Phân dạng và phươg pháp giải bài tập hoá học 11 phần vô cơ phần 1

Phân dạng và phươg pháp giải bài tập hoá học 11 phần vô cơ phần 1 tài liệu, giáo án, bài giảng , luận văn, luận án, đồ á...

P /IJyfD *>6 XUAN HUNG

•— ^ Gi&o vifen c h u y S n l u y § n t h i OH-CD H6a

Pfian ddng&

p n i f o n g p h a p g i d k

HOA HOC^

D3nh cho hoc sinh lop 11 on tap va nang cao la nang lam bai

i THi; VIEW TINHBINH IHUAN

MXUJTBANldNGHQPliNHPHdHdCHlMMI

PHAN DANG VA PHUdNG PHAP GlAi HOA HQC 11 Vd Cd

D 6 X U A N H l / N G

, Chju trach nhi^m xud't ban

N G U Y E N T H | T H A N H HlTCfNG Bien lap : HAI A U

* ChS't di$n li Ih nhifng chat tan trong nurdc phan li ra ion

C6 hai loai: Chat di0n l i manh va chatdi^n li yeu

* DO diOn l i (a): Dp diOn l i (a) cua chat dipn l i la t i so giffa s6' phan tur phan l i

ra ion (n) va tong so phan tur hoa tan (no)

a = J L O ^ a ^ l

«() Neu bieu di^n diTdi dang phan trSm : 0% < a ^ 100%

u a = 0 : chat kh6ng di$n l i

u 0 < a < 1 : cha't di^n li yeu

u a = 1 : cha't diOn li manh

C6 hai l o a i : + Muoi axit: NaHCOj, K H S O 4

+ Muoi trung hoa : KCl, CaC03,

Phan dgng phi^ng phap giai H6a tipc 11 V6 CO - D5 XuSn Hang _ _ _ _ _ _ _

4 Hidroxit li^ng tinh

La hidroxit khi tan trong nifdc vita c6 the phan li nhiT axit vifa c6 the phan 11

nhuf bazd

Vi du : Zn(OH)2, A1(0H)3

III Si; DI5N LI CUA NLTdC pH CHAT CHI TH! AXIT - BAZd

1 Nvldc la cha't di$n li rat yg'u

H2O H * + OH"

* Tich so' ion cua nuTdc (kn^o) la h^ng so cl nhi$t do xdc dinh

k H 2 0 =[H^]-[OH'] = 1-010''^ ( [ H i = [0H-] = I.O.IQ-' (mol//) d 25"C)

* Moi trurdng axit la moi trifdng trong d6 :

[ H 1 > [ 0 H ' ] hay [ H ^ > 1,0.10"^M

* Moi triTcJng kiem 1^ m6i triTdng trong do :

[ H 1 < [ 0 H - ] hay [ H i < 1,0.1 Q-'M

* Moi trirdng trung tinh Ih moi triT&ng trong do: [H^] = [OH"] = 1,0.10"'M

2 Khai ni^m v l pH ChS't chi thj axit - bazof

[ H i = 1,0.10"''" (M)

Ve mat toan hoc : pH = - l g [ H l

+ Khi pH < 7 : moi triTdng axit

+ Khi pH = 7 : moi trifcfng trung tinh

+ Khi pH > 7 : moi tru-dng kiSm

* Thang pH thuTdng dilng c6 gid tri ttt" I -> 14

* Mau ciia hai cha't chi thi axit - bazd la quy tim va phenolphetalein

IV PHAN CfNG TRAO DOI ION TRONG DUNG DjCH C A C CHAT DIEN LI

* PhSn tfng xay ra trong dung dich cdc chat dien li la phan ti'ng giffa cdc ion

* Phan tfttg trao doi trong dung dich cac cha't dien li chi xay ra khi cac ion ke't

hdp dtfdc vdi nhau tao thinh it nhS't mot trong cdc cha't sau :

+ Cha't ket tua;

+ Chat dien li y6'u;

+ Chat khi;

K2SO4 + BaClz > 2KC1 + BaS04>t

PhiTdng trinh ion thu gon : Ba^* + SO^" > BaS04

NaOH + HCl > NaCl + H2O

PhiTdng trinh ion thu gon : H"" + OH" > H2O

K 2 C O 3 + 2HC1 > 2KC1 + C 0 2 t + H2O

PhiTdng trinh ion rut gon : 2H* + CO^" > COzt + H2O

B PHAN L O A I V A PHlTdNG P H A P G I A I C A C D A N G B A I T A P

Dang 1

- Viet phi/cfng trinh di#n li c a c chdt

- Tinh nong dp mol cua tCfng Ion

BAI TAP M A U V A BAI TAP NANG C A O

Bai 1 Viet phifdng trinh dien li cua nhiTng chat sau : a) Cac chat dien l i manh : NaCl, H N O 3 , KOH, Na2S04, Ba(N03)2, H2SO4,

Bai 2 Viet phi/dng trinh c&c chat dien li manh va tinh nong do mol cua tilfng ion

trong cac dung dich sau :

Phan dang va pniiono ph^p giai H6a hpc 11 VP cc - Bi XuSn Hung

a) 0,3 mol dung djch axit sunfuric vdi the tich 2 lit

b) 5,85g dung dich natri clorua vtJi the tich 200nil

c) Dung dich NaOH 0.25M

Bai 5 Hai hdp cha't X va Y khi tan vao nirdc moi chat di^n li ra hai loai ion vdi

nong dp mol nhiT sau :

- Tinh d$ di^n H a h^ng s6 di^n li

- Tinh (H*), (OHl pH cua dung djch

BAI TAP MAU

Bai 1 Tinh nong dp H \, pH ciia dung dich HCl 0.1 OM va dung dich NaOF

jPijilff 4$fi9 ¥^ phuong pWp 9i5i Hda hoc 11 VP CO - B5 Xuan Hung

Uhi I Dung dich axit axetic nong do 0,25M co pH = 2

g) TInh do dien li cua axit axetic trong dung dich tren

Neu hoa tan thtm 0,01 mol HCl vao 1 lit dung dich tren thi do dien l i cua

mit axetic tftng hay giam? GiU thich?

Giai

a) PhiTctng trinh di6n l i : CH3COOH ^ CHiCOO" + H*

Trong 1 lit dung dich c6 0,25 mol CH3COOH c6 pH = 2

=> [ H i = 10"''" = 10"^ = 0,01M

V|y troog 1 lit dung dich c6 0,01 mol CH3COOH phan li ra ion

Bo di^n li cua CH3COOH la: a = — x 100% = 4%

0,25 b) Khi them 0,01 mol HCl vao 1 lit dung dich tren thi nong do H"" tang len, do

d6 cSn bhng dien li chuydn dich sang trai, do do do diSn U giam

Bai 3 Mot dung dich c6 [ H ^ = 0,010M Tinh [OH^ va pH cua dung dich Moi

tri/cfng cua dung dich nay la axit, kiem hay trung tinh? Hay cho bie't mau cua

quy tim trong dung dich nay?

Gidi [HI =0,010M => [0H-] = ' , =1,0.10-'^M

1,0.10"^

pH = - l g [ H l = - l g I 0 - ' = 2

Moi irirftng cija dung djch nay lii axil Quy tim doi sang milu do

Bai 4 Tinh pH cOa cac dung dich sau :

0,004M 0,008M hay [OH-] = 0,008M

N6ng do OH" cua dung dich: [OH'] = 0,008.a = 0,008.0,8 = 0,0064M

=> [ H i = ^'"'^^ = l,5625.10-'2 =:> pH = - l g [ H l = 11,81

6,4.10 ^

d) CH3COOH -> CH3COO- + H^

0,004M 0,004M hay [H1 = 0,004M

=> Nong dp H* cua dung dich:

pH = 9,0 [ H i = lO ^" = lO-^'M =>[0H1 = = lO'^M

VI pH = 9,0 nen dung dich c6 moi tru'dng kiem => phenolphetalein doi sang m^u hong

Ma a = 0,43% = 0,0043 = y =:> x = 0,0043 = 4,3.10"^

^8,57.10-^-[NH3] 1-x 1-x 1-4,3.10"^

Vay Kb = 1,857.10"'

Phan dsino va phiiong phip giai H6a hpc 11 VP c o - D5 XuSn Hung

IH*] = = J O : ! ! = 0,23.10-'' = 2.3.10-'^

[ O H " ] 4,3 10"^

pH = - l g [ H l = -lg(2,3 10-'')= 11,64

Bai 7 Tinh nong dp mol ciSa cdc ion H* va O H ' trong dung djch NaN02 1,0M

Biet n^ng h^ng so phan li bazd cda NOj la Kh = 2 ,5.10'"

Giai

Phuang trinh di§n l i : NaNO: > Na* + N O j

I M I M Phan tfng thuy phan : N O j + H2O -> HNO2 + OH"

0,lM Biet a 25"C cua CH3COOH la 1,75.10-^ va bo qua sy phan li cua nuac

Tinh pH cua dung djch X

'' Trich de thi tuyen sink Dgi hgc khoi B''

Giai

0,1 0,1 Can bang: CH3COOH -> CH3COO +

Bai 9 Cho dung djch HCl c6 pH = 3 Ctin pha loang dung dich axit nay bhng

nurdc bao nhieu Ian dc thu diTtJc dung djch HCl c6 pH = 4?

Giai

Phtfdng irinh diOn l i : HCl > H* + Cl"

Goi V, la the lich dung djch HCl ban dau c6 pH = 3

pH = 3 IH*] = lO'^M mh [H^] = - J l => n^^ ^ = lO-^V, (mol)

Goi V2 la the lich dung djch HCl sau khi pha loang c6 pH = 4

n + pH = 4 => [H*] = lO-'^M ma [H*l = - j ^ - => n^+ = 10-^V2 (mol)

Khi pha loang dung djch so mol H* khong thay d o i : n + = n ^

=> pH = - I g l H * ] = -lg(3.71.10-' + 10-') = 2,33

Bai 11 Trpn 100 ml dung djch hon hrtp gom H2SO4 0,05M va HCl O.IM vdi 100

ml dung djch hon hdp gom NaOH 0.2M va Ba(OH)2 O.IM ihu diTcJc dung djch

X Tinh pH cua dung djch X

{Trich de thi tuyen sinh Dai hoc khoi B)

Phan d^ng phipng ph&p giii H6a hoc 11 VP co - D5 XuSn Hung

Gidi

Ta c6: n = 5.10'Wl; H H C I = 0,01 mol => 2 n + = 0,02 mol

H 2 S O ^ H

" B a ( O H )2 = 0.01 mol; HNaOH = 0,02 mol I n^^ = 0,04 mol

Phan ung trung h6a: H* + OH" -* H2O

0,02 0,02

n d„ = 0,04 - 0,02 = 0,02 mol

O H

=> [OH] = 0,02 : 0,2 = 0,1M => pOH = 1 ^ pH = 13

Bai 12 Trpn Ian V ml dung djch NaOH 0,01M vdi V ml dung dich HCl 0,03M

difcJc 2V ml dung dich Y Dung dich Y co pH la:

0,01V 0,01V

=^ = 0,03V - 0,01V = 0,02V (mol)

H

H^ 0,02 V 2V = 0,01 = 10"2 M =>pH = 2 ^ DapanC

Bai 13 Tron 100 ml dung djch c6 pH = 1 gom HCl va HNO3 vdi 100 ml dung

dich NaOH nong do a (mol/1) thu di/dc 200 ml dung dich co pH = 12 Gia Iri

cua a la: (biel trong moi dung djch [H* ][0H""] = 10"'^)

Dung djch sau phan tfng co pH = 12 (moi triTdng bazcJ) => sau phan \ing tren

OH' dir, H* het

I • Theo phiTdng irlnh phan tfog: n phaming = n 0,01 mol

Bai 14 Tron 250ml dung djch hon hdp HCl 0,08M va H2SO4 0,1M vdi 250ml

dung dich Ba(0H)2 aM thi thu du'dc m gam ke'l lua va 500ml dung dich co

Dung djch sau khi tron co pH = 12 (moi trifdng bazd)

=> Sau phan iJng OH" dif

H* + OH" ^ H2O

0,07 0,07 n,^„_ = (0,5a - 0,07) mol

13

Phan djinfl v i phuang pMp giii H6a hqc 11 VP cO - Dg Xuin Hung

Bai 2 Co hai dung dich sau :

a) CHjCOOH O.IOM (K, = 1,75.10"*) Tinh nong do mol cija ion H\

b) NH3 0,1 OM (Kh = 1,80 lO-') Tinh nong do mol cua ion OH"

a) Them tir tir lOOg dung dich H2SO4 98% v^o niTdc va dieu chinh de diTcJc 1 lit

dung dich A Tinh nong do mol cua ion H* trong dung dich A

b) Phai th6m v^o 1 lit dung dich A tren bao nhieu lit dung dich NaOH 1,8M de

= > n H 2 S 0 4 d ^ = ( l - 0 9 V ) mol

Sau phdn i?ng : Vjj = 1 + V p H = l =>[H1 = 10-'=0,1(M)

H2SO4 > 2H* + S O ^

(1-0,9V) (2-1,8 V)

= > [ H 1 = ^ - ^ = 0.1

1 +V Giai ra ta CO V = 1 (lit)

• pH= 13 => dung dich c6 tinh bazd

Phati djing va phuang phap glSi H6a hge 11 VP c o - D5 Xuan Hi/ng

a) Tinh pH cua dung dich chtfa l,46g HCl trong 400ml

b) Tinh pH cua dung djch tao thanh sau khi tron 100ml dung dich HCl 1,00M

vdi 400ml dung dich NaOH 0,375M

Vdj sau phan i?ng = 0,4 + 0,1 = 0,5 (lit)

PhiTcJng trmh phan tfng : H^ + OH" — - » H2O

0,1 mol 0,1 mol n^ - =0,15-0,1 = 0,05(mol)

"OH dU

[OH-] = - ^ = 0,l(M>

[Hi = h2:lI^=x(r^^.(Pi^-^ 0^110^^ = [§,

Bai 6 Tinh nong do [H*] va [OH"] trong dung dich CH3COOH 2M Neu sau do

them v.^o moi lit dung djch axit tren 0,2 mol muo'i CHjCOONa thi [H*] va [OH"] cua dung dich tang hay giam bao nhieu Ian Bic't hang so' phan 11 axit

la 2.10"'* va the tich dUng dich thay do'i khong dang ke

Gidi

Phircfng trinh dien l i : CH3COOH -> CH^COO" + H* (1)

Bandau: 2 0 0 (M) Can bang: 2 - x x x (M)

nen lam cho nong do CH3COO- trong dung djch tang len do do can bang d

phdn tfng ( i ) dich chuyen theo chieu tOf phai sang irai, nen can b^ng duftJc thanh lap

CH3COOH -> CH3COO- + H*

Bandau: 2 0 0 Canb^ng: ( 2 - y ) (0,2+ y) y

[OH1 = 10 -14 = 0,5.10"'" =5.10-" (M)

2.10 -4

So vdi dung dich CH3COOH ban diu:

[ 0 H 1 tang len = 5.10 - 3 r W e- 4 i ^ 4 f t - v 4 - t t I 1 gi^ni 31,64 Ian

l.:;aH).7'ViEN TINH BIN.H THUAN '

PhSn dgng va phuong phap giii H6a hgc 11 Vfl cO - B5 XuSn Hang

Bai 7

a) Tinh pH ciia dung dich A la hon hcJp gom HF 0,1M va NaF = 0,1M

b) Tinh pH cua 1 lit dung djch A d tren trong hai trUdng hcTp sau :

- Them 0,01 mol HCl vao;

- Them 0,01 mol NaOH vao

Bie't rang hang so axit (hhng so ion hoa) cua HF la = 6,8.10""*

the tich bang nhau, thu dtfcfc dung dich A Lay 600ml dung djch A cho tac

dung vcti dung dich B gom NaOH 0,3M va KOH 0,15M Tinh the tich dung

dich B can dijng de sau khi tac dung vdi 600ml dung dich A thu difdc dung

HNaOH = 0,3.V (mol)

nKOH = 0.l'5.V(mol) NaOH > Na* + OH"

0.3V (mol) 0,3V (mol) KOH > K* + OH"

The tich dung dich sau phan tfng : V j j = 0,6 + V

pH = 3 [ H i = 10-'= 0.001 (M)

[HI = ^ = "'Q^-Q''^^^ = 0.001 V = 0.154 lit

V 0.6 + V Vay the tich dung dich B la 0,154 lit

dich Ba(0H)2 cd nong dp a mol// thu di/dc m gam ket tua va 500ml dung dich

I cd pH = 13 Tinh a va m Cho biet trong cac dung dich vdi dung moi la niTdc, tichso [ H l [ O H - ] = 1 0 - ^

19

Phan dgng va phuong ph^p giii H6a hpc 11 VP CO - 05 Xuan Hi/ng

20H-0,3a mol 20H-0,3a mol 0,6a mol

Khi trpn dung dich gom HCl va H2SO4 vdi dung dich Ba(0H)2 ta c6 cac

phu'dng trinh sau :

a) Dung dich CH3COOH O.IM c6 dp di^n li a = 1% Vifi't phUdng trinh dien li

CHiCOOH va tinh pH cua dung dich nhy

b) A Ih dung djch HCl 0.2M; B la dung dich H2SO4 O.IM Trpn cac the tlch b<^ng

nhau cua A va B diTpc dung dich X Tinh pH cua dung dich X Cho lg4 = 0,6;

Gidi

a) PhiTdng trinh dien li: CH3COOH - » CH3COO- +

BandSu: 0,1 0 0 Can bang: 0 1 - x x x -.100% = 1% =^ X = 10-' hay [H^ = IQ-' (M)

(M) (M)

Bai 11 Trpn dung dich X chuTa NaOH 0,1M, Ba(0H)2 0,2M vdi dung dich Y

(HCl 0,2M; H2SO4 0,1M) theo ti Ip nao ve the tich de dung dich ihu diTpc c6 pH= 13?

Gidi

Taco: nB,(OH)2 = 0,2Vx molj

" N a O H = 0,lVx mol J ~ " ^ ' 0 H

-nH2S04 = 0,1VY mol T

"HCI - 0,2VY mol

Sau khi trpn dung dich c6 pH = 13 moi triTclng bazd => sau phan ifng giiTa axit va bazd thi OH" diT:

Phfln-djino va phuong ph^p gSi HdaJipc 11 VP co - D5 Xuan Hifng

Hkil2 Tron iOOmI dung^ljch g6m Ba(OH)2 O.IM va NaOH 0,1M vdi 400 ml

dung dich gom H2SO4 0/)375M va HCl 0,0125M thu di/dc dung djch X Gia

tri pH cua dung dlchX 1^:

B a i 13 Cho m gam hon hdp Mg va Al vio 250ml dung djch X chiJa hon help

axit HCl I M axit H2SO4 0.5M thu difdc 5,32 lit Hj (dktc) va dung dich Y

(coi the tich dung dich khong doi) Dung dich Y c6 pH la:

= 0.1 = 10"' M pH = 1

KifAwrfiinET

D a n a 3

- Phan ang trao ddi ion

- X a c dinh vol tro axit, bazd, trung tinti hay li/dng tinh

- NhOn bid't dung djch v a ion

BAI TAP M A U Bai 1 Viet phiTdng trinh phan tuf, phifdng trinh ion, phiTdng irinh ion thu gon ciia

cdc p h a n uTng (neu c6) xay ra trong dung dich giffa c a c c a p cha't sau :

a) FeCb + NaOH b) CUSO4 + Ba(N03)2 c) K2CO3 + NaCl d) NajS + HCl e) NazCOj + H2SO4 0 KCl + AgNOj

Gidi

a) FeCb + 2NaOH > Fe(OH )24' + 2NaCl Fe^* + 2Cr + 2Na* + 20H- > FeCOH).! + 2Na* + 2Cr

Fe^* + 2 0 H - ).Fe(0H)2i

b) CUSO4 + Ba(N03)2 > BaS04i + Cu(N0,)2

Cu^* + SO4" + Ba^* + 2NO3 > BaS04i + Cu'^ + 2NO3 Ba^* + SO4- > BaS044'

c) KzCOs + NaCl >• khong xay ra

d) Na2S + 2HC1 > 2NaCl + H2St 2Na* + S^- + 2H* + 2Cr > 2Na^ + 2Cr + H.St

S^- + 2H* j-HzSt

e) NazCOj + H2SO4 > 2Na2S04 + COjt + H2O 2Na* + C 0 | - + 2H* + S O ^ > 2Na" + SO^" + CO2T + H2O 2H* + C O | - • COzt + H2O

0 K C l + AgNOj >-AgCli + KN03 K* + Cr + Ag* + NO3 > AgCli + + NO3

Ag* + cr > AgCli

Bai 2 Chi dung mot thuoc thijr h\, hay trinh bay each nhan bicl

ba dung djch co cClng nong do mol sau : NaOH, H3PO4 va H2SO4

Gidi

~ Cho dung djch phenolphetalcin tifng giot Ian liTdt vao ba mau thi'r mau nao c6 mau hong h\g dich NaOH con lai la H3PO4 va H2SO4

- Lay ba dung dich vdi V b^ng nhau sau do them viio hai dung dich axit Ian

iiTdt vai giot phcnolphctalcin Do tiep V ml dung djch NaOH vao tiTng axit

Phan dgng phuong ph^p giai H6a hgc 11 V6 cO - D8 Xuan Htflig

them mot it dung djch NaOH ntJa neu thay dung djch c6 m^u hong Ih H3PO4,

con hii la dung djch H 2 S O 4 khong hiOn tiTcfng

B a i 3 Rau qua kho dU'cJc bao quan bhng khi SO2 thi/dng chtfa mot liTtJng nho

hdp cha't CO goc S O 3 " De xac dinh suT c6 mat cua cac ion SO3" trong rau

qua, mot hoc sinh ngam mot it qua dau trong niTdfc Sau mot thcli gian Ipc lay

dung dich roi cho tac dung vdi dung djch H2O2 (cha't oxi hoa), sau do cho tac

diing ticp vdi dung djch BaCl2 Viet cac phiTcfng trinh ion rilt gon da xay ra

Gidi

PhiTdng trinh ion rut gon :

S O ^ + H2O2 > SOJ- + H2O

Ba^* + S O 4 " > BaS04>l'

Bai 4 NhuTng hoa chat sau thi/clng di/dc dung trong cong viec noi trcJ : muoi an,

giam, bot nd (NH4HCO3), phen chua (KAI(S04)2.12H20), muoi iot (NaCl +

KI) Hay dilng cac phan iJng hoa hoc de phan biet chiing Viet phiTtfng Irinh

ion riit gon cua cac phan tfng

Giai

Cho cAc chat tren vao nxidc de dufcJc cac dung dich :

* Dung djch muoi an (NaCl)

Cho dung dich AgNOs vao muoi an -> hien ti/cfng A g C l i trang

* Phen chua (KAI(S04)2.12H2O)

Khi hoa tan phen chua v i o niTdc c6 ket tua keo trttng xusi't hien

PhiTcJng trinh ion rut gon:

Bai 5 V i c t cdc phuTcfng trinh phan lit \k ion riit gon cua cac phan iJng (neu c6)

xay ra trong dung djch giiJa cdc cap chat sau :

a) Fe2(S04)3 + N a O H b) NH4CI + AgN03 c) NaF + H C l d) MgCl2 + K N O 3

e) FeS + H C l i") K O H + HCIO

Bai 6 Viet phiTdng trinh ion rut gon cua cac phiin tfug hoa hoc sau:

(1) (NH4)2S04 + BaCl2 (2) CUSO4 + Ba(N03)2 ^ (3) Na2S04 + BaCl2 -> (4) H2SO4 + BaS03 ->

(5) (NH4)2S04 + Ba(0H)2 -> (6) Fc2(S04)3 + Ba(N03)2 ^

GiaM

(1) (NH4)2S04+ BaCl2 ^ 2NH4CI + BaS044'

-> pt ion thu gon: Ba^"" + S04^' -» B a S 0 4 i

(2) CUSO4 + Ba(N03)2 ^ Cu(N03)2 + BaS04 i

-» pt ion thu gon: Ba^* + S04^" -> BaS04 i (3) Na2S04+ BaCl2 - * 2 N a C l + B a S 0 4 i -> pt ion thu gon: Ba^"^ + S04^" BaS04 i

(4) H2SO4 + BaS03 -> SO2 + H2O + BaS04 i

-> pt ion thu gon: 2 H * + S04^" + BaS03 ^ SO2 + H2O + BaS04 i

(5) (NH4)2S04 + Ba(OH)2 -> 2 N H , + B a S 0 4 i + 2H20 -> pt ion thu gon: 2NH4* + S04^"+ Ba'^+ 2 0 H - ^ 2NH3 + BaS04 i + 2H2O (6) Fe2(S04)3 + 3Ba(N03)2 -> 2Fc(N03)3 + 3BaS04 i

pt ion thu gon: Ba^"" + S04^" -> BaS04 i

H6a hQC 11 V6 CO - DS XuSn Hung

Bai 7 Trong cac ion sau: CHjCOO", CO^", HCO^, H S O 4 , CP, N H 4 ,

AKHzO)^", S^', Q H 5 O " , K"" la axit, bazd, trung tinh hay lU3ng tinh? T a i sao?

A 1 ( H 2 0 ) ' V H2O - > H3O* + Al(OH)'^

* Cac ion la bazd : CHjCOO", C O ^ , S^", CfH^Q- v i chung c6 kha nang nhan

proton (theo Bronstet)

* Cac ion la irung tinh : C r , K"^ v l chung khong cho cung khong nhan proton

* Ion la imlng tinh : H C O J v i viTa cho, viTa nhan proton

H C O 3 + HjO^ -> CO2 + 2H2O HCO3 + H2O C 0 | - + HiO\

Bai 8 Viet phiTcJng trinh phan tuT, phU'cJng trinh ion rut gon cua ctic phan tfng sau:

c) BaC03 + 2HNO3 > Ba(N03)2 + CO.T + H2O

BaC03 + 2H* > Ba^* + C02t + H2O

d) CaHP04 + H2SO4 > CaSOA + H3PO4

e) FeCb + 3 K 0 H > F e ( 0 H ) 3 i + K C l

Fe^^ + 3 0 H - > F e ( 0 H ) 3 i

Bai 9

a) Theo dinh nghTa mdi cua Bronstet Cho quy tim vao cac dung djch sau se c6

mau gi? CHjCOONa, Ba(N03)2 NH4CI, K2CO3? Gifii thich?

b) D i / doan dung dich cho difdi day c6 gia tri p H Idn hOn hay nho hcfn 7? NazCOj, K C l C H 3 C O O K , NaHS04 Al2 (S04)3 Na2S, CfiHsONa?

Gidi

a) Theo dinh nghTa m d i cua Bronstet: Axit la cha't nhi/dng proton (H*), bazO Iti chat nhan proton

* Cho quy tim vao dung dich CHsCOONa -> quy tim hoa xanh

V i : CHjCOONa )• CHjCOO" + Na*

* Cho quy tim vao dung dich K2CO3

-V l : K2CO3 > IK" +

COJ-quy tim hoa do

quy tim hoa xanh

C O ^ + H 2 O -> H C O 3 + OH"

b) * Dung djch c6 p H = 7 : K C l

* Dung dich c6 p H < 7 : NaHS04, Al2(S04)3

* Dung djch c6 p H > 7 : C H 3 C O O K Na2C03, NiXjS, CfiHsONa

Bai 10 Chi dung quj' tim hay phan bi?t niim lo mat nhan sau: H C l , H2SO4, K C l ,

NaOH, BaCl2

Gidi

~ L a y m o i l o m o t i t d e l a m m l i u t h i i f

- Khi cho quj'tim vao tifng mSu thuf:

+ M a u lam quy tim hoa do la H C l va H2SO4

+ Mixu Ihm quy tim hoa xanh la NaOH

+ Hai mau khong Ic^m doi mau quy tim la K C l vii BaCb

- Cho mot trong hai mau lam quy tim hoa do vao hin liTtft hai dung djch m u o i :

+ Dung djch H C l khong tac dung difdc \6i hai mu6'i

+ Dung dich H2SO4 tao ket tua triCng vdi BaClj con vdi K C l thi khong hien

tircJng

BaCl2 + H2SO4 > BaS04>l + 2HC1

Phan dgng va phiiong ph^p oi5i H6a hpc 11 VP cO - D5 XuSn Hang

B A I T A P A P D g N G Bai 1 Viet phufdng trlnh ion rut gon cua cac phan iJng (neu c6) xay ra trong

dung djch giiJa cac c$p chaít sau :

e) khong xiiy ra 1) khong X i i y ra

g) Sn(0H)2 + 2 0 H - -> SnÔ" + 2H2O h) Cú* + H.O -> Cu(OH)^ + H ^

Bai 2 V i c t phÚcfng trinh dang phan tur ufng vdi phúdng trinh ion rut gon sau

a) Câ* + COJ~ > C a C O s i b) Pb^^ + SÔ" > PbSOai

a) The nao Ih muói trung h6a, muoi axit? Cho vi diị Axit pholphord H3PO3 la axit

hai lan axit, vay hdp chat Na2HP03 la muói axit hay muoi trung hoả

b) Vic't phuTdng trlnh ion rut gon cua cac phan iJng thuy phan cac muoi :

NaHCOj, NH4CI, H C O O N H 4 , K2SO4 Trong cac phan itng nay núdc dong vai

tro axit hay bazd?

Gidi

a)* M u o i trung hoa \h muói ma anion goc axit khong con hidro c6 kha nang

phan l i ra ion H"^ (hidro co ifnh axit)

NÚdc khong thuy phan nen H2O trung tinh

Bai 4 Cho bon lo di/ng bon dung djch : HCl, BaCb, K O H , Fe2(S04)3 Nhffng cap dung djch nko c6 the ph5n uTng vdi nhaủ V i saỏ Viét phifdng trinh hoa

hoc cua cac phan lirng xay ra dúdi dang phan tuf va ion riit gon

Gidi

* Dung dich H C l phan tfug difdc vdi dung dich K O H

HCl + K O H > K C l + H2O

H * + O H - > H2O

* Dung djch BaCb phan uTng diTdc vdi dung dich Fe2(S04)3

3 B a C l 2 + Fe2(S04)3 > 3BaS044 + 2FeCl3

Bâ* + SỖ > BaS044'

* Dung djch K O H phan uTng diTdc vdi dung djch Fe2(S04)3

6 K 0 H + Fe2(S04)3 > 3K2SO4 + 2Fe(OH)3>l'

Fê* + 3 0 H - > Fe(0H)3>l'

Bai 5

a) Lan liTc^t nhung giay qu>' tim v^o cAc dung djch muoi sau : NH4CI, KNO3, Na2S

thl giáy quy đ co mau gl? Giai thich ng^n gon vil viét phiTdng trinh phAn