Phân dạng và phương pháp giải hóa học 11 phần vô cơ

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i
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In
Ian
thur
1. So
lUcfng
2.000
cuon,
kho
16x24cm.
T^i:
Cong
ty
TNHH
MTV
in an
MAI
THjNH

DLfC.
Dja
chl:
71,
Kha Van
Can,
P.Hi$p
Binh
Chanh,
Q.Thu
Dure,
Tp.HCM.
So
DKKHXB:
1482-12/CXB/12-1SIAHTPHCM
ngay
06/12/2012.
Quyet
djnh
xuat
b5n s6':
1771/QD-THTPHCM-2012
do NXB
T6ng
hap
Thanh
pho Ho Chi
Minh
cS'p
ngdy

27/12/2012.
In
xong
va nop
IIAJ
chieu
qwy I nSm 2013.
A.
T6M
TAT
Lf
THUY€T
I.
SI/DI|NU
*
SI/
di^n
li
1^
qui
trlnh
phan
li
cac chat trong
nifdc
ra ion.
Vf
du :
NaOH
>

Na* +
OH"
*
ChS't
di$n
li
Ih
nhifng
chat tan trong nurdc phan
li
ra ion.
C6
hai
loai:
Chat
di0n
li manh va chatdi^n
li
yeu.
*
DO
diOn
li (a): Dp
diOn
li (a) cua chat
dipn
li la ti
so
giffa
s6'

phan tur phan li
ra ion (n) va tong so phan tur hoa tan (no).
a=JL
O^a^l
«()
Neu
bieu di^n
diTdi
dang phan trSm : 0% < a ^ 100%
u
a = 0 : chat kh6ng
di$n
li
u
0 < a < 1: cha't
di^n
li
yeu
u
a = 1 : cha't
diOn
li
manh.
II.
AXIT-BAZCJ-MU6'l
1.
Axit
*
Theo thuye't
A-r6-ni-ut:

axit
Ift
cha't
khi
tan trong nrfdc phan
li
ra cation
H*.
Vi
du : HCl
>
H*
+
CI"
*
Theo thuyd't Bron-stet:
axit
Ih chat nhtfcJng proton
(H*).
Vi
du :
CH3COOH
+
H2O
U
CH3COO-
+
H3O*
Nhffng
axit

khi
tan trong nurdc ma phan tur phan li nhieu na'c ra ion
H*
la
cac
axit
nhieu nac.
2. Bazcf:
*
Theo thuye't
A-re-ni-ut:
bazd la chat
khi
tan trong nirdc phan
li
ra anion OH'.
Vidu:
KOH J-K^ + OH"
*
Theo thuyet Bron-stet: bazd la cha't nhan proton.
Vidu:
NH3 +
H20->
NH;+0H-
Nhffng
bazd khi tan trong
nxSdc
mi phSn tur phan li nhieu nS'c ra ion OH" la
cdc
bazd nhieu nSfc.

3.
Mu6'i
Muoi
la h0p chSft khi tan trong nurdc phan li ra cation kim
loai
(ho&c
cation
NH4)
va
anion g6c
axit.
C6
hai
loai:
+
Muoi
axit:
NaHCOj,
KHSO4
+
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
kH20
=[H^]-[OH']
=
1-010''^
([Hi = [0H-] = I.O.IQ-' (mol//) d 25"C)
* Moi trurdng axit la moi
trifdng
trong d6 :
[H1>[0H']
hay [H^ >
1,0.10"^M
* Moi triTcJng kiem 1^ m6i triTdng trong do :
[H1<[0H-]
hay [Hi < 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 vl pH. ChS't chi thj
axit
-
bazof
[Hi
=
1,0.10"''"
(M)
Ve mat toan hoc : pH =

-lg[Hl
+ 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 CAC
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
K2CO3
+ 2HC1 > 2KC1 +
C02t
+ H2O
PhiTdng
trinh
ion rut gon : 2H* + CO^" > COzt +
H2O.
B.
PHAN
LOAI
VA PHlTdNG PHAP
GIAI
CAC DANG BAI TAP
Dang 1.
- Viet phi/cfng
trinh
di#n li
cac
chdt
-
Tinh

nong dp mol cua
tCfng
Ion
BAI TAP
MAU
VA BAI TAP NANG CAO
Bai 1. Viet phifdng
trinh
dien li cua nhiTng
chat
sau :
a) Cac
chat
dien li manh : NaCl,
HNO3,
KOH,
Na2S04,
Ba(N03)2,
H2SO4,
[Ag(NH3)2]Cl,
CUSO4.5H2O.
b) Cac
cha't
dien li yeu :
CH3COOH,
Mg(0H)2,
H2S,
HCIO, HCN,
Bi(0H)3.
y

Gidi
a) NaCl
>
Na* +
Cl"
b)
CH3COOH
^
CH3COO"
+ H^
HNO3
> H* + NO3
Mg(0H)2
^ Mg'^ + 20H"
KOH
> K* + OH"
H2S
^ 2H^ + S^"
Na2S04
>2Na^+SO^~
HCIO-> H* + CIO"
Ba(N03)2 > Ba^* +
2N0J
HCN -> H^ + CN"
H2SO4
> 2H^ + SO^-
Bi(0H)3
^ Bi^* + SOH".
[Ag(NH3)2]Cl
>

[Ag(NH3)2]"
+ Cr
CUSO4.5H2O
> Cu^* + S04^' +
5H2O
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 :
Ba(N03)2:0,10M;
HNO3:
0,020M;
KOH : 0,01
OM
HBr04 :
0,025M;
HMn04 :
0,030M;
NaC104
:
0,040M.
Gidi
•

Ba(N03)2
> Ba^* +
2NO3
0,1
OM
0,1
OM
0,20M
=>[Ba^1
= 0,10M
[NO3] =
0,20M
•
HNO3
> H* + NO3
0,020M 0,020M 0,020M
=>[H1=
[NOJ]
=0,020M
• KOH > K* + OH"
0,0
lOM
0,0
lOM
0,0
lOM
=>[K1 = [OH"] =
0,010M
Phan
dang

va
pniiono
ph^p giai H6a hpc
11
VP cc
-
Bi
XuSn
Hung
HBr04
0.Q25M
+
BrO^
0.025M 0,025M
r:>[Hl=
{BrOJ]
=0,025M
•
HMn04
f W *
Mn04
0,030M 0,030M 0.030M
r^iHi
=
IMnO^J =0.030M
•
NaClOa
> Na* + ClO^
(J.040M 0.040M ();040M
==>lNan= lC10;j =0,()40M.

Biii
3.
Vic'l
phifdng irinh di^n li
thco
lifng
na'c
ciia ctic axit saii: H2SO4, H3PO4,
H2S.
H2SO3.
H2SCO4.
Giai
•
H2SO4
>
H*
+
HSOi
HSO4
>
H* + SOi'
•
H3PO4
-> H* +
H2FO4
H2PO;
-> + HPO;'
H2S
->
+

HS-
Hs->ir
+ s^-
H:S03
-> ir +
USO^
H.S05; -> + so;"
H2SCO4
—> H' +
HSco;;
HSc(J4
-> +
ScOj-,
Hsii
4.
Vic't
phiftJng trinh di^n li
va
linh
nong
do
mol
cua cac
ion trong
cac
dung
djch
sau:
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.
cm
a) Nong
do
mol
dung
djch H2SO4:
2H*
+ SO-V
H2SO4
0,15M
n.3M
0,I5M

=>|H*1
=
0.3M;
[S05-1
= 0.15M.
h) So mol NaC!:
HN,,;,
= ~^ = 0.1 (mo!)
NaCl
> Na*
+ Cf
0,5M
0,5M 0,5M
=> [Nal
=
[Cn
=
0,5M.
c) NaOH > Na*
+ OH"
0,250M 0.250M 0.250M
=> [Na*]
=
[OH-]
=
0,250M.
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
:
[Kl
=
O,O50M;[Mg^*]
=
O,20M;
[Cr] =
0,050M; lSOf]=0.20M.
Vie't
cong
Ihtfc
phan
tijr
cua X,
Y
V£k
vie't
phiTdng trinh di^n

li
cua
chung
tronj
dung
djch.
Giai
Theo
de
tW hai hdp
chat
X
va
Y
li
:
KCl
va
MgS04.
Kci
—>
K* + cr
0,050M 0.050M 0,050M
MgS04
>
Mg^*
+ SOj-
0,20M 0,20M 0,20M.
Dqnq
2.

-
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
COIOM.

Giai
•
HCl —• H* + cr
0,10M
0,10M
=>IH*]
=
0,10M; pH
=
-lg[H*J
=
-lgO,l
= l
=>[OH-]=
=1.0.10"'^M.
1.0.10"'
•
NaOH
> Na* + OH"
O.OIOM
O.OIOM
=>[OH-]
=
0.010M
=>[H1=
i:^:H!!l
= 1.0.I0"'2M
1,0.10"^
=>pH
=

-lglH*]
=
-lgl0-'^=12.
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
li cua
mit
axetic
tftng hay
giam?
GiU
thich?
Giai
a) PhiTctng
trinh
di6n
li:
CH3COOH
^
CHiCOO"
+
H*
Trong
1
lit dung dich
c6 0,25
mol
CH3COOH
c6
pH
= 2
=>
[Hi
= 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
=
-lg[Hl
=
-lgI0-'
= 2
Moi
irirftng
cija
dung djch nay lii axil. Quy tim doi
sang
milu
do.
Bai
4.
Tinh pH cOa
cac
dung dich
sau :
a) Dung djch KOH
0,003M.
b) Dung djch H2SO4
0,02M.
c) Dung dich
Ca(0H)2
0,004M
(a = 0,8).
d) Dung djch
CH3COOH

0,004M
(a = 0,7).
Gidi
a) KOH
> K* +
OH-
0,003M 0,003M
[0H-]
=
0.003M
•
I n
10"'"*
^ [Hi
=
ii^ii^
= 0,33.10-'' :^
pH
=
-lg[Hl
= 11,48
3,0.10"^
b) H2SO4
> 2H^ + SO^-
0,02M 0,04M
[HI
=
0,04M
=>
pH

=
-lg[Hl
= 1,4.
KHAWTfi
Vim
c)
Ca(OH)2
> Ca'^ + 20H-
0,004M 0,008M
hay [OH-]
=
0,008M
N6ng do OH"
cua
dung dich: [OH']
= 0,008.a = 0,008.0,8 =
0,0064M
=>
[Hi
=
^'"'^^
=
l,5625.10-'2
=:>
pH
=
-lg[Hl
=
11,81.
6,4.10 ^

d)
CH3COOH
->
CH3COO-
+ H^
0,004M 0,004M
hay [H1
=
0,004M
=>
Nong dp H*
cua
dung dich:
[Hi
= 0,004.a = 0,004.0,7 =
0,0028M
pH
=
-lg[Hl
= 2,55.
Bai
5.
Mot dung dich
c6
pH
= 9,0.
Tinh nong
do
mol
cua cac ion

H""
va
OH"
trong dung dich. Hay cho bid't mau
cua
phenolphetalein trong dung djch nay.
Gidi
lQ-14
pH
= 9,0
[Hi
=
lO ^"
=
lO-^'M
=>[0H1
= =
lO'^M
VI
pH
= 9,0 nen
dung dich
c6
moi tru'dng kiem
=>
phenolphetalein doi
sang
m^u hong.
Bai
6.

a)
Cho hai
chat
NH3
va
C6H5NH2
(anilin)
chat
nao c6
hang
so
bazd
(kb)
Idn
hdn?
Giai
thich?
b) Dung djch NH3 IM
c6 a = 0,43%.
Tinh
hang
so
Kb va pH
cua
dung djch do.
Gidi
a)
Vi
phan tijT anilin
c6 goc

C6H5-
la goc hut
electron
nen lam
giam
mat do
electron
tren nguyen
tijf
N, do do
c6
tinh
bazcJ
yeu hdn phan
tiif
NH3.
•
(NH3)>^b (CfiHs-NHa)-
b)
Phan
tfng
: NH3 + H2O
NH^
+
OH"
1
mol
0 0
(l-x)mol xmol
x

mol
Ma
a = 0,43% = 0,0043 = y =:> x = 0,0043 = 4,3.10"^
Taco:
Kb
=
\^KllO^Kl± jL=
^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
co- D5
XuSn
Hung
IH*]
=
=
JO:!!.
=
0,23.10-''
=

2.3.10-'^
[OH"]
4,3.10"^
pH
=
-lg[Hl
=
-lg(2,3.10-'')=
11,64.
Bai
7.
Tinh
nong
dp
mol ciSa
cdc
ion H*
va
OH' 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
li:
NaNO: > Na*
+ NOj
IM
IM
Phan
tfng
thuy phan
:
NOj
+ H2O
->
HNO2 + OH"
Bandau:
IM 0 0
Canbhng:
(1-x) x x
[HNO,J.[OH-l^^^
[NO2]
(1-x)
„2
hay
—
=
2,5.10-"

1-x
Giai
ra ta
c6:
x =
5.10''^
hay
[OH']
=
5.10-''M
[H*]=
-l^i
= -^^^ =
0,2.10-''(M).
10-'^
_
10-"*
[OH-]
5.10"^
Bui
8.
Cho
dung djch
X
chua
hon hpp gom CH3COOH 0,1M
va
CHjCOONa
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
Phuang
trinh
dien
li:
CH.iCOONa
CH3COO +
Na^
0,1 0,1
Can
bang:
CH3COOH
->

CH3COO
+
Bandau:
0,1 0,1 0
Phan
li:
X x x
Can
bang:
0,1-x
x + 0,1
x
=:»Kc='^'^'^"^"'^^=:
1,75.10-^
(Dieuki|n:0<x<0,l)
0,1-x
=>
X ,=
1,75.10'(nh^n);X2
=
-0,1
(lo^i)
=>
pH =
-
Ig [H=
-
log
(1,75.10') = 4,76
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
li:
HCl
>
H*
+
Cl"
Goi

V,
la the
lich
dung djch HCl ban dau
c6
pH
= 3.
.
pH
= 3
IH*]
=
lO'^M
mh
[H^]
= -Jl
=>
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 doi:
n + =n ^.
H
d H s
=>10-^V,
=
10-^V2 :^

=
^
= 10
=:>V2=10V,
V$y
pha
loang dung djch HCl 10 Ian nghla
la
phai pha loang 1
the
tich
dung
djch
HCl vdi
9
the
lich
niTdc
nguycn cha't.
Bai
10. Dung djch
X
gom
CH3COOH
IM
(Ka
=
1,75.lO')
va
HCl

0.001
M.
Tinh
pH
cAa dung djch X.
(Tn'ch
de thi
tuyen
sink Dai hoc khoi
A
nam 2011)
Giai
Ta
c6: HCl
->
H*
+ Cr
10-^
10-^
CH3COOH
<
>
CHjCOO
+ H*
Bandau
1 0 lO'
Phanli
x x x
Can
bhng

1-x x
x +10"^
Taco:
iLi^tlE!!
=
1,75.10-^ (*)
1-x
VI
X
< <
1 =>
1
-
X => 1, do do ttr (*) => x^ +10-\-5
X,
=-4,71.10-'(loai)
X2
=
3.71.10' (nhan)
=>
pH
=
-IglH*]
=
-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;
HHCI
= 0,01 mol => 2 n + = 0,02 mol
H2SO^
H
"Ba(OH)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
OH
=> [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:
A. 4 B. 3 C. 2 D. 1
(Trich de thi tuyen sinh Ccio ddn^ khoi A, B)
Gidi
Ta CO :
nNaOH
= 0,0l.V mol

nHci = 0,03.V mol
n =0,01.Vmol
OH
n ,-0,03.Vmol
Phurong trinh ion: H^ + OH' H.O
0,01V 0,01V
=^ = 0,03V - 0,01V = 0,02V (mol)
H
H^
0,02 V
2V
= 0,01 =
10"2M
=>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"'^)
A. 0,15 B.0,30 C. 0,03 D. 0,12.
(Trich de thi tuyen sinh Dai hoc khoi B)
Gidi
Ta co: pH =
1
:^ [H^ =
IQ-'M
=^ n ^ = 0,1.0,1 = 0,01 mol
H
Va:
VNaOH

= 200 - 100 = 100 ml
fiNaOH =
0,1a mol
n . = 0,1a mol
OH
Phifdng irinh phdn tfng:
H* + OH" -> H2O
0,01 0,01
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
OH H
=> n
_j^
= (0,1a-0,01) mol
OH
Mat kh^c, ta co: pH = 12 => pOH = 2
0,1a-0,01
0,2
= 0,01 a = 0,12
=> [OUT ] = 10-^ = O.OIM
Dap an D.
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
pH =12. Tinh a va m.
Gidi
Tacd:
nH2S04
= 0,025 mol
"HCI - 0,02 mol
nBa(0H)2
= 0'25a mol
n ,=0,07 mol
H+
n
2
=0,025 mol
•n = 0,5a mol
OH"
= 0,25a mol
Ba
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
pH = 12 => pOH = 2 => [OH"] = 10"'M
n . = 0,01.0,5 = 0,005 mol
OH

Ta co: 0,5a - 0,07 = 0,005 ^ a = 0,15
Ba^* + S04^- BaS04 i
0,025 0,025
mi = 0,025.233 =
5,825
(g) Dap an B.
m
BAITAPAPDyNG
Mi 1. Tinh nong dp mol cua ion H* trong dung dich
HNO2
0,10M bie't ring
hlng so phan li axit cua
HNO2
la Ka = 4,0.10-^
Gidi
I
PhiTdng trinh di^n li:
HNO2
^ H^ + NO2
Bandau: 0,1 OM 0 0
Canbkng: (0,1-x) x x
13
Phan
djinfl
vi phuang pMp
giii
H6a hqc 11 VP cO - Dg Xuin
Hung
[HNO2]
(0,1-X) 0,1-X

Vi
HNO2
la
axit
ycu nen x « 0,1
nen — =
4.10"*
=> x =
6,3.10"^
0,1
Vay
[H1 = 6,3.10-^M.
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".
Giai
a)
CH3COOH
CH,COO-
+ H*
Bandau: 0,1M 0 0
Canb^ng: 0.1-x x x
^
_ [CH3COO-].[H-] X.X _ x^ 5
[CH3COOH]
0,1-X 0,1-X ' •
Vi
CH3COOH
la
mOt
axit
yeu nen
X
« 0,1
nen — =
1,75.10"^
0.1
Giai
ratac6x=
1,32.10-^M
hay [H*] =
1,32.10'^M.

b) NH3 + H2O -> NH; + OH-
Bandau: 0.1 0 0 (M)
Canbkng: 0,1-x x x (M)
[NH;].[0H-]_
X.X _ x^ 3
[NH3]
-0,l-x-0,l-x-^'^°-^°
VI
NH3 la mpt
bazd
yeu nen x « 0,1
—
=
1,80.10"^
=>x=
1,34.10"^
Vay
[0H-] =
1,34.10-^M.
Bai 3.
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
thu dU'dc:
-
Dung djch c6 pH = 1;
-
Dung dich c6 pH = 13.
14
Giai
a) Khoi liTdng chat tan H2SO4 :
C%.m,,d 98%. 100 , 98 , . ,.
'"^••=
-^
=
-lB5^
=
''^^^'"^=^
"" ^
=^-Un.ol)
H2SO4 > 2H* + SO^
1 mol 2 mol
=>[H1=

Y =
2(M).
b) Goi V la the tich
dung
djch NaOH c6
nong
do 1,8M
=>nNaOH=
1.8.V(mol)
2NaOH + H2SO4
)•
Na2S04
+ 2H2O
1,8V 0,9V
=>nH2S04d^=(l-0.9V)
mol
Sau phdn i?ng : Vjj = 1 + V
.
pH=l =>[H1 =
10-'=0,1(M)
H2SO4 > 2H* + SO^
(1-0,9V)
(2-1,8
V)
=>[H1=^-^
=
0.1
1
+V
Giai

ra ta
CO
V = 1 (lit).
•
pH= 13 =>
dung
dich c6 tinh
bazd
=> [H*] =
lO-'^M
=> [0H-] =
10-'M
= 0,1M
H2SO4 + 2NaOH >
Na2S04
+ 2H2O
I
mol 2 mol
=>nNaOHdif=(1.8V-2)moI
NaOH
> Na* + OH"
(l,8V-2)mol (1,8V-2) mol
1 8V-2
=> [0H-] = ' = 0.1 V =
1.2353
(lit).
1
+V
Bai 4.
Tinh

the tich
dung
dich Ba(0H)2 0,025M can cho vao 100ml
dung
djch
hSn hdp gom HNO3 v^
HCl
c6 pH =
1
de tao th^nh
dung
dich c6 pH = 2.
Giai
•Dung djch hon
hcJp
gom HNO3 v^ HCl c6 :
lpH=
1 =>[H1 = 10"' =0.1Mii> n ,
=0,1.0,1
= 0,01 (mol)
[Goi
V
\h.
th6 tich
dung
dich Ba(OH)2 0,025M ^
15
Phati
djing
va

phuang
phap
glSi
H6a
hge
11 VP co- D5
Xuan
Hi/ng
nBa(OH)2
=
0,025.V
(mol)
Ba(OH)2
>
Ba^*+
lOW
0,025V 0,025V
(mol)
Phifdng
trlnh
phin
tfng
:
H*
+
OH"
—»
H^O
0,05V 0,05V
Dung

dich tao thanh
c6
pH
= 2
=> [Hi
dir
=
10"''"
=
10-'
=
0,01M
Vjj
sau phan
iJng
=
V
+
0,1
^ Vd>/=^M-V
=
0,01.(V4-0,l)
oO,01
-
0,05V
=
0,01V
+
0,001
o

V
=
0,15
(lit).
Bai
5.
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.
Gidi
1
46
a)
nHci=
—- =
0,04 (mol)
36,5
HCl
—>
H*
+ cr
0,04
0,04 (mol)

^
[HI
=^
=
^
=
0,im=>pli
=
-IgO,
1
=
1.
V
0,4
b) Ta CO
:
nNaOH
=
0,4.0,375
=
0,15 (mol)
NaOH
>
Na*
+
OH"
0,15 mol 0,15 mol
nHci
= 0,1.1 =0,1 (mol)
HCl

—>
H^
+ cr
0,1 mol 0,1 mol
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
li:
CH3COOH

->
CH^COO"
+
H*
(1)
Bandau:
2 0 0 (M)
Can bang:
2-x x x (M)
K.=
[CH3C00-j.[H"]^
x.x ^ ^2
10-5
[CH3COOH]
2-x
2-x
Vi
la axit yeu nen
x « 2.
2
=^^
=
2.10-5
2
X
=
6,324.10"
=> [Hi
=
6,324.10'^

{M)=> [0H-]
=
10
-14
=
1,58.10"'^
(M).
(2)
6,324.10"
Neu them vao moi lit dung djch axit 0,2 mol muoi CHsCOONa
:
Phi/dng
trinh
dien
li:
CHjCOONa
>
CHjCOO"
+
Na*
0,2 mol// 0,2 mol//
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
Vi
Ih axit yeu ndn y
« 2
=> [CH3COOH]
=
2
-
y
= 2
(M)
[CH3COO-]
=
0,2
+
y
=
0,2 (M)

tCH,COO-l.[H'1^0^^^,^
[CH3COOH]
2
=> y
=
2.10"*
hay
[Hi
=
2.
lO""
(M)
(M)
(M)
[OH1
=
10
-14
=
0,5.10"'"
=5.10-"
(M)
2.10
-4
So vdi dung dich CH3COOH ban diu:
[0H1
tang len
=
5.10
-3rWe-4i^4ft-v4-ttI1

gi^ni
31,64 Ian.
l.:;aH).7'ViEN TINH BIN.H
THUAN
'
JO
17
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""*.
Cho
log6,8
=
0.83.
Gidi
a) Phtfdng trinh dien

11:
HF ->
H^
+ F
Trong
dung
dich
c6
F"
nen
lam cho
can
bang
it bi
chuyen
dich
nen c6 the
coi
[HF]
=
0,1M=>[F]
=
0,1M
k^=mlLlE:i.6,8.10-^M)
[HF]
[H1=^'^-^Q"'-^'^=6.8.10-"(M)
0,1
pH
=
-lg[Hl

=
-lg(6,8.10"')
=
3,17.
b)
*
Khi
them
0,01 mol HCl vao thitac6:H^+
F > HF
[HFJ = 0,1
+
0,01 = 0,11 (mol)
[F]
=
0,1-0,01
=0,09
(mol)
[HI
=
6,8.10-r
=
6.8.10"^
—
-
8,31.10-^
[F-]
, 0,09
=>
pH =-lg(8,31.10"")

=
3,08.
* Khi
them
0,01 mol NaOH vao
ta c6 :
HF + OH"
>
F +
H2O
=:>[F]
=
0,1+0,01
=0,11 (mol)
[HF]=
0,1-0,01
=0,09
(mol)
[H^]
=
6.8.10-*.
= 5,56.10"^(M) pH
=
-lg[Hl
=
3,25.
Mi
8.
Tron
ba

dung
dich
H2SO4
0,05M;
HNO3
0,1M
va
HCl 0,15M vdi nhffng
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
dich
CO
pH
= 3.
Gidi
Vi
the
tich blng
nhau
nen the tich moi
dung
dich axit trtfdc khi trpn Ian
la:
— = 200 ml = 0,2
(lit)
So'mol m6i axit trong 600ml

dung
dich A
:
nH2S04
=0,05.0,2 = 0.01 (mol)
nHN03
=0,1.0,2 = 0,02
(mol)
n
HCl
=0.15.0.2
=
0.03 (mol)
•
Phifdng trinh dien li
cdc
axit:
H2SO4
—^ 2H* + SO^~ •
0.01 mol 0,02 mol
. .
HNO3
> H^ + NO3 .
0,02 mal 0,02 mol
HCl
—> H^ + cr
. 0,03 mol 0,03 mol
=>
I =
0.02 + 0.02 + 0.03 = 0.07 (mol)

Goi
the
tich
dung
djch B can
diing
la
V.
HNaOH
= 0,3.V (mol)
nKOH
= 0.l'5.V(mol)
NaOH
> Na* + OH"
0.3V (mol) 0,3V (mol)
KOH
> K* + OH"
0,15V (mol) 0,15V (mol)
=> Z n_.
=
0,3V + 0,15V = 0,45
V
(mol)
OH
Phan
tfng trung
hoa
dung
dich
A va

dung
dich B
:
H*
+ OH- > H2O .
. 0,45V 0.45V
(mol)
Dung dich thu
dufdc
cd pH
= 3
=> moi trtfcfng axit
^^H""
di/.
The tich
dung
dich
sau
phan
tfng
:
Vjj
=
0,6
+ V
pH
=
3 [Hi
=
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.
Bai
9.
Trpn 200ml
dung
dich
g6m
HCl O.IM
va
H2SO4
0.05M vdi 300ml
dung
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
[Hl.[OH-]
= 10-^
19
Phan
dgng
va
phuong

ph^p
giii
H6a
hpc
11 VP CO - 05
Xuan
Hi/ng
Gidi
Ta CO:
nHci
= 0,1.0.2
=
0,02 (mol); nH2S04 -0,05.0,2 = 0,01 (mol)
HCl
—> H^ + cr
0,02 mol 0,02 mol
H2SO4
> 2H^ + SO4"
0,01 mol 0,02 mol 0,01 mol
=>In + =0,02+ 0,02 = 0,04 (mol)
Vataco nB^(OH)2 =0'3.a(mol)
Ba(0H)2
>
Ba^* + 20H-
0,3a mol 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
:
+
OH" > H2O (1)
0,04
0,04
Ba^*
+ SO^- >
BaS04i
(2)
0,01
0,01 0,01
Dung dich
CO
pH= 13 [Hi
=
lO^'^M
=>[0H-]= -i—= 10"' =0,1M
io~'-^
=>
n _
=0,1.0,5 = 0,05 (mol)
OH
M
Tac6:
n _
= 0,6a-0,04
OH

dir
o 0,6a
-
0,04 = 0,05
o a
= 0,15 (M)
n„
2+ = 0,3a = 0,3.0,15 = 0.045 (mol)
Bu
Ttr(2)=:>n
2+
= 0.045-0.01 = 0,035(mol) => ng^so. =0.01 mol
=> m =
mB„s04
= 233.0,01 = 2.33 (gam).
B^i 10.
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;
lg2 = 0.3. (DHQG.d0l)
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)
a
=
0,1
=>pH = -lg[Hl = -lgl0-' = 3.
b) The tich dung djch sau khi trpn : Vjj =
VA

+
VB
= 2.V (lit)
OHci
= 0,2.V (mol)
HCl
—> H^ + cr
0,2V
0,2V
nH2S04
=0,l.V(mol)
H2SO4
> 2H* + SO^"
0,1.V
0,2.V
=>In„.
= 0,2V + 0,2V = 0,4V (mol)
H
[Hi
=
0,4V
= 0,2(M)
V
2V
Vay pH = -lg[Hl = -lgO,2 = 0,7.
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
"NaOH =
0,lVx mol
J~"^'0H-
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:
H^
+ OH" -> H2O
0,4VY 0,4VY
= 0,4VY
mol
n^,^.u.
=
(0,5VX-0,4VY)
mol

Tac6:pH= 13 =>
[Hi =10
1-13
[0H1
= 10
,-1
0,5VY -0,4VX
VX _ 4
Vx
+ VY
91
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^:
A.
2 B. 1 C.6 D. 7.
(Trich
de thi
tuyen
sink Dai hoc khoi B)
Gidi
Ta c6:
nBa(OH)2
= O'O^ mol
^=>
> I.
OH-
"NaOH
= 0,01 mol
nH2S04
= 0.015 mol -j
"HCI
=
0,005
mol
Z
n
= 0,03 mol
Tn. . =

0,035
mol
Khi
tron Ian hon hdp 2 axit 2 bazd xay ra phan
xing
trung hoa:
H*
+ OH" H2O
0,03 0.03
n^^.
=
0.035
- 0,03 =
0,005
mol
H
•HI
0.005
0,5
=
0,01 = 10-2 =>
pH
= 2 =>
Dap an
A.
Bai
13. Cho m gam hon hdp Mg va Al vio 250ml dung djch X
chiJa
hon help
axit

HCl
IM 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:
A.
7
B. 1
C.
2 D. 6.
(Trich
de
tlii
tuyen
sinh Bai hoc khoi A)
cm
Ta c6:
HHCI
= 0,25 mol
"H2S04
= 0,125 mol
5,32
nH2
=
22,4
=
0,2375
mol
mol
Khi

cho Mg, Al tac dung vdi hon hdp 2 axit
HCl
va
H2SO4,
ta c6 sd do phan
«?ng:
2H*
-> H2
0,475
,
0,2375
=> = 0,5 -
0,475
=
0,025
mol
n
+
-1
0,025
H^
0,25
Dap
an
B.
=
0.1 = 10"' M pH = 1
KifAwrfiinET
Dana
3.

-
Phan
ang
trao
ddi ion
- Xac
dinh
vol tro
axit,
bazd,
trung
tinti
hay
li/dng
tinh
-
NhOn
bid't
dung
djch
va ion
BAI
TAP MAU
Bai
1. Viet
phiTdng
trinh
phan
tuf,
phifdng

trinh ion,
phiTdng
irinh
ion thu gon
ciia
cdc
phan
uTng
(neu c6) xay ra
trong
dung dich giffa cac cap
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^*
+ 20H-
).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* + C0|- +
2H*
+ SO^
>
2Na" +
SO^"
+
CO2T
+
H2O
2H*
+ CO|- •
COzt
+
H2O.
0
KCl
+ 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

H2SO4
khong hiOn tiTcfng.
Bai 3. Rau qua kho
dU'cJc
bao quan bhng khi SO2 thi/dng chtfa mot liTtJng nho
hdp
cha't
CO
goc SO3". 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 :
SO^ +
H2O2
> SOJ- +

H2O
Ba^* + SO4" >
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 AgCli trang.
Ag*
+ Cr > AgCli
* Giam
(CH3COOH)
Cho
CaC03
vao c6 hipn tifdng sui bot khi.
2CH3COOH
+
CaCOj
> (CH3COO)2Ca
+ CO.t + H2O
* Bot nS
(NH4HCO3)
Cho dung djch NaOH vao -> khi, mui khai.
NH;
+ OH- >
NHjt
+
H2O
*
Phen
chua

(KAI(S04)2.12H2O)
Khi
hoa tan phen
chua
vio niTdc c6 ket tua keo trttng xusi't hien.
PhiTcJng
trinh
ion rut gon:
Al'*
+
3H2O
^
Al(0H)3i
+
3H*
* Muoi iof (NaCl+ KI)
Cho
chat
oxi hoa
H2O2
vao -> c6
hi<?n
ti/cJng
h sinh ra lam cho ho
tinh
bot c6
mau xanh.
Phi/cJng
trinh
ion

rial
gon:
2r +
H2O2
> h + 20H
Bai 5. Vict 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
+ NaOH b)
NH4CI
+ AgN03
c) NaF + HCl d) MgCl2 +
KNO3
e) FeS + HCl i") KOH + HCIO.
Gidi
a)
Fe2(S04)3
+ 6NaOH > 2Fe(OH)3i +
3Na2S04

Fe^* + 30H- >Fe(OH)3i
b)
NH4CI
+ AgN03 > AgCli +
NH4NO3
Cr + Ag* ).AgCl4^
c) NaF + HCl > NaCl + HFt
F + H* >HFt
d) MgCl2 +
KNO3 >
khong c6 phan tfng xay ra
e) FeS
(r,
+ 2HC1 >
FeCh
+ H2St
FeS + 2H* >Fe^^ + H2St
0
HCIO + KOH > KCIO + H2O
HCIO
+ OH" > CIO" + H2O.
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^' -»
BaS04i
(2)
CUSO4
+ Ba(N03)2 ^ Cu(N03)2 +
BaS04

i
-» pt ion thu gon: Ba^* + S04^" ->
BaS04
i
(3)
Na2S04+
BaCl2
-*2NaCl +
BaS04i
-> pt ion thu gon: Ba^"^ + S04^"
BaS04
i
(4)
H2SO4
+
BaS03
->
SO2
+
H2O
+
BaS04
i
-> pt ion thu gon: 2H* + S04^" +
BaS03
^ SO2 + H2O +
BaS04
i
(5)
(NH4)2S04

+ Ba(OH)2 -> 2NH, +
BaS04i
+
2H20
-> pt ion thu gon:
2NH4*
+
S04^"+
Ba'^+
20H-
^
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^,
HSO4,
CP, NH4,
AKHzO)^",
S^',
QH5O",
K"" la axit, bazd, trung
tinh
hay lU3ng tinh? Tai
sao?
Gidi
* Cac ion la axit :
HSO4, NH4,
AKHjO)^* vi chung c6 kha ming cho proton
(theo
Bronstet).
HSO; +
H2O
^ HjO^ + SO^-
I
*
NH^
+ H2O HjO^ +
NH3
A1(H20)'V
H2O

->
H3O*
+
Al(OH)'^
* Cac ion la bazd : CHjCOO", CO^, S^", CfH^Q- vi chung c6 kha nang nhan
proton
(theo
Bronstet).
CH3C06-
+ H2O
CH3COOH
+ OH"
CO3"
+ H2O
HCO3
+ OH"
S^'
+
H2O
->
HS"
+
OH"
CfiHjO"
+ H2O -> CfiHjOH + OH"
(phenol)
* Cac ion la irung
tinh
: Cr, K"^ vl chung khong cho cung khong nhan proton.
* Ion la imlng

tinh
:
HCOJ
vi viTa cho, viTa nhan proton
HCO3
+ HjO^ ->
CO2
+ 2H2O
HCO3
+
H2O
C0|- + HiO\
Bai 8. Viet
phiTcJng
trinh
phan tuT,
phU'cJng
trinh
ion rut gon cua
ctic
phan tfng sau:
a) CaCl2 +? >
CaCOji
+?
c)
BaC03+?
>Ba(N03)2+?
e)
FeCl3+?
>KC1

+?
b)
Pb(N03)2
+? > PbCbi +?
d) HPO^" +?
>-H3P04+?
Gidi
a)
CaCl2
+
Na2C03
> CaCOji + 2NaCl
Ca^*+CO|-
^CaCOji
b)
Pb(N03)2
+ 2KC1 >
PbCl2>t
+
2KNO3
Pb^* + 2Cr >PbCl24'
c) BaC03 + 2HNO3 >
Ba(N03)2
+ CO.T + H2O
BaC03 +
2H*
>
Ba^*
+
C02t

+
H2O
d)
CaHP04
+
H2SO4
> CaSOA +
H3PO4
e)
FeCb
+ 3K0H > Fe(0H)3i + KCl
Fe^^ + 30H- > Fe(0H)3i.
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) Di/ doan dung dich cho
difdi
day c6 gia tri pH Idn hOn hay nho hcfn 7?
NazCOj, KCl.
CH3COOK,
NaHS04. Al2(S04)3.

Na2S,
CfiHsONa?
Gidi
a)
Theo
dinh
nghTa mdi 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.
Vi:
CHjCOONa
)•
CHjCOO" + Na*
CH3COO
+
H2OCH3COOH
+ OH"
* Cho quy tim vao dung dich
Ba(N03)2
-> quy tim khong doi mau do khong bi
thfiy
phan.

* Cho quy tim vao dung djch
NH4CI
-
VT :
NH4CI
> NH| + Cr
NH;^
+
H2O
NH3 + HjO^
* Cho quy tim vao dung dich
K2CO3
-
Vl: K2CO3
> IK" + COJ-
quy tim hoa do.
quy tim hoa xanh.
CO^
+H2O
-> HCO3
+ OH"
b)
* Dung djch c6 pH = 7 : KCl
* Dung dich c6 pH < 7 :
NaHS04,
Al2(S04)3
* Dung djch c6 pH > 7 :
CH3COOK.
Na2C03,
NiXjS,

CfiHsONa.
Bai 10. Chi dung quj' tim hay phan bi?t
niim
lo mat nhan sau: HCl,
H2SO4,
KCl,
NaOH,
BaCl2.
Gidi
~ Laymoilomotitdelammliuthiif.
- Khi cho quj'tim vao
tifng
mSu
thuf:
+ Mau lam quy tim hoa do la HCl va H2SO4.
+ Mixu
Ihm quy tim hoa xanh la NaOH.
+ Hai mau khong Ic^m doi mau quy tim la KCl vii
BaCb.
- Cho mot trong hai mau lam quy tim hoa do vao hin
liTtft
hai dung djch muoi:
+ Dung djch HCl khong tac dung difdc \6i hai mu6'i.
+ Dung dich
H2SO4
tao ket tua triCng vdi
BaClj
con vdi KCl thi khong hien
tircJng.
BaCl2

+
H2SO4
>
BaS04>l
+ 2HC1.
Phan
dgng
va
phiiong
ph^p
oi5i
H6a hpc 11 VP cO - D5
XuSn
Hang
BAITAPAPDgNG
Bai 1. Viet phufdng
trlnh
ion rut gon cua cac phan
iJng
(neu c6) xay ra trong
dung
djch
giiJa
cac c$p
chai't
sau :
a) NaHSO, + NaOH b) Cu(0H)2 + HCl
c) NajHPOa + HCl d) Sn(0H)2 ,r, + H2SO4
e)
MgS04

+ NaNO, f) KNO, + NaCl
g)
Sn(OH)2+
NaOH h) Cu(N03)2 +
H2O.
Gidi
a)
HSO3
+ OH- SO^~ +
H2O
b)
Cu(0H)2
+
2H^
-> Cu^* +
2H2O
c)
HPO4"
+ 2H* ^
H3PO4
d)
Sn(0H)2
+
2H"
-> Sn'^ +
2H2O
e) khong
xiiy
ra 1) khong Xiiy ra
g)

Sn(0H)2
+
20H-
-> SnO^" +
2H2O
h) Cu'* + H.O -> Cu(OH)^ + H^
Bai 2. Vict phU'cfng
trinh
dang phan tur
ufng
vdi phu'dng
trinh
ion rut gon sau
a) Ca^* + COJ~ > CaCOsi b) Pb^^ + SO^" > PbSOai
c) S^- +
2H^
> HjSt d) H^ + OH" > H2O
c) NH; + OH" > NHjt +
H2O
1)
CO2
+ 20H- -—> CO^ +
H2O.
cm
a)
CaCl.
+ Na.COj > CaCO.ii +
2NaCl
b) Pb(N0.,)2 +
Na2S04

>
PbS04i
+
2NaN03
c) Na.S +
2HC1 > 2NaCl
+ H.St
d)
KOH + HCl > KCl +
H2O
c)
NH4CI
+ NaOH > NaCl +
NH3T
+
H2O
1) CO2
+
2KOH
vKiCOj + HjO.
Bai 3.
a) The nao Ih muo'i trung h6a, muoi
axit?
Cho vi
dii.
Axit
pholphord
H3PO3
la axit
hai

lan axit, vay hdp
chat
Na2HP03
la muo'i axit hay muoi trung
hoa?
b) Vic't phuTdng
trlnh
ion rut gon cua cac phan
iJng
thuy phan cac muoi :
NaHCOj,
NH4CI, HCOONH4,
K2SO4. Trong cac phan
itng
nay nu'dc dong vai
tro
axit hay
bazd?
Gidi
a)*
Muoi
trung hoa
\h.
muo'i ma anion goc axit khong con hidro c6 kha nang
phan li ra ion
H"^
(hidro co
ifnh
axit).
Vi

du : KCl, NajCOj,
*
Muo'i
axit la muo'i mh anion goc axit con hidro co kha niing phan li ra ion H*.
Vi
du :
NaHS04,
KH2PO4,
: KHAlgfiVngr
*
Na2HP03
la
muo'i
trung hoa.
b)
NaHC03
> Na* +
HCO3
HCO3
+
H2O > H2CO3
+ OH"
_
H2O
dong vai lr5 axit.
NH4C1—>
NH^ +cr
NH;+H20
>NH3 + H30"
_

H2O
dong vai tr6
bazd.
HCOONH4
> NH: +
HCOO"
HCOO"
+
H2O
>
HCOOH
+ OH"
NH;
+
H2O > NH3
+
H3O*
_
H2O
dong vai tro krSng
tinh.
K2SO4 >
2K* + SO^-
- NU'dc khong thuy phan nen
H2O
trung
tinh.
Bai 4. Cho bon lo di/ng bon dung djch : HCl, BaCb, KOH, Fe2(S04)3. Nhffng
cap dung djch nko c6 the
ph5n

uTng
vdi
nhau?
Vi
sao?
Vie't phifdng
trinh
hoa
hoc cua cac phan lirng xay ra du'di dang phan tuf va ion
riit
gon.
Gidi
* Dung dich HCl phan
tfug
difdc vdi dung dich KOH
HCl
+ KOH > KCl +
H2O
H*
+ OH- >
H2O
* Dung djch
BaCb
phan
uTng
diTdc vdi dung dich
Fe2(S04)3
3BaCl2
+
Fe2(S04)3

>
3BaS044
+
2FeCl3
Ba^* + SO^~ >
BaS044'
* Dung djch KOH phan
uTng
diTdc vdi dung djch
Fe2(S04)3
6K0H +
Fe2(S04)3
>
3K2SO4
+
2Fe(OH)3>l'
Fe^* + 30H- >
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
gia'y
quy dd co mau gl? Giai thich ng^n gon vil vie't phiTdng
trinh
phAn
tJng
minh hoa.
(DH Buu

chlnh
Vien
thon^
TP.HCM)
b)
Trinh b^y phiTdng phdp h6a hoc nhan bi6't suf c6 m&t cua cac ion trong dung
dich
chtfa
h5n h(?p ba mu6'i: FeCb, CuCh,
AICI3.
(DHAit
ninit)
29
PhSn
djng
va
phuonq
phap
giai
H6a hpc 11 VP
cO
- D8
Xuan
Hung
Gidi
a) Nhung giS'y quy tim v^o Ian liTcft cac muo'i:
* NH4CI Ihm cho quy tim h6a hong.
Vi dung djch NH4CI c6 moi triTdng axit, pH < 7.
NH4C1
—> NH^ + cr

NH^ + H2O -> NH3 + H3O*
* Dung dich
KNO3:
quy tim khong doi mau.
Vi dung dich KNO3 c6 moi triT^ng trung tinh, pH = 7.
KNO3
>K^+ NOJ
Ion K*,
NO3
khong c6 kha nang cho vtl nhan proton.
* Dung dich
Na2S
: quy tim hoa xanh.
Vi dung dich NajS c6 moi triTcfng kiem, pH > 7.
NajS
>
2Na* + S^"
S^- + H2O -> HS- + OH"
HS" + H2O-> H2S + OH".
b) Cho dung dich AgN03 v^o hon hcfp gom cac dung dich thay xua't hien ket tua
tring AgCl chiJng to c6 ion CI".
Ag^ + cr
>
AgCli
* Cho dung dich
NH3
diT vao hon hcJp dung dich, ta thay xuat hien dung
dieh
mau xanh tham va ke't tua -> chtfug to c6 ion Cu^""; cho dung dich NaOH du"
vao ket tua thu dtfdc thay xua't hien ke't tua nau do chiJng to c6 ion Fe^"",

sue khi
CO2
dir vao dung djch con lai, ta tha'y xua't hien ke't tua keo trfng
A1(0H)3
chu-ng to c6 ion AI^^
FeCb +
3NH3
+ 3H2O >
Fe(OH)34'
+ 3NH4CI
(nau do)
AICI3
+
3NH3
+ 3H2O
>
Al(OH)34'+ 3NH4CI
(keo tr^ng)
CUCI2
+
2NH3
+ 2H2O >
Cu(0H)2>l'
+ 2NH4CI
Cu(0H)2
+
4NH3
>
[Cu(NH3)4](OH)2
(dung dich xanh tham)

A1(0H)3
+ NaOH >
NaA102
+ 2H2O
NaA102 + CO2 + 2H2O > Al(0H)3>t + NaHC03
NaOH +
CO2
>
NaHC03.
Bai
6.
a) Cho a moi
NO2
hap thu ho^n toan vao dung dich chiJa a moi NaOH. Dung
dich thu diTdc c6 gii tri pH Idn hdn hay nho hdn 7? Tai sao?
b) Hoan thanh cac phU'ctng trinh sau
vh.
cho biet cha't nao la axit, bazcJ?
C2H5NH2 + H2Q
C6H5OH
+ H2O
CH3COO-
+ H2O
CH3O-
+ H2O.
Gidi
a) Phifdngtrinh phan tfng :
2NO2
+ 2NaOH > NaNQz + NaNOj + H2O
NaN02

—^ Na* + NO2 .
NO2 + H2O -> HNO2 + OH""
Do do
NO2
la mot bazd p pH > 7.
b) . C2H5NH2 + H2O . CzHjNH^ + OH- •
bazd axit
.
CH3COO-
+ H2O ^
CH3COOH
+ OH-
bazd axit .
• C6H5OH+ H2O -> C6H5O- + H3O*
axit bazcf
•
CH3O-
+ H2O ->
CH3OH
+ OH-
bazd axit.
Bai 7. Vie't phiTdng trinh phan
tijT
va phiTdng trinh ion rut gon cua cac phan
vCng
sau :
a) FeS +?
>
FeClz +? b) HNO3 +? >
CH3COOH

+?
c)
Ba(HC03)2
+?
>
BaC03 +? d) AlBr3 +?
>
A1(0H)3 +?
e) K3PO4 +? > Ag3P04 +?
f)
Ca(0H)2
+ NaHC03
)-?
+ CaC03+?
Gidi
a) FeS + 2HC1 > FeCh + H2St
FeS + 2H*
>
Fe^*
+ HjSt
b) HNO3 + CH3C00Na >
CH3COOH
+ NaN03
H* +
CH3COO-
>
CH3COOH
c) Ba(HC03)2 + Ba(0H)2 > 2BaC03i + 2H2O
Ba^* +
HCO3

+ OH"
>
BaC03i + H2O
d) AlBrj + 3NaOH
>
Al(OH)34' + 3NaBr
Al'*
+ 30H-
>
Al(0H)3i
e) K3PO4 + 3AgN03
>
Ag3P04^ + 3KNO3
3Ag* + P0|-
>
Ag3P04i
f) Ca(OH)2 + 2NaHC03
>
Na2C03 + CaC03^ + 2H2O
Ca'* + 20H- + 2HCO3 > COJ- + CaC034< + 2H2O.
Phan d?ng
va
phtfOng phap
g\i\a hQC
11
V6
cd-
D5
Xuan
Hang

Bai
8. Cho
mot
it
phenolphetalein
vao
dung djch
amoniac
loang
chiJa a
mol NH,
dxidc
dung dich
A c6
mau. Hoi m^u
cua
dung dich bie'n ddi nhU'
the nao
trong
til'ng
trUdng hdp.
a) Them
a
mol HCl
vi\
dung djch
A.
b) Them
^
mol

AICI3
v^o
dung djch
A.
Viet
cac
phuTcfng
trinh
phan uTng
xdy ra
(dang ion thu
gon)?
Giai
thich?
(TS
DH Van
Lana,
khoi B)
Gidi
a) Phifdng tnnh phan ufng
:
NH3
+
HCl
>
NH4CI
NHj
+
H^
> NHj

Do do dung djch
A
chuy6'n
sang
khong mau vi muoi
NH4CI
la
mot axit
c6
pH
< 7.
b)
AICI3
+
3NH3
+
3H2O
>
Al(0H)3i
+
3NH4CI
Al^^
+
3NH3
+
3H2O
>
Al(0H)3i
+
3NH^

Do
do
dung djch
A
chuyen
sang
khong mau.
Vi NH4CI
va
AICI3
dcu la
nhufng axit
c6
pH
< 7.
Dang
4.
-
SU ton tqi dong
thdi
ion
trong
dung djch
-
Bai t6p dUa vdo
ptiUdng
trinh
Ion thu gon.
BAI TAP MkU
Bail.

a) Trong dung djch
A c6 cac ion K"",
Mg^",
Fe^* va
CI".
Neu c6 can
dung dich
thu
duTdc
hon hop gom
nhurng muoi
n^o?
Viet cong thtfc
h6a hoc v^ ten cua
muoi.
b)
Can lay
nhi?ng muoi
nao de pha
chd' duTcfc dung dich
c6 cic ion Na*, Cu^\
SO^"
va
NO3.
Gidi
a) Neu CO
can
dung djch
A
thu di/dc muoi:

KCl:
kali clorua
MgCIj:
magie
clorua
FeClj:
s^t
(III) clorua.
b)
Can lay cdc
muo'i tfng vdi
cdc
ion trSn
:
Na2S04
va
Cu(N03)2
ho^c
hai muoi CUSO4
va
NaNOj.
32
Bai
2.
Trong dung dich
c6 the
ton
tai
dong
thcJi

cac
ion
sau day
dufc^c
khong?
a)
K.\
OH-
va Cr b)
Na^
Fe'*,
CI"
va SO^"
c)
Ba^
Na^ CI"
va SO^" d)
Ag^
H\ vh CV
e) Na*. NH^
SO^" va C0|
Gidi
a) Khong ton t^i di/dc vi
c6
sU"
tao
thanh ke't tua Mg(0H)2.
Mg^*
+
20H- >Mg(OH)2i

b)
T6n tai
difcfc
cdc
ion tren U-ong mpt dung djch.
c) Khdng ton
tai
diTcJc
\\
s\i
tao
thanh ke't tua
BaS04.
Ba^*+
SO^" )-BaS04>l'
d) Kh6ng
ton tai
di/dc
vi c6 sif tao
thanh kd't
tua
AgCl trong dung dich
va
siji
bpt khi.
Ag^
+ Cr
>AgCl4'
2H*
+ CO^- >

COzt
+
H2O
e)
T6n tai
dufdc
cac
ion tren trong mot dung dich.
Bai
3. C6 ba ong
nghiem, moi
ong
chtfa
hai
cation
va hai
anion (khong triJng
lap
giffa
cdc 6ng
nghiem) trong
so cdc
cation
va
anion
sau :
Na\;, Ag^
Ba^^
Mg^^ A1^
cr, Br". NO3 va SO^",

PO^-, CO^".
Xic dinh
cic
cation
va
anion trong tuTng 6'ng nghiem.
Gidi
Muon
tao
thanh trong dung dich
thl cic cap
cation
va
anion phai khong
tao
thanh ke't tua,
cha't
khi,
chat
di?n
li yeu
trong
til'ng
dung dich.
Do d6 cac ong
nghi^ra
Ian
li/dt
la :
Ong

1
g6m
:
Na^ NH^,
PO^", C0|".
6ng
2
g6m
:
Ag^ Mg^ NO3
. SO^.
6ng
3
g6m
: Ba^
Al^
cr. Br".
Bai
4.
a) Mpt dung dich chtfa
a
mol Na*,
b
mol
Ca^\
mol HCOi
va d
mol
Cr. Lap
bieu

thiJc
li6n
he
giffa
a, b, c, d va
c6ng
thuTc
tinh
t6ng
khS'i
luTcJng
muoi trong
dung
dich.
^) K6't qu^
xic
dinh nong dp mol ciia
c^c
ion trong
mOt
dung dich
nhU"
sau
Na*
: 0,05; Ca^^ : 0.01; NO3 : 0,01; Cr : 0,04 va
HCO3
: 0,025.
H6i
ke't
qua do

dung
hay sai? Tai sao?
PhSn
d?ng
va
phuang
phAp
g\i\a hqc 11 Va cO - D5
Xuan Hung
Gidi
a) Vi irong mgt dung djch Irung
hoa vc
dicn
nen
theo
djnh luat
biio
totin
dicn
tich thi tong dien tich dU'dng
bang
tdng dicn tich
am.
Ta
CO
: a +
2b
= c + d
Kho'i lifdng muoi trong dung dich chinh
la

tong khoi
lufdng
cac
ion.
Ta
CO
:
m
= m ,
+
m
7+
+
m
_
+
m
_
Na+
Ca'^*
HCO3
CI
m
= 23a +
40b
+ 61c +
35,5d.
b)
Tong
dicn tich

dU'dng
:
Sdt(+)
=
0,05+
2.0,01
=0,07
(1)
Tong
dien
tich
am
:
Idt(-)
=
0,01+0,04
+
0,025
=
0,075
(2)
Tir
(1) va (2) ta c6 :
Idt(+)
^
Zdt(-)
hay 0,07 ^
0,075
Nen
ket

qua phan tich sai.
Bai
5.
Hoa tan
4
gam
CUSO4
vao mot
liTcfng
niTdc
viTa du
250ml
dung dich.
a) Tinh nong do mol
cua
cac
ion trong dung dich.
b) Tinh
the
tich dung djch NaOH
0,25M
du
de
lam
ket tua het
ion difdng trong
dung dich.
c) Tinh
the
tich dung dich BaCl^

0,2M
du
de
lam ket tua
het
ion
804".
Gidi
CUSO4
>
Cu^^
+ S04~
0.025 0,025 0,025
(mol)
_4_
160
a)
Nong
do mol
cac
ion trong
dung
dich
:
[Cu^1=[SOr]
=
^
=
0,l(M)
b) NaOH

—-> Na* + OH"
0,05
mol
0,05
mol
Cu^^
+
20H-
)•
Cu(OH)24'
0,025
0,05 (mol)
The tich dung dich NaOH
0,25M:
V=JL.M5.o,2(lit)
CM
0,25
BaCl2
>Ba^-^
+ 2Cr
0,025 0,025
(mol)
"CUS04
=T7r
=
0.025
(mol)
34
Ba'"
+

SO^
>
BaS04i
0,025
0,025
0,025
(mol)
=
0,125
(lit).
Vdd BaCl2
~ 0 2
Bai
6.
Mot dung djch
co
chiJa
hai
cation
Fe^""
(0,1
mol)
va Ar^"" (0,2
mol)
va
hai
anion
: Cr (x
mol)
va SO4' (y

mol). Tinh
x va y,
biet
rang khi
c6 can
dung
dich thu
du-dc
46,9
gam
chaft
ran khan.
(TSDHQGTP.HCM,c1(/t2)
Gidi
Dung djch
c6
Fe^^
:0,1
mol
Al''^
:0,2"mol
cr
:
X
mol
SO4" :
y mol
Djnh luat
biio
toan dien tich

:
0,1.2+
0,2.3 = x +
2yii>x
+ 2y = 0,8 (1)
Khoi lU'dng
chat
ran
:
m
=
56.0,1
+
27.0,2
+
35,5x
+ 96y = 46,9
=>
35,5x
+ 96y = 35,9 (2)
Ta CO
he
phiTdng
trinh
:
x
+ 2y =
0,8 fx
=
0,2 mol

'
35,5x + 96y
=
35,9
[y =
0,3mol.
Bai
7.
Dung dich
A
chiJa
cac
ion
:
Na\,
SO4" va CO]~.
a) Dung dich tren
diTcJc
dicu
che
tuf hai muoi trung hoa
nao?
b)
Chia
dung dich
A
l^m
hai phan b^ng nhau
:
-

Phan
1 : tac
dung
\6i
dung djch
Ba(0H)2
duT dun ndng, thu di/dc
4,3g
kcl tua
X
va 470,4ml
khi
Y d
13,5"C
va 1
atm.
-
Phan
2 :
tiic dijng vcKi dung dich HCl diT thu diTdc
235,2ml
khi
d
13,5"C
va
1 atm.
Tinh tdng khoi lifcing
cac
muoi trong
^

dung dich
A.
Gidi
a) Dung dich
A dMc
dieu
che
tif hai muoi:
Na2S04
va
(NH4)2C03
hoSc Na2C03 va
(NH4)2S04.
b)
Phan
1 : tac
dung vdi
Ba(0H)2
dM
35
Phan
dgng
va
phuong
phAp
giai
H6a hpc 11 VP co - D5
Xuan
Hiing
Ba(OH)2

>
Ba'* + 20H-
Ba'^ +
COJ-
>
BaCOji
0,01 molj
Ba'"+ SO^-J
0,01 moli
NH; + OH- H
0,02 mol I
Somolkhi Y: i
0,01 mol
-> BaS044'
0,01 mol
-> NHjt + H2O
0,02 mol
PV
1.0,4704
.
,„„
NH3
= — = 0,082.(13,5 + 273)
m=rn
BaC03 + m BaS04 = 4,3 (g)
Phan 2
:
tac dung vdi dung dich HCl
d\i
Hci

—>
+ cr
2H^ + CO^" > CO2T + H2O
0,01 mol 0,01 mol
''''' PV _
1.0,2352
-0,02 mol
(I)
nco2
=•
RT 0.082.(13,5
+
273)
= 0,01 (mol)
Kh6i
hfang
BaCOj :
m^^coj
= 197.0,01 = 1,97 (g)
2,33
mBaS04
= 4,3-1.97 = 2,33(g) =^
03,504
= ^ = 0,01 (mol)
Theo djnh luat Ho
tokn
dien tich : n ^
CO:
2-
+ 2.n

so
2-
n^^,
=
2.0,01
+
2.0,01
- 0,02 = 0,02 (mol)
Tong khoi liTdng cdc muoi trong dung dich A :
m = 0,02.23 + 0,02.18 + 0,01.60 + 0,01.96 = 2,38 (g).
Bai 8. Mot dung dich X c6 chura cdc ion Zn^^, Fe'*
\i
S0^~. Biet r^ng diing het
350ml dung dich NaOH 2M thi lam k6't tua het ion
Zn^*
va ion Fe^* trong
100ml dung dich X, neu do tiep 200ml dung dich NaOH thi mot chat ket tua
vCra tan het, con lai mpt chat ke't tua m^u nau do. Tinh nong dp mol cua moi
muo'i trong dung djch X.
Gidi
100ml dung dich X
Zn^^: a mol
Fe^^:
b mol
36
RHAJPG VIET
NaOH > Na*+ OH"
0,7 mol 0,7 mol
HNaOH = 0,7 (mol)
PhiTcfng trinh ion rut gpn :

Zn^* + 20H- > Zn(OH)2i (1)
a mol 2a mol a mol
Fe^*
+ 30H-
>
Fe(0H)3i
(2)
b mol 3b mol b mol
Neu do' tiep dung dich NaOH
Zn(OH)2
+ 20H- >
ZnOJ'
+
2H2O
(3)
a mol 2a mol
Theo(l)va(2)tac6: 2a + 3b = 0,7 (I)
So mol NaOH trong 200ml dung dich NaOH 2M
HNaOH
= 2.0,2 = 0,4 (mol) => n_= 0,4 mol
OH
Theo
(3):
2a = 0,4
i=>
a = 0,2 (mol)
The a = 0,2 vao (I) =>b = 0,l(mol)
So'
mol muo'i :
ZnS04 > Zn^^ + S04~

0,2 mol 0,2 mol
Fe2(S04)3
>
2Fe'*
+ 3S0^~
0,05 mol 0,1 mol
Nong dp mol moi muoi trong dung dich X
[ZnS04]=
—
= —-2M
V 0,1
[Fe2(S04)3]= ^ = 0,5M.
Bai 9. Cho dung dich A chiJa cac ion AP^
Fe^*,
SO4-
va Br'.
- Trpn Ian 100ml dung dich A vdi dung djch AgNOs
d\i
thu diTpc 75,2g ke't tua.
- Trpn Ian 100ml dung dich A vdi dung djch KOH
dU"
thu diTpc m gam ke't tua
X, nung nong X den khoi li/png khong doi thu di/pc 16g chat ran Y.
- Trpn Ian 100ml dung djch A vao dung dich BaClj diT thu du-pc 93,2g ke't tua.
a) Tinh nong dp mol moi ion trong dung dich A.
b) Neu thoi mpt luTdng khi
NH3
vao 100ml dung dich A. Sau do Ipc, tach lay ket
tua. Dem nung ke't tua den kho'i liTpng khong doi thi thu diTpc bao nhieu gam
cha't r^n?

37
Phan
dgng
phuong
phip
gi3i
H6a
hpc 11 VP co- D5
XuSn
Hung
a)
100ml
dung
dich A
Gidi
AP^:X
mol
Fe''+
:
y mol
SOJ z
mol
Br":
t mol
v
I
* Tron
dung
dich A vdi
dung

djch AgNOi dif:
Ag*
+ Br" >
AgBri
0,4 mol 0,4 mol
"AsBri
=-^ = 0,4(mol)
1
OO f
i
=> n„ , = t = 0,4 mol
*
Tron
dung
dich A vdi
dung
djch KOH dU":
Al^*
+ 30H~
>
Al(OH)3i
A1(0H)3
+ OH" >
AIO2
+
2H2O
i|
Fc^* + 30H- >Fc(0H)3i (ketluaX)
'
0,2 mol 0,2 mol

2Fe(OH)3 -A FejOj +
3H2O
(chat
nln Y la FcjOj)
0,2 mol 0,1 mol
I ,
nFc203
=1^
= 0,1 (mol) => np^3^ =y = 0,2mol
*
Tron
dung
djch A vdi
dung
dich BaCh di/ '
Ba'*
+ SO^-
>
BaS04i
0,4 mol 0,4 mol .
I 93 2
I
"BaS04i
= ^ = 0,4(mol) n^^2- = z = 0,4 mol ,
I
* Djnh luat bao
toan
dien tich :
5.
Ta CO : 3x + 3y = 2z + t

'
2z + t-3y 2.0,4 +
0,4-3.0,2
'„ '
=>
X =
=
— '-
—
=
0,2 mol)
I Ndng do mol cac ion trong
dung
djch A :
I
[AP1=^
=
2(M)
(FC'1=^-2(M)
;
I
0,1 0,1
i
•> 0 4 0 4 ^
i
lSOl-J ^-4(M) [Br-J=-^ =
4(M).
,j
t,)
Thdi khi NHj v^o

dung
djch A :
Al'^
+
3NH3
+
3H2O
>
Al(OH)3>l'
+ 3NH*
0,2 mol 0,2 mol
Fe^* +
3NH3
+
3H2O
> Fe(OH)3i + 3NH^
0,2 mol 0,2 mol
Dem kct tua
nung
:
2Al(OH)3
AI2O3
+
3H2O
0,2 mol 0,1 mol
2Fe(OH)3 Fe203 +
3H2O
0,2 mol 0,1 mol
Khoi
liTcJng

AI2O3:
m^i,^o.^
=
102.0,1
= 10,2 (g)
Khoi
li/dng
Fe203:
mFe^Os =
160.0,1
= 16(g)
Khoi
liTrtng
cha't
ntn
:
m = 10,2 + 16 = 26,2 (g).
Bai
10. Hoa tan hoiin
toan
0,1022g
mot muoi kim loai hoa trj (11)
MCO3
trong
200ml
dung
djch
HCl 0,080M. De trung hoa
liWng
HCl

diTcan
5,64ml dung
djch
NaOH 0,10M. Xac
dinh
kim loai M.
Gidi
PhuTdng
trinh
phan
u^ng :
MCO3
+ 2HC1 >
MCI2
+ CO.t +
H2O
(1)
0,000518
0,001036
(mol)
NaOH + HCl > NaCl + H20 (2)
0,000564 0,000564
(mol)
nHci
=
0,08.0,02
=
0,0016
(mol)
HNaOH =

0,1.0,00564
=
0,000564
(mol)
HHCKD
=
0,0016
-
0,000564
=
0,001036
(mol)
0.1022
0,000518
M
= 60 = 197 => M = 137 (Bari)
Vay kim loai M la Ba.
BAITAPAPOgNG
Bai
1: Dung dich X c6
chiJa:
0,07 mol Na"^; 0,02 mol
SO4'"
va x mol OH'.
Dung dich Y c6
chiJa
CIO4";
NO3"
va y mol H^ Tong so mol C104' va NO, la
0,04. Tron X va Y

diTdc
100 ml
dung
dich Z. Tinh pH cua
dung
dich Z (bo
qua sif dicn li cua
H2O).
39
PhSn
djng
va
phuong
ph^p giii H6a hpc
11
VP co
-
D5
XuSn
Hung
Giai
- Ap dung dinh luat bao toan dien tich cho dung dich X ta diTdc:
X
=
0,07
-
0,02.2 = 0,03 mol
- Ap dung dinh luat bao toan dien tich cho dung dich
Y ta
dxXdc:

y =
0,04 mol
- Tron dung dich
X
va dung dich Y xay ra phan ilng:
+
OH"
>H20
0,03
0,03
=>
n^^
= 0,04-0,03
=
0,01 mol
^
[Hi
=
IO'M
=> pH =
1.
Bai
2:
Hoa
tan
hoan toan 8,94 gam hon hdp gom Na,
K va Ba
vao niMc,
thu
dMdc

dung dich
X va
2,688
lit
khi H2 (dktc). Dung dich
Y
gom HCl
va
H2SO4,
ti
le
mol
tu-Ong
ij-ng
la
4:1. Trung
hoa
dung djch
X bdi
dung dich
Y,
tdng
kho'i liTtJng cac muoi di/cfc tao
ra la
A. 13,70 gam.
B.
18,46 gam.
C.
12,78 gam.
D.

14,62 gam.
Gidi
Tac6:
nH2
=
0,12 mol
Sddophanurng:
H2O
>
OH"
+
-H2
2
0,24
0,12
Goi nH2S04
= ^
'"ol
-> HHCI =
4x mol
-> =
6x mol
Phan u-ng trung hoa:
+ OH^ ->
H2O
0,24
0,24
6x
=
0,24 => X

=
0,04 mol
=>
nH2S04
=
0,04 mol;
UHCI =
4.0,04
=
0,16 mol
Ta c6:
Kho'i-lu'dng
muoi = kho'i
liTdng
kim loai
+
kho'i
liTpng
go'c axit
-> Kho'i liTcJng muoi
=
8,94
+
0,04.96
+
0,16.35,5
=
18,46g Dap an
B.
Bai 3: Cho dung dich chtira 0,1 mol

(NH4)2C03
tac dung vdi dung djch chiJa 34,2
gam
Ba(0H)2.
Sau phan u-ng thu difdc
m
gam kct tua. Tinh
m.
Gidi
Ta c6: n(NH4)2C03
=
O-lmol =>
n^^^j-
=
0,1
mol
, nBa(OH)2
=
0,2mol n^^2+
=
^,2 mol
Pu: Ba^^
+
COj^-
^
BaCOj
i
0,1
0,1
Vay:

mi =
197.0,1
=
19,7 (gam)
40
»
Bai 4: Nho tCr
tCr
0,25 lit dung dich NaOH 1,04M
\ko
dung dich gom 0,024 mol
FeCls; 0,016 mol
Al2(S04)3
vk 0,04 mol H2SO4 thu dUOc
m
gam ket tua. Tinh
m.
Gidi
Taoo:
UNaOH
=
0,26 mol
=> n . =
0,26 mol
OH
"FeCis
=
0'024mol
^ Up
3^.

=
0,024 mol
nAi2(S04) ,
=
0,016mol n^^,^
=
0,032 mol
n = 0,04mol =>
n .
= 0,08mol
H2SO4
H+
Khi
cho dd
NaOH
vao dd
gom FeCls;
Al2(S04)3;
H2SO4
; cac
phan I'mg
Ian
li/dt xay
ra :
H^
+ OH" ->
H2O
0,08
0,08
Fe^*+ 30H"

->
Fe(OH)3
0,024 0,072 0,024
Al^^
+ 30H- ->
Al(OH)3
0,032 0,096 0,032
I
n .
= 0,08 + 0,072 + 0,096 = 0,248 mol
OH
=>
n
.j„ = 0,26-0,248 = 0,012 mol
OH
Do OH" du nen tiep tuc xay ra phan ung hoa tan ket tua
A1(0H)3.
A1{0H)3
+
OH"
->
AIO2"
+
2H2O
0,012
0,012
=> so mol Al(OH)3c6ni,i
=
0,032
-

0,012
=
0,02(mol)
Vay:
mi =
mA,(OH)3 c6n„i
+
mp,(OH)3
=
0,02.78
+
0,024.107
=
4,128(g)
Bai 5: Cho 200ml dung dich
A
chiJa HCl
IM vil
HNO3
2M tac
dung
vdi
300ml
dung dich chuTa NaOH 0,8M
va
KOH (chifa bict nong do) thi thu du^dc dung
dich C. Biet rhng
de
trung hoa dung dich
C

can 60ml HCl IM. Tinh nong do
mol ciia dung dich KOH.
Gidi
Gpi x M
la
nong do mol cua dung dich KOH:
Taco: nNaOH
= 0,24 mol
"KOH
=
^'•^^
"^o'
nHN03
= 0,4 mo
"HCI
0,2 mol
=>
Yn =
(0,3x
+
0,24)
mol
^
OH"
nHN03
= 0.4 mol
Tn
+ = 0,6
H+
mol

PhSn
dgng
vci
phudng
ph^p
giSi
H6a hpc 11 VP cO -
D5
Xufln
Hi;ng
w
Khi
tron iSn hon hrfp 2 axit
vil
2
bazcJ
xay ra
phan
i^ng triing h6a:
+
OH- ^ H2O (1)
0,6 0,6
Sail
phan
iJng (1) can
dung
HCl de trung hoa, chifng to OH" dir.
=>
n . = (0,3x + 0,24) - 0,6 = (0,3x - 0,36) mol
OH

Ta c6:
UHCI
= 0,06 mol => n^. = 0,06 mol
H
Phiin
xing trung hoa giffa HCl vii
dung
djch C:
H^
+ OH" -> H2O
0,06 (0,3x - 0,36)
=>
0,3x - 0,36 = 0,06 => x = l,4M
Bai
6: Cho
dung
djch Ba(0H)2 den dU vao 50ml
dung
dich X c6
chiJa
cac ion
NH4'';
SO4'"
; NO," thl c6 11,65 gam kcl lua dmc tao ra va dun
nong
c6 4,48
lit
khi bay ra (dktc). Tinh
nong
do mol moi muo'i trong

dung
djch X.
Gidi
Ta c6: n^^.^ = -^-^ = 0,05 mol; n^H, = — = 0,2 mol
Bab04 ' 22 4
Ba'* + S04^- -> BaS04>l'
0,05 0,05
NH4* + OH" NH3 + H2O
0,2 0,2
Dung dich X c6 cdc ion
NH4^;
504^";
NO,r
CO
2 muoi la (NH4)2S04 va
NH4NO3.
Ta c6: n,^^^,^,^, - ^of = ^'^^ "^"^ =^ [(NHJ.SO,
nNH4No,, = 0,2 -
0,05.2
= 0,1 mol => [NH^NOJ
0,05
0,05
0,1
0,05
=
1M
=
2M
Bai
7: Dung djch A

chiJa
HCl IM va
H2SO4
0,6M. Cho 100ml
dung
djch B gom
KOH
IM va NaOH 0,8M vao 100ml
dung
djch A, c6
ciin
dung
djch sau
phan
u'ng thu diWc m(g) chift
r^Cn.
Tinh m.
Gidi
Dung dich A c6: n^^^o^ = 0,06 mol •
"Hci^O.lmol
Dung dich B c6:
n^^oH
= 0,08 mol
n^,
= 0,22 mol
H
"KOH
= 0,1 mol
n^^.
= 0,18 mol

OH
Dd
sau
phan
u'ng gom
H'
+ OH" ^ H2O
0,18 0,18
•
H*
dif: 0,22 - 0,18 = 0,04 mol
:
0,1 mol
Na^ : 0,08 mol
cr : 0,1 mol
•
SO4-
:
0,06 mol
Khi
CO
can
dung
djch HCl bay hd. Vay kho'i lifdng
chat
ran thu difdc c6
giii
tri:
m =
m,„,i.,„

+
m,„i.,n
=
0,1.39
+
0,08.23
+
0,06.35,5
+
0,06.96
= 13,63 (g)
Dana
5. Bai t^p ap
dung
djnh
lugt bao
loan
di^n tich.
BAI
TAP
MAU
VA
BAI
TAP AP
Dl^NG
Bai
1. Mot
dung
djch
chuTa

0,02 mol Cu^*, 0,03 mol K*, x mol CI" va y mol SO^".
Tong khoi lifdng ciic muo'i tan c6 trong
dung
dich la
5,435
gam. Tim x va y.
Gidi
Khi
tron
dung
dich A \di
dung
dich B xay ra
phan
Ofng:
/^p
dung
djnh luat bao loan dien tich, ta c6:
X
+ 2y =
0,02.2
+ 0,03 = 0,07 (1)
Ta lai c6: mmu,-,: =
Sm,„,i n
+
Ima„K,n
=>
35,5x
+ 96y =
5,435

-
(0,02.64
+
0,03.39)
=
2,985
(2)
Tir(l)va(2)=>^
= °'''
[y
= 0,02
Bai
2. Mot
dung
dich c6 chtfa cac ion: x mol 0,2 mol Mg^\3 mol Cu^^
0,6 mol
S04^",
0,4 mol NOf. Co can
dung
dich nay thu
du'rtc
116,8 gam hon
hcJp
cac muo'i khan. Tim M.
Gidi
Ap
dung
dinh luat bao loan dien tich, ta c6:
3x + 0,2.2 + 0,3.2 = 0,6.2 + 0,4 x = 0,2 mol
Ta c6:

m^^,;=
m^.,
+ni^^,
+m^^,. +m^^,^ +m^^_
=>
116,8 =
0,2.MM
+
0,2.24
+
0,3.64
+
0,6.96
+
0,4.62
=> MM = 52 => M la Cr
^ai
3. Dung djch A c6
chiJa
5 ion: Mg^^ Ba'\\1 mol CI" va 0,2 mol
NO3'.
Them dan V lit
dung
dich NajCOj IM vao A den khi
du^dc
lifdng ket tua Idn
nhat.
Tim V.
Phan
dgng vh phuong ph^p

giSi
H63
hoc 11 V6
cd
- D5
XuSn
Hung
Gidi
PhifcJng trinh ion rut gon:
Mg'*
+ CO^- > MgCOjl
Ba^*+
CO3-
>
BaCOji
Ca^*
+
CO3"
>
CaCOjJ-
Khi phan i?ng ke't thilc, cac ke't tua tach khoi dung dich, phan dung dich chiJa
Na"",
cr
va
NO3".
De trung hoa dien
thi: n^^^^
=
n^^.
=

0,3 mol
=> V,,,,^co,
=^^=
°;^=0,15(l)=150ml
I
Na 2
Bai
4.
Dung dich
A
chiJa cac ion
COj^",
SOj^",
SO^^',
0,1 mol HCO3"
va
0,3 mol
Na^ Them
V lit
dung dich Ba(0H)2
IM
vao dung dich
A thi thu
diTdc Ming
ket tua
Idn
nhat. Tim V.
Giai
Nong
do cac ion

[Ba^^]
= IM;
[OH"]
=
2M.
De thu
di/cJc liTdng
ket tua Idn
nhat, can 0,1 mol OH"
de
tac dung
het
vdi HCO3".
HCO3'
+0H"
->C03^"+
H2O
Mat khac, can 0,3 mol OH' de trung hoa Na"".
Vay tdng
so
mol OH' can
la
0,1
+
0,3
=
0,4 mol
=>
ny^joH,^
= 0,2

mol
The tich dung dich
Ba(0H)2
la: V =
^
= 0,2
lit
I
Bai
5.
Dung dich
X
chtfa
cac
ion:
Fe'\, NH4\.
Chia dung dich
X
thanh hai phan bang nhau:
- Phan
mot tac
dung
vdi
liTcJng
du"
NaOH, dun nong
thu
di/dc 0,672
lit
khi

d
(dktc)
va
1,07 gam ket tua.
- Phan hai tac dung vdi liTdng du'dung dich BaCl2, thu du'dc 4,66 gam ke't tua.
Co can dung djch X thu diTdc m gam muo'i khan (qua trinh c6 can chi cd nU'dc
bay hdi). Tinh m.
Gidi
- Phdnl:
n,H,
= ^-0,03 mol;
np,,o„,,
=
1^
= 0,01 mol
Fe^^
+ 30H- ->
Fe(0H)3
0,01
0,01
NH4*
+ OH" -> NH3 + H2O
0,03
0,03
KWAWfi VIET
Phan 2:
nB„so4
= 4^ =
0,02 mol
Ba

2+
233
S04^"
-»
BaS04
0,02
0,02
Vay trong dung dich X cd: 0,02 mol
Fe^* ;
0,06 mol
NH4^ ;
0,04 mol
S04^"
va
X
mol
cr.
Ap dung dinh luat bao toan dien tich cho dung dich X, ta cd:
0,02.3 + 0,06 = 0,04.2 + x x = 0,04 mol
m^utfi
=
m^^,,+m^^j+m^^,.+m^,.
= 0,02.56 + 0,06.18 + 0,04.96 + x.35,5 = 7,46 (g)
B^i 6. Dung dich X cd chtfa
c&c
ion Ca^*,
Af
CI'. De k^t tiia het ion
Cr
trong

100 ml dung dich
X
c^n dting 700 ml dung dich chtfa ion Ag* cd nong d6 1^
IM.
Co can
dung dich
X
thu diTdc 35,55 gam muoi. Tinh nong
dp
mol
cac
cation tiTdng tfng trong dung dich X.
Gidi
Gpi a, b
lin
li/dt
Ih
s6 mol ciia
Ca^*
va Al^*.
Ta cd:
n =n
. =0,7mol
CI
Ag J
Ap
dung djnh lu|t bSo to^n di^ tich, ta cd:
2a
+
3b = 0,7

(1)
Ta cd: mmirifi
=
nicaUon
+
manion
=> 40a + 27b = 35,55
-
0,7.35,5 = 10,7
(2)
a = 0,2
J[Ca^^]
= 2M
b = 0,l |[A1^^] =
1M
Tiir(l)va
(2):
=>
Dap D.
B^i 7. Dung dich X chiJa
cic
ion: Ca^*, Na*, HCOJ
v^
CI" ,
trong dd so mol cua
ion
cr la
0,1.
Cho 1/2 dung dich X ph^n i^ng vdi dung dich NaOH
(du"),

thu
diTdc
2
gam
ket
tiia. Cho
1/2
dung dich
X c6n lai
phan iJng
vdi
dung dich
Ca(0H)2
(dir), thu di/dc 3 gam ket tua. M^t khac, neu dun soi den can dung
dich X thi thu dtfdc m gam cha't r^n khan. Tim m.
Gidi
* 1/2 dung dich X
tic
dung vdi dung dich NaOH
diT:
Ptf: HCOj
+
OH"
^
C03^'
+
H2O
Ca
2+
C03^-

->
0,02
0,02
CaC03>l'
0,02
45
Phan
dgng
va
phudng
ph^p
giai
H6a hpc 11 VP ca - P5
XuSn
Hung
2
Taco:
iicco, = = 0.02 mol => = 0,02 mol
100
*
1/2
dung
djch X tac
dung
vdi
dung
dich Ca(OH)2 diT:
PiJ:
HCO3
+ OH" -> COj^" + H2O

0,03 0,03
Ca^^ + COj^"
CaC03
i
0,03 0,03
3
Ta c6: nr.rn = = 0,03 mol
CaCO,
'
Ca(0H)2 dir => COj^" hct
theo
pu" ta c6: n^^^,^^ = 0,03 mol
Ap
dung
djnh luat bao
toan
dicn tich cho
dung
djch X ta c6:
n
, = n
_
+ n
_
- 2n = 0,1 +
0,03.2
-
2.0,02.2
= 0,08 mol
Na CI HCO^ Ca*"

Khi
CO
can
dung
dich X xay ra
phan
tfug:
2HCO-3
—^ COj^- + CO2 T + H2O
0,06 0,03
V|iy:
sau khi c6 can trong m(g)
cha't
ran c6: Ca"*; Na*;
CI";
CO^^'
=>
m =
0,04.40
+
0,08.23
+
0,1.35,5
+
0,03.60
= 8,79 (g)
Bai
8. 100ml
dung
djch X

chiJa
cac ion Ca^*: O.lmol; NO^: 0,05mol; Br":
0,15mol;
HCO3
: 0,1 mol mot ion cua kim loai M. Co can
dung
djch thu
difdc 29,Ig muoi khan. Tim ion kim loai M va tinh
nong
do cua no trong
dung
dich.
Gidi
Goi
n la dien tich cua ion kim loai M; x la so mol cua
M"*
Ap
dung
DLBTDT, ta c6: 0,1.2 + xn
=0,05
+ 0,15 + 0,1 xn = 0.1
Matkhac: m.„,i =m^^,. +m_^„. +m^^_ +m^^_ +m^^.^_^
^0,1.40
+
1
M —
n
^
M
+

0,05.62
+
80.0,15+.0,1.61
= 29,1<^ — = 39c:>M = 39n
n
Trong so cac dap an da cho thl la dap an phii hdp va C . = — = 1(M>
0,1
=>
Dap an D.
Bai
9. Cho 100ml
dung
dich A chiJa
Na2S04
0,1M va
NasCOj
0,2M tac duni-
vira
du vdi 100ml
dung
dich B chiJa Ba(N03)2 va Pb(N03)2 0,05M tao kc'i
tia. Tinh
nong
do mol ciia Ba(N03)2 va khoi liTdng
chung
cua cdc ket tiia?
46
Gidi
Theo
DLBTDT cac ion S04~, C03~phan

ilng
vdi Ba^* va Pb^*
theo
ti le mol
1:1. Dien tich cac ion iren
bhng
nhau
ve
giii
tri tuyct do'i. Ncn ta c6:
^
n(SO^"
+
CO3")
= ^ n(Ba'* + Pb'*)
=:>0,1(0,1+0,2) = 0,l(x +0,05)
=>x = C
J,
=0,25(M)
Ba
Khoi
lUdng
chung
cua cac ket
tiia
(BaS04
+
PbS04
+
BaCOs

+ PbCOj) la:
=
0,1.0,1.96
+
0,1.0,2.60
+ 0,25.0,1. 137 +
0,05.0,1.207=
6,62 (g)
BAI
TAP
TRAC
NGHI^M
CSu
1. Co mot
dung
dich
chat
dien
li
yeu, khi
tang
nong
do
cha't
dien
li
thi:
A.
Do dicn li
tang.

B.
Do dien li khong
thay
doi.
C.
Do dien
H
giam.
D.
Dp dien li va
hang
so' dien li deu
tang.
C&u
2. Chpn
phat
bieu sai:
Trong ciic tieu
phan
NH4*,
C03^-, S^', HS", Na*, FeCHjO)'*,
H2O
theo
li
thuye't
Bronsted.
A.
Axit
la cac tieu
phan

NH4*,
Fe(H20)'*
B.
Bazd la cac tieu
phan
CO^^", S'~
C.
LirOng tinh la tieu
phan
HS"
D.
Trung tinh la cic tieu
phan
H2O,
Na*.
Cfiu
3. Phat bieu n^o sau day
khong
dung?
A.
Cdc
dung
dich
NH4CI.
A1(N03)3,
NaHS04 deu c6 pH < 7.
B.
Cac
dung
dich NaHC03, KHS, NaHS04 deu c6 pH < 7.

,
C. Cdc
dung
djch
KHSO4,
CHjCOONa
c6 pH = 7.
D.
Dung djch
NaN02
co pH > 7.
Cdu
4. cap
chat
nko sau day khi cho vao niTdc khong lam lhay doi dp pH ciia
dung
djch?
A.HCI,H2S04 B. KCLNaNOj
C.
NH4CI,
AICI3
D.
NaHS04,
Na2C03.
Cfiu
5. Cho ba
dung
dich c6 cOng
nong
d6 mol//:

NH3
(1); NaOH (2); Ba(0H)2
(3).
pH ciia
dung
dich nay diTpc xe'p
theo
day:
A.(l)<(2)<(3)
B.(3)<(2)<(1)
C.(2)<(3)<(1) D. (2) < (1) < (2).
47
Phan
djng
vA
pHuong
ph^p
giSl
H6a hoc 11 VP cd - B6
Xuan
Hung
CSu 6. Nho tir
tiif
dung NaOH vao dung dich
AICI3
cho ddn
dir.
Hi^n Wdng quan
sdtdi/cJcia:
A.

Xua't
hi$n ket tiia mau
vSng
B.
Xua't
hien ket tua keo tr^ng tang dan den
cifc
dai, sau d6 tan dung dich
trd
nen trong
suo't
C. Ket tiia xuat hi^n roi tan
ngay
D.
Khong hi^n ti/dng.
Cfiu
7. Nh6
tilf tilf
dung dich
AICI3
v^o ong nghi^m duTng dung dich NaOH cho
den
duT.
Hipn
tufdng quan sdt difdc la:
A.
Xuaft
hi0n
ket tua m^u keo tr^ng
B.

Xuat
hien ket tua keo tr^ng va tang dan den
ciTc
dai, sau do tan dung dich
trd
nen trong
suo't
C. Ke't tua xua't hi?n roi tan
ngay
D.
Xua't
hien ket tiia xanh.
Cfiu
8. Nho
tilf
tCf dung HCl v^o dung dich NaAlOz cho den
di/.
Hien tiTcfng quan
sdtdiTdc la:
A.
Xua't
hi0n
ke't tua mau keo tr^ng
B.
XuS't
hi$n ke't tua keo tr^ng va t3ng dan de'n
ctfc
dai, sau d6 tan dung djch
trd
nen trong

suo't
C. Ke't Ilia xua't hidn r6i tan
ngay
D.
Khong hi^n tiTdng.
Cfiu
9. Mot dung djch
nvtdc
cua natri
cacbonat
c6 pH > 7 1^ do:
A.
Natri
cacbonat
phan li
hokn
tohn
B. S6' mol ion Na* nhieu hdn ion COa^"
C. So mol COi^' nhieu hdn so mol niTdc
D.
Ion COi^~ tham gia ph^n
iJng
thiiy
phan vdi niTdc.
Cfiu
10. Cho cic dung djch
NH4CI
(1); CHjCOONa (2),
NazCOj
(3),

NaHS04
(4), Cu(N03)2 (5), KCl (6), Ba(N03)2 (7),
K2S
(8).
Day cdc dung dich c6 pH < 7 1^:
A.(l),(2),(4),(5)
B.(I),(4),(5)
C.(2),(4),(5),(7)
C. (2), (4), (6). (7).
Cfiu
11. Phdt bieu nao sau day kh6ng chinh
xdc?
A.
Cdc dung djch
NH4CI,
CuCb,
NaHS04
deu c6 pH < 7
B. Cdc dung dich
NaHC03,
KHS.
NaHS04
d6u c6 pH < 7
C. C&c dung dich NaCl,
KNO3,
H2O d^u c6 pH = 7
D.
Cic dung dich
A1(N03)3.
FeCU,

NaHS04
deu c6 pH < 7.
48
CSu 12. De nhan
bie't
drfdc dung dich dufng trong 4 Ip
khac
nhau la: KOH,
NH4CI,
Na2S04,
(NH4)2S04 ta chi can dOng mot trong 4
cha't:
A.
Dung dich AgN03 B. Dung dich
BaCb
C. Dung dich NaOH D. Dung dich Ba(OH)2.
Cfiu
13. Co 4 muoi
FeClj,
CuCb,
AICI3,
va ZnCh. Neu them tit tir dung djch
NaOH cho den dir vao 4 muoi tren. Sau do them tiep
NH3
dir thi so ket tiia
thu
du'dc la:
A.
1 B. 2 C. 3 D. 4.
Cfiu

14. Mot dung dich gom x mol Na'^' y mol NO3", z mol
HCO3",
t mol Ca^*.
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,iM, dung djch HCl
0,01M.
Vay pH cua hai dung
dich tren Ian
liTdt
la:
A.
Iva2
B.
13va2
C.
2val3
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:
A.

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:
A.
1:10 B. 1:1
C.10: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:
A.
2% B. 3% C.5% D. 6%
Cfiu
19.
Phai
pha loang dung dich KOH 0,00
IM
vcti
niTctc
bao nhieu Ian de diTdc
dung dich c6 pH = 9.
A.
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
nufdcca't
can diingia: .
A.
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
I9
the tich cua
dung dich
HCIl
Q-'M
vdi dung dich KOH 10'^M m:
A.
1:2 B. 11:9 C. 9:11 D. 2: 15
49
limmiiiii
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