48
CSu 12. De nhan biét drfdc dung dich dufng trong 4 Ip khac nhau la: KOH, N H 4 C I , Na2S04, (NH4)2S04 ta chi can dOng mot trong 4 chát:
Ạ Dung dich AgN03 B. Dung dich BaCb
C. Dung dich NaOH D. Dung dich BăOH)2.
Cfiu 13. Co 4 muoi FeClj, CuCb, A I C I 3 , va ZnCh. Neu them tit tir dung djch
NaOH cho den dir vao 4 muoi tren. Sau do them tiep N H 3 dir thi so ket tiia thu dúdc la:
Ạ 1 B. 2 C. 3 D. 4.
Cfiu 14. Mot dung dich gom x mol Ná^' y mol N O 3 " , z mol H C O 3 " , t mol Câ*.
He thiJc lien he giffa X, y, z, t la:
A . x + 2y = t + z B. x + 2t = y + z C. X + 2z = y + z D. X + 3y = t + 2.
Cfiu 15. Dung dich NaOH 0 , i M , dung djch HCl 0,01M. Vay pH cua hai dung
dich tren Ian liTdt la:
Ạ I v a 2 B. 13va2 C. 2 v a l 3 D. 0,1 va 0,01.
Cfiu 16. Tron 100 ml NaOH 0,1M vdi dung djch HCl 0,01M. Vay pH cua dung
dich sau khi trpn la:
Ạ 7 B. 12,69 C. 13 D. 2.
Cfiu 17. De thu diTdc dung dich c6 pH = 7 thi ti le the tich dung dich NaOH 0,1M vaHC10,01Mcanla'ym:
Ạ 1:10 B. 1:1 C . 1 0 : l D. 2 : 5.
Cfiu 18. Dung dich X ^ dung dich HCl, dung dich Y la dung dich NaOH. Lay 10
ml dung dich X pha loang bkng nirdc thanh 1000 ml dung dich thi thu dirdc
dung dich HCl c6 pH = 2. De trung hoa 100 g dung dich Y can 150 ml dung
dich X. Vay C% cua dung dich Y Ih:
Ạ 2% B. 3% C.5% D. 6%
Cfiu 19. Phai pha loang dung dich KOH 0,00 I M vcti niTctc bao nhieu Ian de diTdc
dung dich c6 pH = 9.
Ạ 100 Ian B. 110 Ian C. 99 Ian D. 80 Ian
Cfiu 20. De thu diTdc 1 lit dung dich HCl c6 pH = 5 tir dung dich HCl c6 pH = 3
thi the tich nufdccát can diingia: .
Ạ 900 ml B. 990 ml C. 1000 ml D. 110 ml.
Cfiu 21. De thu dirdc 1 lit dung dich c6 pH = 4 sau khi trpn thi ti I 9 the tich cua dung dich H C I l Q - ' M vdi dung dich KOH 10'^M m:
Ạ 1:2 B. 11:9 C. 9 : 1 1 D. 2 : 15 (adsbygoogle = window.adsbygoogle || []).push({});
49
Phfln dgnq vy phuong phAp giat H6a hgc 11 VP c o - 05 XuSn Hung
CSu 22. t h e tich dung dich KOH 0.001 M can d l pha thSnh 1.5 lit dung dich c6 pH = 9m: " pH = 9m: "
Ạ 0,015 lit B. 0,02 lit C. 0,0015 lit D. 0,15 lit
Cau 23. Can phai them v^o 1 lit dung dich H2SO4 IM bao nhieu lit dung dich NaOH 1,8M thu di/dc dung dich c6 pH = 1. NaOH 1,8M thu di/dc dung dich c6 pH = 1.
Ạ 2 lit B. 1,5 lit C. 1 lit D. 0,5 lit
CSu 24. Can phai them vao 1 lit dung dich H2SO4 IM bao nhieu lit dung dich NaOH 1,8M de thu difdc dung dich c6 pH = 13. NaOH 1,8M de thu difdc dung dich c6 pH = 13.
Ạ 1,235 lit B. 1,25 lit C. 1 lit D. 0,9 lit
Cflu 25. Tron 250 ml dung dich hSn hdp gom HCl 0,08M va H2SO4 0,0IM vdi
250 ml dung dich Bă0H)2 a mol//, thu diTdc m gam ket tua va 500 ml dung dich c6 pH = 12. Vay a CO gia tri la: dich c6 pH = 12. Vay a CO gia tri la:
Ạ 0,05M B.0,04M C. 0,06M D. 0,05M
cau 26. Tron 250 ml dung dich h6n hdp gom HCl 0,08 moVl vh H2SO4 0,01 mol// vdi 250 ml dung dich NaOH a mol//, diTdc 500 ml dung dich c6 pH = 12. mol// vdi 250 ml dung dich NaOH a mol//, diTdc 500 ml dung dich c6 pH = 12.
Vay a c6 gii tri 1^:
Ạ0,12M B.0,13M C. 0,14M D.0.15M. cau 27. Mpt dung dich A g6m h6n hdp 2 axit HCl v^ H2SO4. De trung h5a 10 ml cau 27. Mpt dung dich A g6m h6n hdp 2 axit HCl v^ H2SO4. De trung h5a 10 ml
dung dich A cin dilng 40 ml dung dich NaOH 0,5M. Mat khac, neu ISy 100 ml
dung djch A cho tic dung vdi mot IvTdng vifa du, roi c6 can dung djch thu difdc
13,2 gam mu6^i khan. V$y nong dp mol// cua 2 axit trong h5n hdp la:
Ạ0,8Mva0.6M B.0,6va0,8M C.O,5vaO,7M D.0,9val,5M. C.O,5vaO,7M D.0,9val,5M. (Jfiu 28. Cho a mol khi NO2 sue v^o dung dich chtfa a mol KOH. Vay dung dich
thu di/dc CO moi triT&ng l^:
Ạ axit B. bazd C. trung tinh D. Iur3ng tinh.
CfiU 29. Axit CH3COOH 0,1M c6 hiing s6'di$n li = 10""". V$y pH cua dung dich la: dich la:
Ạ 1 B.2,1 C.2,88 D.2,78. Can 30. Biet hkng so phan li ciia ion NlitMa K» = lÓ'^-^^ Vay de thu dtfdc 250 Can 30. Biet hkng so phan li ciia ion NlitMa K» = lÓ'^-^^ Vay de thu dtfdc 250
ml dung dich c6 pH = 5 thi s6' gam NH4CI la:
Ạ 2,32 g B.2,56g C.4,64g D. l,16g.
cau 31. Axit nitrd c6 hlng so phan li axit la Ka = 10"^'^^. Vay pH cua dung dich
HNO2 0,01Mia:
Ạ 2,35 B. 2,69 C. 2,61 D. 3,01. 50 50
(^fiu 32. Trpn 2 0 0 ml dung dich NaOH 0 , 1 M \di 3 0 0 ml dung dich CH3COOH
0,075M. Biet hang sódien Vay pH cua dung dich thu drfcJc Ih:
^.5,66 B . 5 , 1 4 C. 5,76 D. 5,71. (adsbygoogle = window.adsbygoogle || []).push({});
C&u 33- Dp tan trong nvtdc cua CaF2 d 25"C Ih 2,14.10-''M. Vay tich só tan ciia CaF21^: CaF21^:
Ạ 3,9.10-" B. 4,1.10-" C. 2.14.10-" 0.3,5.10"".
CfiU 34. Tich tan cua BaS04 5 25"C m IQ-'". Vay HQ tan cua BaS04 trong
dung dich H2SO4 0 , 1 M la (chap nhan H2SO4 dipn li hoan tôn)
A . 1 0 - ' M B.10-'"M C . I O - ' M D. IQ-'M.
HUdNG DAN GIAI CfiU 1. Chpn C.
TCfcdngthiJc: a = K C
Vdi K la h^ng so di#n lị C la nong dp khi C tSng thi a giam. Cau 2. Chpn D. H2O la chát lU3ng tinh Cau 2. Chpn D. H2O la chát lU3ng tinh
V( du: CH3COO- + H2O < > CH,COOH + OH"
HSO4- + H2O < > H3O + S04^-
CSu 3. Cac phtfdng trlnh phan vlng:
NH4CI -> NH-4. + c r NH4% 2H2O NH4OH + H.,0* Al(N03)-> Al'^+BNOj- Al'^HjO < = ^ Al(OH)^%H30* NaHS04->Na%HS04- HSO4- + H2O ( > H30*+ S04^- => Chpn Ạ cau 4. Chpn B
Muoi KCl \k NaNOj khi dipn li deu cho cac ion trung tinh
cau 5. Chpn A NH3 + H2O <r=± NH4V + OH- (1) NH3 + H2O <r=± NH4V + OH- (1) a a a NaOH^Na*+OH- (2) a a BăOH)2 ^ Bâ* + 20H- (3) a 2a 51
d phdn iJng (1) nong do OH = a a , nho nhát. 6 (3) la Idn nhát nen
p H ( l ) < p H ( 2 ) < p H ( 3 )
C a u 6. Cac phiTdng trlnh phan tfng:
3NaOH + AlCị, 3NaCl + Al(OH),, i
Khi NaOH dif: NaOH + Al(OH)3 - > NaAlOz + H2O
=> Chon B
C a u 7. LiTdng NaOH luon d\i so vdi AlCi, nen két tua tao ra tan ngay trong NaOH dir. Co the viét phuTdng trinh tong cong
4 N a O H + AlCl., -> NaAIOa + 3NaCi + 2H2O
:=> Chon C
CSu 8. PhUcJng trinh phan iJng:
HCl + NaAlOj + H 2 O -> NaCl + Al(OH),, + H2O
khidUHCl:
Al(OH).i + 3HC1 -> A i d , + 3H2O
=:>Ch9nB.
C a u 9. C h o n D
C O / ' + HzQ < > HCỌr + OH" (adsbygoogle = window.adsbygoogle || []).push({});
OH" tao ra tCf phan uTng tren l^m cho moi triTcfng c6 tinh bazd nen pH > 7
C S u 10. NH4C1->NH4*+Cr
NH% + 2 H 2 O < > N H 4 O H + H, 0 * CHjCOONa CH,,COO- + Na*
C H 3 C O O - + H 2 O < > C H 3 C O O H + OH" Na2CO,->2Na*+2CO,-
cor + H 2 0 < ) Hcor + O H -
HCOj'+HzO < > H2CO3 + OH- NaHS04->Na*+HS04- H S O 4 " + H 2 O < > H.,0 * + S04^" Cu(N03)2->Cu^*+2NO,- C u^ V 2 H 2 0 < r = ± Cu(0H)*H30* K 2 S- ^ 2 K % S ^ - S^- +H 2 O < = = ± HS" + OH" HS- + H 2 O < > H2S + OH-
Cac dung dich KCl,BăN03)2 chi dỉn l i tao ra cdc ion trung tinh => Chon B.
C f l u l l. C h o n B .
Chii y: Cdc dung dich NaHC03, KHS dcu la muoi axit tuy nhidn trong dung dich
chung lai thuy phan tao ra moi trtfdng bazd.
w
^ K l i A I g f i V I i i T
Cflu 12. Chon D.
BăOH)2 + N H 4 C i -> BaClj + 2 N H 31 + 2 H2O
Bă0H)2 + Na2S04 B a S 0 4 i + 2NaOH
Bă0H)2 + (NH4)2S04 -> BaS04 i + 2 N H 31 + 2H2O
=> Chon D.
Cflu 13. Chon Ạ
Cho NaOH vao: Fé* + 3 OH " FeCOH), i
Cu*' + 2 0 H - Cu(0H)2 i
A l' * + 3 0 H - ^ A l ( 0 H ) 3 i A1(0H)3 + OH"-)• AIO2" + H2O
Z n ' * + 2 0 H " ^ . Z n ( O H) 2 i '
Zn(OH)2 + 2 0 H " Zn02^" + HjO Khi cho NH3 vao:
Cu(0H)2 + 4 NH3 < > [Cu(NH3)4]' + 20H"
=> Chi con 1 ket tua la Fe(0H)3
=> Chon Ạ
Cfiu 14. Trong dung dich thi tong dỉn tich (+) = tdng dien tích (-) = > y + z = x + 2t
=> Chon B .
C f l u l S . NaOH - > Na* + OH" 0,1M 0,1M HCl - > H^+ CI" 0,01 0,01 = > p H ( A ) = 1 3 pH(B) = 2 =i>ChonB.
Cau 16. Tron 100 ml dung dich A + 100 ml dung dich B (adsbygoogle = window.adsbygoogle || []).push({});
CU><100.0.05M ^ 2 0 0
0,01x100
CR = — = 0,005 M ^ 200 ^ 200
Phan iJng trung hoa H* + OH" H 2 O
0,05M 0,005M
C = 0,05 - 0,005 = 0,0495 M
O H' D i f
pOH = 1,305 => pH = 12,69 =:> Chon B .
Cflu 17. Goi VA va V B la the tich cua dung dich A va B can lay de dung dich sau phan iJng c6 pH = 7
Phfln djng va phuong phAp giai H6a hpc 11 VP cO - D5 Xuan Hmg K h i d 6: C A V A = C „ V B => 0,1 X V A = 0,01 x V „ V,^ 0 , 0 1 1 ^ - A = = _ =>chonẠ V B 0,1 10 C a u 18. Ta c6: pH = 2 [ H* ] = 10'^ M 1 0 - ' = > C M ( X ) = = I M " 0 , 0 1 Khi (rung hoa
C,x, XVx = C,Y, xV,Y, = 0.15 X1 = 0,15 mol
= • mN,()H = 0 , 1 5 x 4 0 = 6 g = > C % ( Y ) = 6% = > C h o n D . C a u 19. PhiTdng trlnh div*n l i KOH ~> K* + OH' 10"'M 1 0 " 'M pH = 9 lOH"J = IQ-' M T a c 6 C , x V , = C 2 X V : = . 10-' X V , = 10"' x\/2=> — = = 100 10-^ => Pha loang 100 Ian => Chon A . Cau 20. Chun B.
C, X V , = C: X Vz
V , 10"^ LL - J
= — = 100 => V2 = lOOV,
" V , 10- , .„
=> can lay 10 ml HCl pH = 3 roi pha loang vdi 990 ml niTdc. => Chon B.
Cau 21. Goi V , va V2 la the tich HCl va KOH can lay pH = 4 HCl dir v,+v, io-^v,-io-^v, , . O 9 V | = I I V 2 v,+v, => l O V , - 10V2 = V | + V2 V, 11 V i - V , 1 V , + V 2 10 V , 9 Chon B. Cfiu 22. m KOH - > K* + O H " I Q - V l O - ' M => p H = 9 => 1 0 H - J = l O - ' M C, xV, = C2 XV2 ^ V . = ^ ^ l ^ = i ^ = 0 , 0 1 5 1 i t C , 1 0 - ' => Chon Ạ c a u 23. Ta CO pH = 1 => (H^J = 0,1M => axit diT Goi V la the tich dung dich NaOH can dung
H*+ O H " - > H2O 2 mol 1.8.V ThcodC-ra: ^ " ^ ' ^ " " ^ = 0 . 1 1 + V => 2 - 1,8 V = 0.1 + 0,1V => 1,9= 1,9 V => V = 1 lit => Chon C.
Cau 24. Khi pH = 13 = > O H - diT
1 + V o 1,8 V - 2 = 0,1 V + 0 , 1 o 1,7 V = 2,1 o 1,8 V - 2 = 0,1 V + 0 , 1 o 1,7 V = 2,1 => V = 1,235 lit => Chon Ạ Cau 25. n„(., = 0,25 X 0,08 = 0,02 mol "H^SO, =0,0025 mol => Z n ^ i , = 0,025 mol
"BăOHi2 = 0,25a => n^^_ =0,5a
Sau phan iJng: pH = 12 => [OH") = 0 , 0 I M Phan u-ng trung hoii: H^ + OH" - > HjO
0,5 => 0,5 a - 0,025 = 0,005 => 0,5a = 0,03 => a = 0,06M nBaso4 = " s o i = 0 ' " " 2 5 m o l . niaaso, = 0,5825 g =>Chon C. 55
Phan d?ng va phuong pMp g'tii H6a hpc 11 VP co - D5 Xuan Hang Cflu 26. H H C I = 0,25 X 0,08 = 0,02 m o l A r>r>ic I r ^ i^ i i . = 0 , 0 2 5 mol n H 2 S 0 4 = 0 . 0 0 2 5 m o l J " l^NaOH = 0,25a m o l p H = 12 => O H " dir so v d i H"" 0,25a-0,025 _ 0,5 => 0.025a - 0,025 = 0,005 => 0.25a = 0.03 a = 0 , I 2 M => Chon A . - C S u 2 7 .
G o i a,b la nong dp mol// cua H C l va H2SO4 "NaOH = 0,5 X 0,04 = 0,02 m o l
H C l + N a O H NaCl + H 2 O
0,01a 0,01a 0,01a
H2SO4 + 2 N a O H - > Na2S04 + H 2 O
0,01b 0,02b 0,01b (adsbygoogle = window.adsbygoogle || []).push({});
rO,01a + 0 , 0 2 b = 0,02 (1) j a ^ O . S m o l
jo.Ola X 58,5 + 0 , 0 1 b X142 = 1,32 ( 2 ) ^ [ b = 0.6 m o l
Chii y: K h d i liTOng m u d i ci (2) bang 1/10 so v d i de chọ
=> C h o n A . C a u 28. 2NO2 + 2 K 0 H K N O 2 + KNỌ, + H 2 O K N O 2 ^ + N O 2 " N O 2 " + H 2 O < > H N O 2 + O H " => Chon B . c a u 29. CH.,COOH < > C H , C O O - + H * Trirdc k h i d i ^ n l i : 0,1 0 0 D i 0 n l i : x S a u d i ^ n l i : 0 , 1 - x x x Ta CO — = lO-"''** 0 , 1 - x x=10-^''*''M r:>pH = 2,88. Chon C.
C a u 30. G o i so gam NH4CI can lay la m gam.
N 6 n g dp NH4CI = N 6 n g dp N H 4 " = 53 5^250 ^ ^ " ^ " ^ ^ ^ ^ Trifdc k h i di§n H NH4C1 ^ NH4* + c r NH4' - > N H , + H ^ 4 m 53:5 4 m Sau k h i d i e n l i — - x x Theo dinh luat tac dung k h d i Iifdng ta c6:
^ — = l O - ' ^ - ' % d i x = io-^ 53,5 Ta CO m = 2,32 gam = > C h p n A . c a u 31. HNO2 < > + NO2' Trifdc k h i dipn l i 0,01 0 0 Di§n l i X S a u k h i d i ^ n l i 0,01 - x x x x^ Theo dinh luat tac dung k h d i luTdng la c6: K^- = — —
U, U 1 X Thay sd la co: x ' + KaX - 0,01Ka = 0 Thay sd la co: x ' + KaX - 0,01Ka = 0
x^+10-^-^'^x-0,01.10-«'-' = 0
= > [ H * ] = X = 2 , 0 2 . 1 0 - 'M
=> p H = 2,69
Vay chon B
C§u 32. N o n g dp cac chát sau k h i trpn:
^ 200.0,1 n n ^ x y r
^ - - = 2 0 0 ^ 3 0 0 = ' ' ^ ' ^
C c H c o o H = 0 . 0 4 5 M
CH,c(>..H 2 0 0 + 3 0 0 Phtfdng trinh phan lifng:
C H , C O O H + N a O H C H j C O O N a + H2O
0,045 0,04
thu diTdc g o m : C H 3 C O O H dirO,005M, CH,,COONa 0,04M C H j C O O N a - > C H 3 C O O - + N a *
0,04 0,04 Phtfdng trinh di0n l i ciia axit: Phtfdng trinh di0n l i ciia axit:
Phan dgng va phuong phap gi&i H6a hpc 11 V6 CO - P5 XuSn Hung
CH,COOH < » CHjCOO" + H* TruOc khi di^n l i 0,005 0,04 0 Di$n li X
Sau khi dien li 0,005 - x 0,04 + x x Thco dinh luat tac dung khoi liTcJng la c6: K,. = ^ ' ^
0,0()5-x Gia sur X < < 0,005, khi do: 0,005 - x « 0,005 va 0.04 + x « 0,04
[ H -] = X = [ H n = X = Q ' Q O ^ -I P " ' ' ' ^2.17.10-^-M 0,04 => pH = 5,66 => Chon Ạ Cdch 2: pH = pK„ + ig = 4,76 + Ig = 5,66 CcHjCOOH 0,005
CSu 33. Xet can bang sau:
F 2 <
CaF, < > CâV2F-
S 2S Bieu thiJc tinh tich so tan: Bieu thiJc tinh tich so tan: (adsbygoogle = window.adsbygoogle || []).push({});
T = [Câ*][F]= T = S(2S)^ = 4.S' =4.(2,14.10-Y = 3,9.10-" Vay chon Ạ Cau34. H2S04-> 2H* + SO/- ' 0,1M • 0,IM Xet can bling sau:
BaS04 < > Ba-^ + 804^" Trirdckhilan 0 0,1 Tan S S S Khicanbling S S + 0,1 Taco: T = S(S + 0,1). Gid s i i r S < < 0 , l . l h i T = 0,lS. : ^ S = 10-'^ V 9 y S « 0 , l Chon C. Qiam^ N T T O - P H O T P H O A . T6M T A T L f T H U Y E T
* Nh6ni nitd gom cac nguyen to: N (nild); P (photpho); As (Asen); Sb (anlimon); Bi (Bitmut) -> (deu thuoc cac nguyen to p).
* Ldp electron ngoai cDng ns' np^.
Ị NITCJ U=^2s*2p'
C T C T : N ^ N
Tinh chcít luki hoc :
1. Tinh oxi hoa : 3Mg + ' > Mg^N^
3 H 2 + N , . • 2 N H ,
2. Tinh khuT: N , + Ọ < » 2 NO
2 N O + 0 2 > 2 N O 2
Dieu die : Trong phong thi nghiem :
NH4NO2 — ^ N 2 + 2 H 2 O
NH4CI + NaN02 — ^ NaCl + N 2 + 2 H 2 O .
IỊ AMONIAC VA MUOI AMONI
1. Amoniac (NH3)
* Tinh chat hoa hoc :
+ Tinh bazd yeu : NH., + H.O -> NH;; + OH"
Dung giay quy tim am de nhan biet khi amoniac -> quy tim h6a xanh. - Tiic dung vcfi dung dich muoi -> i hidroxit ciia cac kim loaị
V i dii : AICI3 + 3 N H 3 + 3 H 2 O > A1(0H).,I + 3NH4CI
- Tiic dung vtJi axit -> muoi amonị
NH3 + H C I > NH4CI
+ Kha nang tao phtfc: Dung dich amoniac c6 khii nang hoa lan hidroxit hay muoi it tan cua mot so kim loai tao thanh cac dung djch phtjTc chát.
Cu(OH)2 + 4 N H 3 > |Cu(NH3)4J(OH)2
C u ( 0 H)2 + 4 N H 3 > ( C u( N H 3) 4 l ' " + 2 0 H -
(dung djch xanh tham) + Tinh khiV : 4 N H 3 + 3 0 . 2 N 2 + 6 H 2 O