zn2+ + 2OH- Zn(OH)2(
(4)
Zn(OH)2 + 2OH- Zn(OH)42-hoc Zn(OH)2 + 2OH- ZnO22- + 2H2O(5)
Zn2+ + 4OH- Zn(OH)42-hoc Zn2+ + 4OH- ZnO22- + 2H2O(6)
t
Zn(OH)2Zn(OH)42- hoc ZnO22-
2
4
Nhn xt:
- T = 2 => = 2: Lng kt ta cc i tnh theo (4)
- T ( 4 => ( 4: Lng kt ta cc tiu tnh theo (6)
- T < 4 => < 4: iu kin c kt ta
Nu < : ng vi mi gi tr c th c tng ng 2 gi tr khc nhau.
+ Trng hp 1: Kt ta ng vi gi tr cc i, ch xy ra phn ng (1): = 2 (Lng OH- tiu tn t nht).
+ Trng hp 2: Kt ta cn li sau khi b ha tan mt phn, xy ra phn ng (4) v (5): (lng OH- tiu tn nhiu nht).CC CCH GII1. Cch gii thng thng (theo phng trnh ion).Zn2+ + 2OH- Zn(OH)2(
Zn(OH)2 + 2OH- Zn(OH)42- hoc Zn(OH)2 + 2OH- ZnO22- + 2H2O2. S dng s v p dng bo ton nguyn t vi Al v OH-Zn2+ + OH- ( Zn(OH)2( + Zn(OH)42- hoc Zn2+ + OH- ( Zn(OH)2( + ZnO22- + H2O- Bo ton nguyn t Zn:
=
- Bo ton nhm OH-:
=
3. Theo cng thc tnh nhanh:
- Lng OH- tiu tn t nht:
- Lng OH- tiu tn nhiu nht:
Ch : Nu cho NaOH vo hn hp gm (mui Zn2+ v axit H+) th cng thm s mol H+ vo 2 cng thc trn, tc l:
- +
- +
4. Phng php dng th:Rt t t dung dch kim n d vo dung dch cha a mol mui Zn2+. Sau phn ng thu c b mol kt ta.
S mol Zn(OH)2
S mol OH- x 2a y 4a
S mol OH-(min) phn ng l: x = 2b (mol)
S mol OH-(max) phn ng l: y = 4a - 2b (mol)MT S LU :1. Al(OH)3 ni ring v hiroxit lng tnh ni chung ch tan trong axit mnh v baz mnh, khng tan trong axit yu (NH4+ hoc H2CO3) v baz yu (NH3, amin, CO32-), do :- Khi cho t t kim vo mui Al3+ th lng kt ta tng dn n cc i sau s gim dn v tan ht nu kim d.
- Khi thay kim bng dung dch NH3 th lng kt ta tng dn n cc i v khng b ha tan khi NH3 d (ring Zn(OH)2 l hiroxit lng tnh nhng tan c trong NH3 l do to phc tan [Zn(NH3)4](OH)2).
2. Khi cho kim tc dng vi dung dch hn hp gm H+ v Al3+ th cc phn ng xy ra theo th t:
OH- + H+ H2O
3OH- + Al3+ Al(OH)3(
OH- + Al(OH)3 Al(OH)4-
2. Dng 2: Thm dung dch axit (H+) vo dung dch aluminat Al(OH)4- (AlO2-) hoc dung dch zincat Zn(OH)42- (ZnO22-).a. Dung dch aluminat Al(OH)4- (AlO2-):
Hin tng: u tin c kt ta keo trng Al(OH)3 xut hin. Khi lng Al(OH)4- ht, lng H+ d ha tan kt ta:
Al(OH)4- + H+ Al(OH)3( + H2O hoc AlO2- + H+ + H2O Al(OH)3((7)
Al(OH)3 + 3H+ Al3+ + 3H2O
(8)
Al(OH)4- + 4H+ Al3+ + 4H2O hoc AlO2- + 4H+ Al3+ + 2H2O(9)t
Al(OH)3 Al3+
1
4Nhn xt:- T = 1 => = : Lng kt ta cc i, tnh theo (7)
- T ( 4 => (: Lng kt ta cc tiu, tnh theo (9)
- T < 4 => < : iu kin c kt ta.
ng vi mi gi tr c th c tng ng 2 gi tr khc nhau.
Cch gii tng t nh dng 1, ch c 1 s lu sau:
- S dng s :
Al(OH)4- + H+ Al(OH)3( + Al3+
Thng bi cho bit s mol Al(OH)3, p dng phng php bo ton nguyn t Al s tnh c:
Bo ton cation H+:
- S dng cng thc tnh nhanh:
+ Lng H+ tiu tn nh nht: =
+ Lng H+ tiu tn ln nht:
Ch : Nu cho dung dch axit (H+) vo hn hp gm mui Al(OH)4- v baz OH- th cng thm mol OH- vo 2 cng thc trn, tc l:
=+
v +
b. Dung dch zincat Zn(OH)42- (ZnO22-).Hin tng: u tin c kt ta keo trng Zn(OH)2 xut hin. Khi lng Zn(OH)42- ht, lng H+ d ha tan kt ta:
Zn(OH)42- + 2H+ Zn(OH)2( + 2H2O hoc ZnO22- + 2H+ Zn(OH)2((10)
Zn(OH)2 + 2H+ Zn2+ + 2H2O
(11)
Zn(OH)42- + 4H+ Zn2+ + 4H2Ohoc ZnO22- + 4H+ Zn2+ + 2H2O(12)
t
Zn(OH)2 Zn2+
2
4Nhn xt:
- T = 2 => = : Lng kt ta cc i, tnh theo (10)
- T ( 4 => (: Lng kt ta cc tiu, tnh theo (12)
- T < 4 => < : iu kin c kt ta.
ng vi mi gi tr c th c tng ng 2 gi tr khc nhau.
Cch gii tng t nh dng 1, ch c 1 s lu sau:
- S dng s :
Zn(OH)42- + H+ Zn(OH)2( + Zn2+
Thng bi cho bit s mol Zn(OH)2, p dng phng php bo ton nguyn t Zn s tnh c:
Bo ton cation H+:
- S dng cng thc tnh nhanh:
+ Lng H+ tiu tn nh nht: =
+ Lng H+ tiu tn ln nht:
Ch : Nu cho dung dch axit (H+) vo hn hp gm mui Zn(OH)42- v baz OH- th cng thm mol OH- vo 2 cng thc trn, tc l:
= 2+
v +
MT S LU :
- Khi cho t axit H+ vo dung dch mui aluminat (hoc mui zincat) th lng kt ta tng dn n cc i, sau s tan mt phn hay hon ton ty thuc vo lng H+ d.
- Khi thay axit bng mui NH4+ hay sc kh CO2 d th lng kt ta tng dn n cc i v khng b ha tan. Sc kh CO2 d s to mui HCO3- ch khng phi mui CO32-.
Khi cho axit H+ tc dng vi dung dch hn hp gm OH- v Al(OH)4- th cc phn ng xy ra theo th t:
OH- + H+ H2O
Al(OH)4-+ H+ Al(OH)3( + H2O
Al(OH)3 + 3H+ Al3+ + 3H2O
- Cc cng thc tnh nhanh ch p dng gii hn cho mt s bi, cn nm vng c bn cht, th t phn ng c th gii tt c cc bi tp v Al(OH)3 mt cch linh hot.
B. BI TP MINH HA
Bi 1. Trn dung dch cha a mol AlCl3 vi dung dch cha b mol NaOH. thu c kt ta th cn c t l:A. a:b < 1:4B. a:b = 1:4C. a:b ( 1:3D. a:b > 1:4Phn tch, hng dn gii:Al(OH)3Al(OH)4- hoc AlO2-
3
4T s : thu c kt ta th => b:a < 4 ( a:b >1:4Vy p n l D.Bi 2. Cho 200 ml dung dch AlCl3 1,5M tc dng vi V lt dung dch NaOH 0,5M thu c 15,6 gam kt ta. Gi tr ln nht ca V l:A. 1,2B. 1,8C. 2,0D. 2,4Phn tch, hng dn gii:
EMBED Equation.DSMT4 - Cch 1: Gii thng thng theo phng trnh ion:V(max) (s mol NaOH ln nht) xy ra trng hp 2: Kt ta Al(OH)3 b ha tan 1 phn cn li 15,6g kt ta.Al3+ + 3OH- Al(OH)3((1)0,3 mol 0,9 mol 0,3 mol
Al(OH)3 + OH- Al(OH)4-(2)
(0,3-0,2) mol 0,1 molTheo (1) v (2): (nNaOH = 0,9 + 0,1 = 1 mol
Vy V(max) = => p n C.- Cch 2: S dng s v bo ton nguyn t Al, bo ton nhm OH-:V(max) (s mol NaOH ln nht) xy ra trng hp 2: Kt ta Al(OH)3 b ha tan 1 phn cn li 15,6g kt ta.
S :
Al3+ + OH- ( Al(OH)3( + Al(OH)4- + Theo bo ton nguyn t Al:
= => = 0,3 0,2 = 0,1 mol
+ Bo ton nhm OH-:(== 3.0,2 + 4.0,1 = 1,0 mol
Vy V(max) = => p n C.- Cch 3: S dng cng thc tnh nhanh:Lng OH- tiu tn nhiu nht: = 4.0,3 0,2 =1,0 molVy V(max) = => p n C.- Cnh 4: S dng th:S mol Al(OH)3
S mol OH- x 3a=3.0,3 y 4b=4.0,3 T th: Lng OH- tiu tn nhiu nht: y = 4a b = 4.0,3 0,2 = 1 molVy V(max) = => p n C.Bi 3. Cho a mol AlCl3 vo 1 lt dung dch NaOH c nng c (mol/l) c 0,05 mol kt ta, thm tip 1 lt dung dch NaOH trn th thu c 0,06 mol kt ta. Gi tr ca a v c ln lt l:A. 0,15 v 0,06B. 0,09 v 0,18C. 0,09 v 0,15D. 0,06 v 0,15Phn tch, hng dn gii:- Cch 1: Gii thng thng theo phng trnh ion:+ Khi thm tip NaOH thu c thm kt ta => trc khi thm Al3+ cn d, NaOH ban u ht:Al3+ + 3OH- Al(OH)3(
c molc/3 mol
=> c/3 = 0,05 => c = 0,15 M+ Khi thm tip 1 lt NaOH cM m lng kt ta thu c ch tng 0,01 mol kt ta b ha tan mt phn:Al3+ + 3OH- Al(OH)3(a 3a
a
Al(OH)3 + OH- Al(OH)4-
(a-0,06)(a-0,06)(nNaOH p = 3a + (a 0,06) = 2c = 2.0,15 = 0,3
=> a = 0,09 mol
Vy p n C.- Cch 2: S dng s v bo ton nguyn t Al, bo ton nhm OH-:+ Khi thm tip NaOH thu c thm kt ta => trc khi thm Al3+ cn d, NaOH ban u ht:
Al3+ + 3OH- Al(OH)3(
c molc/3 mol
=> c/3 = 0,05 => c = 0,15 M+ Khi thm tip 1 lt NaOH cM kt ta b ha tan mt phn. Theo bo ton nguyn t Al:
= => = a 0,06
+ Bo ton nhm OH-:(== 3.0,06 + 4.(a 0,06) = 2c = 0,3
=> a = 0,09 Vy p n C.
- Cch 3: S dng cng thc tnh nhanh:+ Lng OH- tiu tn t nht: => c = 3.0,05 =0,15 M
+ Lng OH- tiu tn nhiu nht:
( 2.0,15 = 4a 0,06
=> a = 0,09 molVy p n C.- Cch 4: S dng th:S mol Al(OH)3
S mol OH- x=c 3a y=2c 4a + S mol OH- phn ng to ra b = 0,05 mol kt ta l: x = 3b ( c = 3.0,05 = 0,15
+ Khi thm NaOH, tng s mol NaOH phn ng to ra b = 0,06 mol kt ta l y = 4a b
=> 2c = 2.0,15 = 4a 0,06 => a = 0,09Vy p n C.Bi 4. X l dung dch AlCl3, Y l dung dch NaOH 2M. Cho 150 ml dung dch Y vo cc cha 100 ml dung dch X , khuy u n phn ng hon ton thy trong cc c 7,8 gam kt ta. Thm tip vo cc 100 ml dung dch Y khuy u n khi kt thc phn ng thy trong cc c 10,92 gam kt ta. Nng mol ca dung dch X l:
A. 3,2MB. 2,0MC. 1,6MD. 1,0MPhn tch, hng dn gii:Ta c n((1) = 0,1 mol; n((2) = 0,14 mol
(= 0,3 + 0,2 = 0,5 mol => n((2) > 0,14
Trc khi thm NaOH th cn d Al3+, sau khi thm NaOH kt ta b ha tan 1 phn. Gi a l s mol AlCl3 ban u.
- Cch 1: Gii thng thng theo phng trnh ion:Al3+ + 3OH- Al(OH)3(a 3a
a
Al(OH)3 + OH- Al(OH)4-
(0,5-3a) (0,5-3a)
S mol kt ta cn li = a (0,5-3a) = 0,14 => a = 0,16 mol
=>
Vy p n C.Cch 2: S dng s v bo ton nguyn t Al, bo ton nhm OH-:Theo bo ton nguyn t Al:
= => = a 0,14
+ Bo ton nhm OH-:
(== 3.0,14 + 4.(a 0,14) = 0,5
=> a = 0,16 =>
Vy p n C.- Cch 3: S dng cng thc tnh nhanh:Lng OH- tiu tn nhiu nht:
( 0,5 = 4a 0,14
=> a = 0,16 mol=>
Vy p n C.- Cch 4: S dng th: S mol Al(OH)3
S mol OH- x 3a y 4a Sau khi thm OH-, lng kt ta thu c l b = 0,14S mol OH- phn ng l: y = 4a b
( 0,5 = 4a 0,14=> a = 0,16 mol=>
Vy p n C.Bi 5. (H 2009A). Ho tan ht m gam ZnSO4 vo nc c dung dch X. Cho 110 ml dung dch KOH 2M vo X, thu c a gam kt ta. Mt khc, nu cho 140 ml dung dch KOH 2M vo X th cng thu c a gam kt ta. Gi tr ca m lA. 20,125.B. 22,540.C. 12,375.D. 17,710.Phn tch, hng dn gii:
Ta c: nKOH(1) = 0,22 mol; nKOH(2) = 0,28 mol
Trong c 2 trng hp u thu c a gam kt ta => trng hp 1 cn d Zn2+, trng hp 2 kt ta Zn(OH)2 b ha tan 1 phn.
Gi s mol ban u ca Zn2+ l x- Cch 1: Gii thng thng theo phng trnh ion:+ Trng hp 1: Zn2+ d, OH- phn ng htZn2+ + 2OH- ( Zn(OH)2(
0,22 mol 0,11 mol+ Trng hp 2: Zn2+ phn ng ht, kt ta tan 1 phn:
Zn2+ + 2OH- ( Zn(OH)2(
x mol 2x mol x mol
Zn(OH)2 + 2OH- ( Zn(OH)42-
(0,14-x)(0,28-2x)Ta c: = x (0,14 x) = 0,11
=> x = 0,125 mol
=>
Vy p n A.
Cch 2: S dng s v bo ton nguyn t Zn, bo ton nhm OH-:Zn2+ + OH- ( Zn(OH)2( + Zn(OH)42- Trong trng hp 1: => = 0,22/2 =0,11 molTrong trng hp 2:Bo ton nguyn t Zn:
=( = x 0,11Bo ton nhm OH-:
=
( 0,28 = 2.0,11 + 4(x 0,11)
=> x = 0,125
=>
Vy p n A.- Cch 3: S dng cng thc tnh nhanh:- Lng OH- tiu tn t nht: => = 0,22/2 =0,11 mol- Lng OH- tiu tn nhiu nht:
0,28 = 4x 2.0,11
=> x = 0,125 mol
=>
Vy p n A.- Cch 4: S dng th: S mol Zn(OH)2
S mol OH- x 2a y 4a
S mol OH-(min) phn ng l: x = 2b (mol) ( = 0,22/2 =0,11 mol
S mol OH-(max) phn ng l: y = 4a - 2b (mol) ( 0,28 = 4x 2. 0,11
=> x = 0,125 mol
=>
Vy p n A.Bi 6. (H 2010A). Ho tan hon ton m gam ZnSO4 vo nc c dung dch X. Nu cho 110 ml dung dch KOH 2M vo X th thu c 3a gam kt ta. Mt khc, nu cho 140 ml dung dch KOH 2M vo X th thu c 2a gam kt ta. Gi tr ca m l
A. 17,71.B. 16,10.C. 32,20.D. 24,15.Phn tch, hng dn gii:
- TN1: nKOH = 0,22 mol => 3a gam kt ta
- TN2: nKOH(2) = 0,28 mol => 2a gam kt ta=> Khi thm 0,28 0,22 = 0,06 mol KOH th s mol kt ta gim 3a 2a =a gam
=> nZn(OH)2 b ha tan = 0,06/2 = 0,03
=> nZn(OH)2 trong TN1 = 0,03.3 = 0,09 < 0,22/2 => TN1 OH- d, Zn2+ phn ng ht.Gi s mol ban u ca Zn2+ l x
- Cch 1: Gii thng thng theo phng trnh ion:Trong TN1: Zn2+ phn ng ht, kt ta tan 1 phn:
Zn2+ + 2OH- ( Zn(OH)2(
x mol 2x mol x mol
Zn(OH)2 + 2OH- ( [Zn(OH)4] 2-
(0,11-x)(0,22-2x)
Ta c: = x (0,11 x) = 0,09
=> x = 0,1 mol
=>
Vy p n B.Cch 2: S dng s v bo ton nguyn t Zn, bo ton nhm OH-:Zn2+ + OH- ( Zn(OH)2( + [Zn(OH)4] 2-Trong TN1: => = 0,09 molBo ton nguyn t Zn:
=( = x 0,09Bo ton nhm OH-:
=
( 0,22 = 2.0,09 + 4(x 0,09) => x = 0,1 mol
=>
Vy p n B.- Cch 3: S dng cng thc tnh nhanh:Trong TN1: => = 0,09 molLng OH- tiu tn nhiu nht:
( 0,22 = 4x 2.0,09 => x = 0,1 mol
=>
Vy p n B.- Cch 4: S dng th: S mol Zn(OH)2
S mol OH- x 2a y 4a
Trong TN1: Lng kt ta thu c l b = 0,09 mol
=> S mol OH- phn ng l:
y = 4a - 2b (mol) ( 0,22 = 4x 2. 0,09
=> x = 0,1 mol
=>
Vy p n B.Bi 7. Thm HCl vo dung dch cha 0,1 mol NaOH v 0,1 mol NaAlO2 (hay Na[Al(OH)4]). Khi kt ta thu c l 0,08 mol th s mol HCl dng l bao nhiu ?
A. 0,18 hoc 0,26B. 0,18C. 0,26 D. 0,14 hoc 0,24Phn tch v hng dn gii:
- Cch 1: Gii thng thng theo phng trnh ion:Cc phn ng c th xy ra theo th t sau:
H++ OH-(H2O
(1)
[Al(OH)4]-+H+(Al(OH)3( +H2O(2)
Al(OH)3( +3H+(Al3++3H2O
(3)
ng vi 1 gi tr kt ta s c hai trng hp:
+ TH1: 0,8 mol kt ta l gi tr cc i, tc l H+ thiu, phn ng (3) cha xy ra.
T (2): = = 0,08 mol
T (1) v (2): = + =0,1 + 0,08 = 0,18 mol
+ TH2: 0,08 mol kt ta l gi tr min, tc l H+ d ha tan 1 phn kt ta, phn ng (3) xy ra.
H++ OH-(H2O
(1)
0,1 mol 0,1 mol
[Al(OH)4]-+H+(Al(OH)3( +H2O(2)
0,1 mol
0,1mol 0,1 mol
Al(OH)3( +3H+(Al3++3H2O
(3)
(0,1-0,08)
0,06
T (1), (2), (3) => = 0,1 + 0,1 + 0,06 = 0,26 mol. Vy p n A
- Cch 2: S dng s v bo ton nguyn t Al:Al(OH)4- + H+ Al(OH)3( + Al3+
+ Trng hp 1: = + = 0,08 + 0,1 = 0,18 mol+ Trng hp 2:
Theo bo ton nguyn t Al: = 0,1 0,08 = 0,02 mol
Theo bo ton cation H+: + = 4.0,02 + 0,08 +0,1 = 0,26 mol
=> Vy p n A.
- Cch 3: S dng cng thc tnh nhanh:+ Trng hp 1: + => 0,08 + 0,1 = 0,18 mol+ Trng hp 2: + = 4.0,1 3.0,08 + 0,1 = 0,26 mol
=> Vy p n A.
- Cch 4: Dng th:S mol Al(OH)3
S mol H+
x a y 4a S mol H+(min) phn ng l: x = b (mol)
S mol H+(max) phn ng l: y = 4a - 3b (mol).
Vi bi ton trn th b = 0,08 mol, a = 0,1 mol
- S mol H+(min): x = b + nOH- =0,08 + 0,1 = 0,18 mol
- S mol H+(max): y = 4a 3b + nOH- = 4.0,1 3.0,08 + 0,1 = 0,26 mol
Vy p n A.Bi 8. Mt dung dch X cha NaOH v 0,3 mol NaAlO2 (hay Na[Al(OH)4]). Cho 1 mol HCl vo X thu c 15,6 gam kt ta. S mol NaOH trong dung dch X l:
A. 0,2 hoc 0,8B. 0,4 hoc 0,8C. 0,2 hoc 0,4D. 0,2 hoc 0,6Phn tch v hng dn gii:- Cch 1: Gii thng thng theo phng trnh ion:Ta c: = 0,2 molCc phn ng c th xy ra theo th t sau:
H++ OH-(H2O
(1)
[Al(OH)4]-+H+(Al(OH)3( +H2O(2)
Al(OH)3( +3H+(Al3++3H2O
(3)
ng vi 1 gi tr kt ta s c hai trng hp:
+ Trng hp 1: 0,2 mol kt ta l gi tr cc i, tc l H+ thiu, phn ng (3) cha xy ra.
T (2): = = 0,2 molT (1): = = 1 0,2 = 0,8 mol+ Trng hp 2: 0,2 mol kt ta l gi tr cc tiu, tc l H+ d ha tan 1 phn kt ta, phn ng (3) xy ra.
H++ OH-(H2O
(1)
[Al(OH)4]-+H+(Al(OH)3( +H2O(2)
0,3 mol
0,3 mol0,3 mol
Al(OH)3( +3H+(Al3++3H2O
(3)
(0,3-0,2) mol
0,3 molT (1) = = 1 (0,3 + 0,3) = 0,4 mol. Vy p n B.- Cch 2: S dng s v bo ton nguyn t Al:Al(OH)4- + H+ Al(OH)3( + Al3++ Trng hp 1: = + => = - = 1 0,2 = 0,8 mol+ Trng hp 2: Theo bo ton nguyn t Al: = 0,3 0,2 = 0,1 molTheo bo ton cation H+: +
=> = - () = 1 (4.0,1 + 0,2) = 0,4 mol
=> Vy p n B.- Cch 3: S dng cng thc tnh nhanh:+ Trng hp 1: + => = - =1 0,2 = 0,8 mol+ Trng hp 2: +
=> = - = 1 (4.0,3 3.0,2) = 0,4 mol
=> Vy p n B.- Cch 4: Dng th:S mol Al(OH)3
S mol H+
x a y 4a S mol H+(min) phn ng l: x = b (mol)
S mol H+(max) phn ng l: y = 4a - 3b (mol).
Vi bi ton trn th b = 0,2 mol, a = 0,3 mol
- S mol H+(min): x = b + nOH- => nOH- = 1 0,2 = 0,8 mol
- S mol H+(max): y = 4a 3b + nOH- => nOH- = 1 (4.0,3 3.0,2) = 0,4 mol
Vy p n B.
C. BI TP RN LUYN
1) Mt cc thu tinh cha 200ml dung dch AlCl3 0,2M. Cho t t vo cc V ml dung dch NaOH 0,5M. Tnh khi lng kt ta nh nht khi V bin thin trong on 200mlV280ml.
A.1,56gB.3,12gC.2,6gD.0,0g
2) Ho tan hon ton 8,2 gam hn hp Na2O, Al2O3 vo nc thu c dung dch A ch cha mt cht tan duy nht. Tnh th tch CO2 (ktc) cn phn ng ht vi dung dch A.
A.1,12 ltB.2,24 ltC.4,48 ltD.3,36 lt
3) Thm 150ml dung dch NaOH 2M vo mt cc ng 100ml dung dch AlCl3 nng x mol/l, sau khi phn ng hon ton thy trong cc c 0,1 mol cht kt ta. Thm tip 100ml dung dch NaOH 2M vo cc, sau khi phn ng hon ton thy trong cc c 0,14 mol cht kt ta. Tnh x.
A.1,6MB.1,0MC.0,8MD.2,0M
4) Cho m gam hn hp B gm CuO, Na2O, Al2O3 ho tan ht vo nc thu c 400ml dung dch D ch cha mt cht tan duy nht c nng 0,5M v cht rn G ch gm mt cht. Lc tch G, cho lung kh H2 d qua G nung nng thu c cht rn F. Ho tan ht F trong dung dch HNO3 thu c 0,448 lt (ktc) hn hp kh gm NO2 v NO c t khi so vi oxi bng 1,0625. Bit cc phn ng xy ra hon ton. Tnh m.
A.34,8gB.18g C.18,4gD.26g
5) Cho 200 ml dung dch AlCl3 1M tc dng vi dung dch NaOH 0,5M thu c mt kt ta keo, em sy kh cn c 7,8 gam. Th tch dung dch NaOH 0,5M ln nht dng l bao nhiu?
A.0,6 ltB.1,9 ltC.1,4 ltD.0,8 lt
6) Thm NaOH vo dung dch cha 0,01 mol HCl v 0,01 mol AlCl3. Lng kt ta thu c ln nht v nh nht ng vi s mol NaOH ln lt l:
A.0,04 mol v 0,05 mol
B.0,03 mol v 0,04 mol
C.0,01 mol v 0,02 mol
D.0,02 mol v 0,03 mol
7) Ho tan 0,54 gam Al trong 0,5 lt dung dch H2SO4 0,1M c dung dch A. Thm V lt dung dch NaOH 0,1M cho n khi kt ta tan tr li mt phn. Nung kt ta n khi lng khng i ta c cht rn nng 0,51 gam. Gi tr ca V l?
A.1,2 ltB.1,1 ltC.1,5 ltD.0,8 lt
8) Cho m gam Kali vo 250ml dung dch A cha AlCl3 nng x mol/l, sau khi phn ng kt thc thu c 5,6 lt kh (ktc) v mt lng kt ta. Tch kt ta, nung n khi lng khng i thu c 5,1 gam cht rn. Tnh x.
A.0,15MB.0,12MC.0,55MD.0,6M
9) Cho dung dch cha 0,015 mol FeCl2 v 0,02 mol ZnCl2 tc dng vi V ml dung dch NaOH 1M, sau khi phn ng xy ra hon ton tch ly kt ta nung trong khng kh n khi lng khng i c 1,605 gam cht rn. Gi tr ln nht ca V thu c lng cht rn trn l:
A.70mB.100ml C.l40mlD.115ml
10) Ho tan hon ton m gam hn hp Na2O, Al2O3 vo nc c dung dch trong sut A. Thm dn dn dung dch HCl 1M vo dung dch A nhn thy khi bt u thy xut hin kt ta th th tch dung dch HCl 1M cho vo l 100ml cn khi cho vo 200ml hoc 600ml dung dch HCl 1M th u thu c a gam kt ta. Tnh a v m.
A.a=7,8g; m=19,5gB.a=15,6g; m=19,5gC.a=7,8g; m=39gD.a=15,6g; m=27,7g
11) Cho 200ml dung dch KOH vo 200ml dung dch AlCl3 1M thu c 7,8 gam kt ta. Nng mol ca dung dch KOH dng l:
A.1,5M hoc 3,5MB.3MC.1,5M D.1,5M hoc 3M
12) Cho m gam Na vo 50ml dung dch AlCl31M, sau khi phn ng xy ra hon ton thu c 1,56 gam kt ta v dung dch X. Thi kh CO2 vo dung dch X thy xut hin kt ta. Tnh m.
A.1,44gB.4,41gC.2,07gD.4,14g
13) Thm 240ml dung dch NaOH 1M vo mt cc thu tinh ng 100ml dung dch AlCl3 nng x mol/l, khuy u n phn ng hon ton thy trong cc c 0,08 mol cht kt ta. Thm tip 100ml dung dch NaOH 1M vo cc, khuy u n phn ng hon ton thy trong cc c 0,06 mol cht kt ta. Tnh x.
A.0,75MB.1MC.0,5MD.0,8M
14) Trong mt cc thu tinh ng dung dch ZnSO4. Thm vo cc 200ml dung dch KOH nng x mol/l th thu c 4,95 gam kt ta. Tch kt ta, nh dung dch HCl vo nc lc th thy xut hin kt ta tr li, tip tc cho HCl vo n khi kt ta tan ht ri cho dung dch BaCl2 d vo th thu c 46,6 gam kt ta. Tnh x.
A.2MB.0,5MC.4MD.3,5M
15) Cho m gam Na vo 200 gam dung dch Al2(SO4)3 1,71%, sau khi phn ng hon ton thu c 0,78 gam kt ta. Tnh m.
A.1,61gB.1,38g hoc 1,61gC.0,69g hoc 1,61gD.1,38g
16) Dung dch A cha m gam KOH v 40,2 gam K[Al(OH)4]. Cho 500 ml dung dch HCl 2M vo dung dch A thu c 15,6 gam kt ta. Gi tr ca m l?
A.22,4g hoc 44,8gB.12,6gC.8g hoc22,4gD.44,8g
17) Cho 3,42 gam Al2(SO4)3 tc dng vi 200 ml dung dch NaOH, sau phn ng thu c 0,78 gam kt ta. Nng mol/l nh nht ca dung dch NaOH dng l?
A.0,15MB.0,12MC.0,28MD.0,19M
18) Cho V lt dung dch NaOH vo dung dch cha 0,1 mol Al2(SO4)3 v 0,1 mol H2SO4 n phn ng xy ra hon ton thu c 7,8 gam kt ta. Gi tr ln nht ca V thu c lng kt ta trn l:
A.0,9B.0,45C.0,25D.0,6
19) Cho 120 ml dung dch AlCl3 1M tc dng vi 200 ml dung dch NaOH thu c 7,8 gam kt ta. Nng mol/l ln nht ca NaOH l?
A.1,7MB.1,9MC.1,4MD.1,5M
20) Mt cc thu tinh cha 200ml dung dch AlCl3 0,2M. Cho t t vo cc V ml dung dch NaOH 0,5M. Tnh khi lng kt ta ln nht khi V bin thin trong on 250mlV320ml.
A.3,12gB.3,72gC.2,73gD.8,51g
21) Ho tan hon ton 19,5 gam hn hp Na2O, Al2O3 vo nc c 500ml dung dch trong sut A. Thm dn dn dung dch HCl 1M vo dung dch A n khi bt u thy xut hin kt ta th dng li nhn thy th tch dung dch HCl 1M cho vo l 100ml. Tnh nng mol ca cc cht tan trong dung dch A.
A.[Na[Al(OH)4]]=0,2M; [NaOH]=0,4MB.[Na[Al(OH)4]]=0,2M; [NaOH]=0,2M
C.[Na[Al(OH)4]]=0,4M; [NaOH]=0,2MD.[Na[Al(OH)4]]=0,2M22) Cn t nht bao nhiu ml dung dch HCl 1M cn cho vo 500 ml dung dch Na[Al(OH)4]0,1M thu c 0,78 gam kt ta?
A.10B.100C.15D.170
23) Cho V lt dung dch NaOH 0,4M vo dung dch c cha 58,14 gam Al2(SO4)3 thu c 23,4 gam kt ta. Gi tr ln nht ca V l?
A.2,68 ltB.6,25 ltC.2,65 ltD.2,25 lt
24) Rt V ml dung dch NaOH 2M vo cc ng 300 ml dung dch Al2(SO4)3 0,25M thu c mt kt ta. Lc kt ta ri nung n khi lng khng i c 5,1 gam cht rn. V c gi tr ln nht l?
A.150B.100C.250D.200
25) Cho 100 ml dung dch Al2(SO4)30,1M. S ml dung dch NaOH 0,1M ln nht cn thm vo dung dch trn cht rn c c sau khi nung kt ta c khi lng 0,51 gam l bao nhiu?
A.500B.800C.300D.700
26) Cho dung dch NaOH 0,3M vo 200 ml dung dch Al2(SO4)3 0,2M thu c mt kt ta trng keo. Nung kt ta ny n khi lng khng c 1,02 gam cht rn. Th tch dung dch NaOH ln nht dng l?
A.2 ltB.0,2 ltC.1 lt D.0,4 lt
27) Ho tan m gam ZnSO4 vo nc c dung dch B. Tin hnh 2 Th nghim sau:
TN1: Cho dung dch B tc dng vi 110ml dung dch KOH 2M thu c 3a gam kt ta.
TN2: Cho dung dch B tc dng vi 140ml dung dch KOH 2M thu c 2a gam kt ta.Tnh m.
A.14,49gB.16,1gC.4,83gD.80,5g
28) Thm dung dch HCl vo dung dch hn hp gm 0,1 mol NaOH v 0,1 mol Na[Al(OH)4] thu c 0,08 mol cht kt ta. S mol HCl thm vo l:
A.0,16 molB.0,18 hoc 0,26 molC.0,08 hoc 0,16 molD.0,26 mol
29) Cho 250ml dung dch NaOH 2M vo 250ml dung dch AlCl3 nng x mol/l, sau khi phn ng hon ton thu c 7,8 gam kt ta. Tnh x.
A.1,2MB.0,3MC.0,6MD.1,8M
30) Trong 1 cc ng 200 ml dung dch AlCl3 0,2M. Rt vo cc 100 ml dung dch NaOH, thu c mt kt ta, em sy kh v nung n khi lng khng i thu c 1,53 gam cht rn. Nng mol/l ca dung dch NaOH dng l?
A.0,9MB.0,9M hoc 1,3MC.0,5M hoc 0,9MD.1,3M
31) Cho 200 ml dung dch AlCl3 1,5M tc dng vi V lt dung dch Ba(OH)2 0,25M, lng kt ta thu c l 15,6 gam. Gi tr ln nht ca V l?
A.2,4 ltB.1,2 ltC.2 ltD.1,8 lt
32) Thm dn dn Vml dung dch Ba(OH)2 vo 150ml dung dch gm MgSO4 0,1M v Al2(SO4)3 0,15M th thu c lng kt ta ln nht. Tch kt ta, nung n khi lng khng i thu c m gam cht rn. Tnh m.
A.22,11gB.5,19gC.2,89gD.24,41g
33) Thm m gam kali vo 300ml dung dch cha Ba(OH)2 0,1M v NaOH 0,1M thu c dung dch X. Thm t t dung dch X vo 200ml dung dch Al2(SO4)3 0,1M thu c kt ta Y. thu c lng Y ln nht th gi tr ca m l:
A.1,71gB.1,59gC.1,95gD.1,17g
34) Hn hp A gm Al v Al2O3 c t l khi lng tng ng l 1,8:10,2. Cho A tan ht trong dung dch NaOH va thu c dung dch B v 0,672 lt kh (ktc). Cho B tc dng vi 200ml dung dch HCl thu c kt ta D, nung D nhit cao n khi lng khng i thu c 3,57 gam cht rn. Tnh nng mol ln nht ca dung dch HCl dng.
A.0,75MB.0,35MC.0,55MD.0,25M
35) Cho V lt dung dch NaOH 0,1M vo cc cha 200 ml dung dch ZnCl2 0,1M thu c 1,485 gam kt ta. Gi tr ln nht ca V l?
A.1 ltB.0,5 ltC.0,3 ltD.0,7 lt
36) Cho p mol Na[Al(OH)4] tc dng vi dung dch cha q mol HCl. thu c kt ta th cn c t l :
A.p: q < 1: 4B.p: q = 1: 5C.p: q > 1:4D.p: q = 1: 4
37) Cho dung dch A cha 0,05 mol Na[Al(OH)4] v 0,1 mol NaOH tc dng vi dung dch HCl 2M. Th tch dung dch HCl 2M ln nht cn cho vo dung dch A xut hin 1,56g kt ta l?
A.0,06 ltB.0,18 ltC.0,12 ltD.0,08 lt
38) Khi cho V ml hay 3V ml dung dch NaOH 2M tc dng vi 400ml dung dch AlCl3 nng x mol/l ta u cng thu c mt lng cht kt ta c khi lng l 7,8 gam. Tnh x.
A.0,75MB. 0,625MC. 0,25MD. 0,75M hoc 0,25M
39) Ho tan hon ton m gam hn hp K2O, Al2O3 vo nc c dung dch A ch cha mt cht tan duy nht. Cho t t 275ml dung dch HCl 2M vo dung dch A thy to ra 11,7 gam kt ta. Tnh m
A.29,4 gamB.49 gamC.14,7 gamD.24,5 gam
40) Cho 200 ml dung dch NaOH tc dng vi 500 ml dung dch AlCl3 0,2M thu c mt kt ta trng keo, em nung kt ta trong khng kh n khi lng khng i th c 1,02 gam cht rn. Nng mol/l ln nht ca dung dch NaOH dng l?
A.1,9MB.0,15MC.0,3MD.0,2MLI KTQua qu trnh ging dy nm hc va qua cho HSG tnh, c bit l HSG i thi olympic, ti nhn thy:
- Kin thc ca hc sinh ngy cng c cng c v pht trin sau khi hiu nm vng c bn cht v s lai ha v dng hnh hc phn t.
- Hc sinh c th t hc, t lm trn 90% bi tp xc nh trng thi lai ha v dng hnh hc phn t trong cc thi HSG tnh, thi HSG quc gia v olympic ha hc. Do thi gian c hn, ti c th cha bao qut ht c s lai ha ca nguyn t trung tm v dng hnh hc phn t ca ht cc loi phn t. Cc v d c a ra trong ti c th cha thc s in hnh nhng v li ch thit thc ca cng tc ging dy bi dng HSG v hc tp nn ti mnh dn vit, gii thiu vi cc thy c v hc sinh.
Rt mong s ng gp kin b sung cho cho ti thc s gp phn gip hc sinh hc tp ngy cng tt hn.
Xin chn thnh cm n.
Chn Thnh, ngy 06 thng 03 nm 2013
Ngi vit
Trn c PhngTI LIU THAM KHO
1. Sch gio khoa ha hc lp 10 nng cao NXB GD H Ni nm 20072. Nguyn t v lin kt ha hc o nh Thc, Nh xut bn gio dc.3. Bi dng hc sinh gii ha hc THPT tp 1 Qu Sn Nguyn Tr Nguyn 4. Bi tp bi dng HSG ha hc tp 1 - TS Cao C Gic, NXB HQG TP H Ch Minh.5. Ti liu bi dng HSG THPT - PGS.TS Nguyn Xun Trng ThS Phm Th Anh NXB HQG H Ni.6. Ha hc v c tp 1 Nguyn Th T Nga, i hc KHTN, Tp HCM7. Lin kt ha hc trong phc cht L Ch Kin, NXB HQG H Ni8. Tuyn tp thi olimpic 30 thng 4 va thi hoc sinh gioi quc gia KIN CA T CHUYN MN
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KIN CA HI NG KHOA HC S GIO DC O TO TNH
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b
a
b
a
b
a
b
a
b=0,2
a=0,3
a
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a
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