����������Ion equilibriumIon equilibrium
����������
��� ��� ������������� �����
��������������� ����� �����������������
1. strong electrolyte ������������� ��!"#$%&��'(� )*��
HCl HNO HClO NaOH KOH NH Cl NaCl
������������� ����
��� �������� �������� ������������������� ���������������� ���������� !��"�
2
HCl HNO3 HClO4 NaOH KOH NH4Cl NaCl
2. weak electrolyte �������������&��'(� )*��
HNO2 HClO2 H2SO3 CH3COOH NH4OH BaSO4
3. Non electrolyte ��#678���������&��'(� ��9����(��::;� )*�� ��"<=� >"<=#�
�#�-)!� (Acid- Base)
1. ArrheniusG �#� =H� ��#678�9����'(�����&I� H+
��������������� �-���
ClHHCl−+
+→
3
G )!� =H� ��#678�9����'(�����&I� OH-
=����# KL'���!��#������&I������
ClHHCl aqaqaq+→
)()()(
OHKKOH aqaqaq−+ +→ )()()(
#��$����"
♦♦♦♦ ��#��'���� �9����'(����)6����'� ��9��#��'�����7 H+
I#H� OH- &�<�)���� M9�������#N!��������)OP��#�-)!� )*��
NH3 OH-
H2O
4
NH3 OH
NH3 + H
2O NH
4+ + OH-
NH4Cl NH
4+ + Cl-
NH4+ + H
2O NH
3 + H
3O+
2. Bronsted-Lowry
�#� =H� ��#678&I� H+ )!� =H� ��#678#�! H+
="��#�-)!� (conjugate acid-base pairs)
5
HA �����'(��"#�) A- ��� A- �����'(�+�#�) HABH+ �����'(��"#�) B ��� B �����'(�+�#�) BH+
HF + H2O H3O+ + F-
�#� 1 )!� 1�#� 2)!� 2
NH4+ + H2O NH3 + H3O
+
="��#�-)!� (conjugate acid-base pairs)
6
HNO2 + H2O H3O+ + NO2
-
NH3 + H2O NH4+ + OH-
#����)���
•••• �'(�)OP����6�' �#���9)!� (amphoteric substance) ������� amphoteric )*�� NH3 HSO4
- HCO3-
•••• �#���9)!���M��"�&�#"O<�)����I#H������
7
•••• �#���� �7 ="�)!�)OP�)!����� )*�� HClO4 �#���� : ClO4
- )!����� )!���� �7 ="��#�)OP��#����� )*�� NH2
- )!���� : CH3NH2 �#�����
•••• �#���9)!���M��"�&�#"O<�)����I#H������
•••• �#� =H� ��#678����#N#�!="��Y)�Z��#��
•••• )!� =H� ��#678����#N&I�="��Y)�Z��#��
3. Lewis acid
�+���"OHHOHH :: →+ −+
8
8
��#678)OP��#���� Lewis (Lewis acid)1. �=������ : Na+ Be2+ Mg2+ Ag+
2. <�)����678�9������ �7�Y)�Z��#�����=#!�O� BF3
3. �7]��^9="���!�9���678�7=�� EN ��� ��� SO3
�+���"
1. �������� OH-
2. �9������ 678�7�Y)�Z��#��="�<��)�78��
���������+���� Lewis (Lewis base)
9
H2O NH3
H2C=CH2 + 2HBr H3C-CH3 + Br2
3. ��# hydrocarbon (CH) 678�7]��^9="�
Lewis acid - base
10
10
=����# K� �#�-)!���� Bronsted-Lowry
•••• �#�������&I� H+ 100% )OP��#����
HCl + H2O H3O+ + Cl-
�#�������&I� H+ < 100% )OP��#�����
11
CH3COOH + H2O CH3COO- + H3O+
NH2- + H2O NH3 + OH-
•••• )!�#�! H+ ��� 100% )OP�)!����
)!�#�! H+ ��� < 100% )OP�)!�����
=����# K� ="��#�-)!�
12
12
1. Hydro acids O#9��!���� H ��9^����H8��7� 1 *�Y�
�<�I9&�=�!)�7�����N��=�� EN �" =����# �#�M9)]Y8�
=����# K� �#�
PH <<<< H S <<<< HCl
��"� 2 ;�<"=�>(?
13
PH3 <<<< H2S <<<< HCl
NH3 <<<< H2O <<<< HF
�<�I9I�"�)�7����� =����# �#�)]Y8� ������<���K� ]�� ��&���#����]��^9
HF<<<< HCl <<<< HBr <<<< HI(]�� ��&���#����]��^9�" 678���)
2. Oxo acid O#9��!���� O, H ��9^����H8��7� 1 *�Y��#����<>678�7�9������ )I�H��������I�"�678��)��9��"��7M(�������)6����� =����# K� �#�)]Y8�KL'����)�K
���>Y)�*�� K� �9������ +1 +5
=����# K� �#�
14
���>Y)�*�� K� �9������
HClO <<<< HClO2 <<<< HClO3 <<<< HClO4
+1 +3 +5 +7
�9������ ��� *�Y���������"�&�I�"�)�7�������9�7)�K���>Y)�*�� )6����� =����# K� �#�M9)]Y8�KL'���� EN K� �9������ HlO4 <<<< HBrO4 <<<< HClO4
1. �g�#���>�%K� <�I9I�"� IA )OP�)!����
<�I9K���&Ih�KL'� =����# )!�)]Y8� : KOH > NaOH
2. ������!�9���)�78��
=����# K� )!�
15
2. ������!�9���)�78��M(����O#9M������)]Y8�=����# )!�)]Y8� :
N 3- >>>> O2- >>>> F-
&�I�"�)�7����� =�� EN )]Y8�=����# )!�)]Y8� O2- >>>> S2-
&�=�!)�7�����=�� EN ��� M��K���O>���
=����# )!�)]Y8� NH2- >>>> OH- >>>> F-
G �#����-)!���� ������ 100 %KOH K+ + OH-
1 mol 1 mol 1 mol�#���� )!����
��#������K� �#� - )!�
16
�#���� (Strong acids)
)!���� (Strong bases)
Group 1A (hydroxides)(NaOH, KOH, LiOH, etc.)Group 2A (hydroxides)Ca(OH)2, Mg(OH)2, Sr(OH)2, etc.)
)*�� HClO4, HCl, HBr, HI, HNO3, H2SO4
G �#�����-)!����� �������������� 100 %()OP�OkY�Y#Y��l�����!)
��#������K� �#�����
HA + H2O H3O+ + A-
17
][
][][3
HA
AOHK a
−+
=
= =��= 678��#������K� �#�����K a
G�#�<�<�<O#�Y�: �#� 1 <�)����������&I� H+ 1 ���
CH3COOH + H2O H3O++ CH3COO-
18
][
[][
3
33]
COOHCH
COOCHOHK a
−+
=
C�#�<]�Y<O#�Y�: 1 <�)����������&I� H+ >>>> 1 ���
H3PO4 H2CO3 H2S
H3PO4 + H2O H3O++ H2PO4- K1
19
H2PO4- + H2O H3O++ HPO4
2- K2
HPO42- + H2O H3O++ PO4
3- K3
H3PO4 + 3H2O 3H3O++ PO43- Ka
Ka = K1 x K2 x K3
NH3 + H2O NH4++ OH-
��#������K� )!�����
][][4
OHNHK
−+
=
20
][
][][
3
4
NH
OHNHK b
=
��#������K� �'(� (Hydrolysis)
H2O + H2O H3O++ OH-
���� H2O H++ OH-
[ ]OHOH
OHOHK
3
][
][][−+
=
21
[ ]OHOHK
2][ 2
=
[ ] [ ] [ ]OHOHOHK−+=
32
2
[H2O] = � !"#
Kw = =��= 678l�="$�����K� �'(�
][][3
OHOHK w
−+=
= 1.008x10-14 (mol dm-3)2 � 250CKw = 2.95x10-14 (mol dm-3)2 � 400C
678 250C [H3O+] = [OH- ] = 1.0x10-7 mol dm-3 )OP����
��#�9����#� [H O+] >>>> 10-7 mol dm-3
][][3
OHOHK w
−+=
22
��#�9����#� [H3O+] >>>> 10-7 mol dm-3
��#�9���)!� [H3O+] <<<< 10-7 mol dm-3
[OH- ] <<<< 10-7 mol dm-3
[OH- ] >>>> 10-7 mol dm-3
]7)�* (pH)
���#����� pH (pH scale)
10][][
10][][
log
log33
pOH
pH
OHOHpOH
OHOHpH
−−−
−++
=⇒−=
=⇒−=
23
�'(�!#Y��6^Yq [H3O+] = [OH- ] = 1.0x10-7 mol dm-3
pH = - log (1.0x10-7 )
pH = 7 ()OP���� )[H3O
+] >>>> 10-7 mol dm-3 pH <<<< 7 )OP���#�9����#�[H3O
+] <<<< 10-7 mol dm-3 pH >>>> 7 )OP���#�9���)!�
=������]�� %̂K� pH ��9 pOH
log
log
100.1
100.1][][
][
14
14
3
3
−=
==
−=
−
−
−−+
+
×
×
pH
OHOHK
OHpH
w
24
14
14
loglog
log
][10
][
100.1
14
=+
−=
−−=
−=
−−
−×
pOHpH
pOHpH
OHpH
OHpH
��#=(���$I� pH������� ��#�9��� CO2 �Y8�����7 [H3O
+] = 1.3x10-4molL-1
M =(���$ pH K� ��#�9���
�Y 7̂6(� ][ 3log OHpH +−=
25
103.1log4× −
−=pH
89.3
411.0
43.1log
10log43.1log
=
+−=
+−=
+−=
1. M I� [H3O+] K� ��#�9���678�7 pH = 4.4(antilog 0.6 = 4 antilog 0.4 = 2.5)
2. M =(���$ [H3O+] ��9 [OH-] K� ��#�9���678�7
pH = 4.5 (antilog 0.5 = 3.2)
�!!st�I��
26
3. M I� pH K� ��#�9��� 0.01 M NaOH
4. M I� pH K� ��#�9��� 0.001 M HCl
5. M I� pH K� ��#�9��� 0.2 MCH3COOH Ka=1.8x10-5
6. M I� pH K� ��#�9��� 0.2 M NH4OH Kb=1.8x10-5
pH = 4.5 (antilog 0.5 = 3.2)
�� ���������������� �������
� ��#=(���$=�� pH K� ��#�9����#���� ��9)!���� � � � �
[H+] ��9 [OH-] )6����!=���)K��K��)#Y8����K� �#������9)!������'�
������� HCl )K��K�� 1.0 x 10-3 M M =(���$ [OH-].
27
HCl = strong acid : HCl →→→→ H+ + Cl-
[H+] = [HCl] = 1.0 x 10-3 M
Thus (1.0 x 10-3 )[OH- ]= 1.0 x 10-14
[OH- ] = 1.0 x 10-11
�������� Calculate pOH and pH of 5.0 x 10-2 M NaOH.
[OH- ] = 5.0 x 10-2 M
�! ����� NaOH = strong base
pOH = - log (5.0 x 10-2 ) = 1.30
28
pOH = - log (5.0 x 10 ) = 1.30
[H+] = = 2.0 x 10-131.0 x10-14
5.0 x 10-2
pH = -log (2.0 x 10-13) = 12.70
pOH + pH = 1.30 + 12.70 = 14.00
�������� Calculate pH of solution prepared by mixing 2.0 mL of a strong acid of pH 3.00 and 3.0 mL of a strong base of pH 10.0
[H+] of strong acid solution = 1.0 x 10-3 M
mmol H+ = 1.0 x 10-3 M x 2.0 mL
29
mmol H+ = 1.0 x 10-3 M x 2.0 mL
= 2.0 x 10-3 mmol
pOH of strong base solution = 14.00 - 10.00 = 4.00
[OH- ] = 1.0 x 10-4 M
mmol OH- = 1.0 x 10-4 M x 3.0 mL
= 3.0 x 10-4 mmol
excess acid ,
mmol H+ = 0.0020 - 0.0003
30
mmol H = 0.0020 - 0.0003
= 0.0017 mmol
[H+] = 0.0017 mmol / 5.0 mL = 3.4 x 10-4 M
pH = -log ( 3.4 x 10-4 ) = 3.47
��#=(���$678)�78����! �#����� )!����� ��9)��H�K� �#�������9)��H�K� )!�����
Weak Acids and Bases.
- weak acids (or bases) are only partially ionized.
- amount of ionization , pH can be calculated from
31
- amount of ionization , pH can be calculated from
Kaor Kb.
������� ��#�9����#� HCN 0.2 M ������&I�[H3O
+] 4.0x10-3 M M I�=�� Ka
HCN + H2O H3O+ + CN-
)#Y8���� 0.2 M
678�����- -
4.0x10-3M 4.0x10-3M(0.2- 4.0x10-3 M)
32
678����� 4.0x10-3M 4.0x10-3M(0.2- 4.0x10-3 M)
][
][][ 3
HCN
CNOHK a
−+
=
1016.8]004.02.0[
004.0004.0 5×−× −==K a
��#=(���$)�78����!=�� Ka ��9 Kb
������� M I� [H+] &���#�9��� CH3COOH)K��K�� 1.0 mol dm-3 678 250C Ka = 1.8x10
-5
CH3COOH + H2O H3O++ CH3COO-
- -)#Y8���� 1.0 M
33
- -x M x M(1.0- x M)
)#Y8���� 1.0 M678�����
Ka =
1.8x10-5 =(x) (x) (1.0 - x)
[H3O+] [CH3COO- ]
[CH3COOH]
M9��� x = -b ±±±± b2 y 4ac 2a
x2 + (1.8x10-5)x - 1.8x10-5 = 0
34
2a
�6�=�� a = 1, b = 1.8x10-5, c = - 1.8x10-5
M9��� x = 4.2 x 10-3 mol dm-3
[H+] = 4.2 x 10-3 mol dm-3
)�H8� M�� [H+] �7=���������)�H8�)67�!��!=��� )K��K��)#Y8���� ��M���=�� x &�)6�� 1.00-x ������
1.8x10-5 =(x) (x) (1.0 - x) =
(x) (x) (1.0)
x2 = 1.8x10-5
35
x2 = 1.8x10-5
x = √√√√1.8x10-5 = 4.2 x 10-3 mol dm-3
% ��#������ = 4.2 x 10-3 x100 = 0.421
������������������� ��(� 5% ������ +��+�����#��#����<���� ���� [HA]/Ka > 1000=����"��<��S��������� (x) $�� HA �"�
CH COOH + H O H O++ CH COO-
�%&'(�)*
36
CH3COOH + H2O H3O++ CH3COO-
)#Y8���� 1.0 M - -
678����� x M x M(1.0- x M) ≅≅≅≅ 1.0
7. Salts of Weak Acids and Bases.
- salt of weak acid is a strong electrolyte.
- salt of weak acid is a Bronsted base.
Example NaA Na+ + A-
37
Example NaA Na + A
A- + H2O HA + OH- ---------- 1
( Bronsted base)
Equilibrium for these Bronsted base are identically to the weak base.
So, KH = Kb = ---------- 2
KH = hydrolysis constant of the sale = Kb
K can be calculated form K
[HA] [OH-]
[A-]
38
Kb can be calculated form Ka
The product of Ka of any weak acid and Kb of its conjugate base is always equal to Kw.
KaKb = Kw
So , from 2 = = Kb ---------- 3
: The pH of a salt (Bronsted base) is calculated as same as weak base.
Kw
Ka
[HA] [OH-]
[A-]
39
weak base.
If concentration of A is CA- .
A- + H2O HA + OH-
(CA- - X) X X
: The quantity X can be neglected compared to CA- if CA- > 100 Kb.
from 3 = = Kb
[OH-] [OH- ]
CA-
Kw
Ka
40
[OH- ] = . CA- = Kb . CA- -------- 4
4 holds only if CA- > 100 Kb
Kw
Ka
�������� Calculate the pH of 0.10 M solution of Sodium acetate.
NaOAc Na+ + OAc- (ionized)
OAc- + H2O HOAc + OH- (hydrolysis)
41
= Kb = = [HOAc] [OH- ]
[OAc- ]
Kw
Ka
1.00 x 10-14
1.75 x 10-5
= 5.7 x 10-10
[HOAc] = [OH- ] = X
[OAc- ] = COAc - X = 0.10 - X
since COAc >>> Kb neglected X compared to COAc , Then
= 5.7 x 10-10 (X)(X)
42
= 5.7 x 10-10
X = 5.7 x 10-10 x 0.10 = 7.6 x 10-6 = [OH- ]
( compared the solution to equation 4 )
(X)(X)
0.10
[OH- ] = 7.6 x 10-6 M
[H+] = = 1.3 x 10-9
pH = -log 1.3 x 10-9 = 8.89
1.0 x 10-14
7.6 x 10-6
43
Problem : Calculate the pH of a 0.25 M solution of ammonium chloride. ( Kb of NH3 = 1.75 x 10
-5 )
8. Buffer Buffer : a solution that resists change in pH.
add a small amount of acid.
When add a small amount of base.
44
Solution is dilute.
Buffer solution mixture of a weak acid and its conjugate base. mixture of a weak base and its conjugate acid.
Consider an HA / A- buffer . HA H+ + A-
When add A- into system, so, [H+ ] = [A- ]
[H+] = Ka
-log [H+] = -logKa - log
[HA] [A- ]
[HA] [A- ]
45
-log [H ] = -logKa - log
pH = pKa - log [A- ]
[HA] [A- ]
[A- ] [HA]
pH = pKa + log
(Henderson - Hasselbalch equation)
The pH of a buffer is determined by the ratio of the conjugate acid - base pair concentration.
pH = pKa + log
pH = pK + log
[conjugate base] [acid]
[proton acceptor]
46
pH = pKa + log
�������� Calculate the pH of a buffer prepared by adding 10 ml of 0.10 M acetic acid to 20 mL of 0.10 M Sodium acetate.
[proton acceptor] [proton donor]
For HOAc,0.10 mmol / mL x 10 mL = MHOAc x 30 mLMHOAc = 0.033 mmol / mL
For OAc-,0.10 mmol / mL x 20 mL = MOAc- x 30 mL
0 067
47
0.10 mmol / mL x 20 mL = MOAc x 30 mLMOAc- = 0.067 mmol / mL
[HOAc] [OAc- ] at equilibrium 0.033 - X 0.067 + X
=0.033 = 0.067
0 0
pH = -logKa + log
= - log(1.75 x 10-5) + log
= 4.76 + log 2.0
[proton acceptor] [proton donor]
0.067 mmol / mL 0.033 mmol / mL
48
= 5.06
�������� Calculate the pH of a solution prepared by adding 25 mL of 0.10 M Sodium hydroxide to 30 mL of 0.20 M acetic acid.
mmol HOAc = 0.20 M x 30 mL = 6.0 mmolmmol NaOH = 0.10 M x 25 mL = 2.5 mmol
These react as Follows :
HOAc + NaOH NaOAc + H2OAfter reaction,
49
After reaction,mmol NaOAc = 2.5 mmolmmol HOAc = 6.0 - 2.5 = 3.5 mmol
pH = 4.76 + log = 4.612.5 3.5
• pH ��(��� ����)������� ���U���� ��#�9���!�:):�#% 1 �Y�# �7 CH3COOH 0.1 mol ��9CH3COONa 0 .1 mol
+&�(),%+�-.+/-(01.+/23045(5��6
][ COOHCH
50
][
][
3
3log
COOCH
COOHCHpK
apH
−−=
( )( )
745.4
1.0
1.0log745.4
=
−=
�<V�<" HCl 1.0 M 1 cm3 �$�����Y�� = 0.001 mol
CH3COOH + H2O H3O++ CH3COO-
N��)�Y� HCl 1.0 M O#Y���# 1 cm3 M96(�&I���#�9����7 pH )6���#
51
CH3COOH + H2O H3O + CH3COO
H+ + CH3COO- CH3COOH
HCl H++ Cl-
0.001mol �" 0.001mol ��<� 0.001mol
)�H8�)�Y� HCl ������� =Y�=���)K��K��&I��
( )
( )L
mol
L
mol
L
mol
L
mol
COOCH
COOHCH
099.0001.1
001.01.0
101.0001.1
001.01.0
][
][
3
3
=−
=
=+
=
−
52
$��
(pH =����� )�"<�)
][
][
3
3log
COOCH
COOHCHpK
apH
−−=
736.4
009.0745.4
099.0
101.0log745.4
=
−=
−=
•••• N��)�Y� NaOH 1.0 M O#Y���# 1 cm3 M96(�&I���#�9����7 pH )6���#
CH3COOH + H2O H3O++ CH3COO-
NaOH Na++ OH-
OH- + CH COOH CH COO- + H O
53
0.001 mol
)�H8�)�Y� NaOH ����=���)K��K��K� ��#)O�78���O
M(����<��K� NaOH 678)�Y� =
OH- + CH3COOH CH3COO- + H2O
0.001mol �� 0.001mol ����� 0.001mol( )
( ) molmol
L
mol
L
mol
COOCH
COOHCH
101.0001.01.0
099.0001.1
001.01.0
][
][3
=+
=
=−
=
−
OH- + CH3COOH CH3COO- + H2O
54
LLCOOCH 101.0
001.1][
3==
(pH =����� )�"<�)
][
][
3
3log
COOCH
COOHCHpK
apH
−−=
749.4
101.0
099.0log745.4
=
−=
� !�"��#$% &#' !�(� )*+,�,-!.
H2O H++ OH-)�Y��'(�
CH3COOH + H2O H3O++ CH3COO-
55
•••• =���)K��K��K� �������� ~ = )�Y� pH ML = 678
CH3COO- + H+ CH3COOH
CH3COOH + OH- H2O + CH3COO-
� !�"!/)�' !�(� )*+,�,-!.
•••• )�H���#� I#H�)!�678�7 pKa I#H� pKb &���)=7� ��!pH I#H� pOH 678��� ��# I#H���"�&� *�� pKa ±±±± 1 I#H�
pKb ±±±± 1 )*��][
logHA
pKpH −=
56
][
][log
A
HApK
apH
−−=
I#H�[ ][ ]�����
��" [ ][ ]��"
�����1
10
10
1−≈
������)������ �+� � ��Z$����"�(��Y" =��������� � [H3O+] = 3.3 x 10-2 M�(��$��������";�<"=" �(�����
HNO , K = 4.6x10-4
)47�'/& +&�()�"'%1&�898&'045(5��6
57
HNO2, Ka = 4.6x10-4
H3PO4, Ka = 7.5x10-3
HSO4-, Ka = 1.2x10-2
�#�678)�H��&*�=H�.........................?
HSO4- + H2O H3O+ + SO4
2 -
•••• M��=�� Ka )�H��&*� HSO4- ��9O#�!���#�����
#9I��� �#���9)��H�
[ ]=+
−HSO
58
��� &*� HSO4- M(���� 2.7 <�� l����!)��H� SO4
2- M(����1 <��
[ ][ ]
[ ][ ] 1
75.2
102.1
103.3][
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2
23
2
4
24
4
3
4
=××=
+=
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−
−
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−
−
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K a
OH
SO
HSO
SO
HSOK aOH
������� M I����#�����M(����<��K� �#��9>Y�Y�(CH3COOH) ��9<>)�7���9>Y)�� (CH3COONa) )]H8�)�#7��!�:):�#%678�7 pH = 5.7 (Ka = 1.8x10-5)
loglog7.5
log
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3
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3
−−=
−=
−
−
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pH
COOCH
COOHCH
COOCH
COOHCHpK
a
101.8 5
59
)95.0log(
95.07.575.4log
log75.47.5
][
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3
3
3
3
3
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−=
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−−=
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antiCOOCH
COOHCH
COOCH
COOHCH
COOCH
COOHCH
COOCH101.8
11.0
101.11)05.01log(
=× −=+−= anti
������� M =(���$ pH K� !�:):�#% 0.4 M CH3COOH +0.4 M CH3COONa (Ka =1.8x10-5)
�Y 7̂6(� ][3
logCOOHCH
pKpH −=
60
�Y 7̂6(�][
][
3
3log
COOCH
COOHCHpK
apH
−−=
4.75101.8
101.85
5
=
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×
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−
log
4.0
4.0loglog
������� M =(���$ pH K� !�:):�#%#9I��� 0.1 M NH3
400 cm3 ��9 0.2 M NH4NO3 400 cm3 (Kb=1.8x10-5)
0.1 M NH3 400 cm3 �$�����Y��
0.2 M NH4NO3 400 cm3 �$�����Y��
��<������� = 400+400 = 800 cm3
mol1000
4001.0 ×=
mol1000
4002.0 ×=
61
��<������� = 400+400 = 800 cm3
[ ]
[ ] 01.0800
1000
1000
4002.0
05.0800
1000
1000
4001.0
34
3
=××
=
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=
NONH
NH �6�=��&�����#(Henderson - Hasselbalch equation)
�Y��Y)=)��#% (Indicator)
•••• ��#�Y�6#7�%678�7<=# �#�� >�!>�����9)O�78���7���
)�H8� pH K� ��#�9���)O�78��HIn (+�-�/�3) H+ + In -
�� �'(�) Y�
62
HIn (+�-�/�3) H + In�� �'(�) Y�
][
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HIn
InHK In
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��#�9����7�'(�) Y�
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10][=
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63
��#�9����7�'(�) Y�
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1][=
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1
][
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In
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pH ≤≤≤≤ 5 �7�7�� pH ≥≥≥≥ 8 �7�7�'(�) Y�
pH 5 - 8 �7�7l��#9I��� �� ��!�'(�) Y�&I� KIn K� �Y���� = 10-7
−+
64
&I� KIn K� �Y���� = 10
][
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7
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N�� pH = 5, [H+] = 10-5 mol dm-3
)�H8� pH<5, [HIn] >>>> [In- ] 100 )6�� ML �7�7�� ][
][
100
1
]10[
105
7
HIn
In−
−
−
==
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65
)�H8� pH<5, [HIn] >>>> [In ] 100 )6�� ML �7�7��
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10
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108
7
HIn
In−
−
−
==
N�� pH = 8, [H+] = 10-8 mol dm-3
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���� (�)�<�"<������ZIndicator pH �7678)O�78��
Thymol blue 1.2-2.8 �� -)I�H�
Bromphenol blue 3.0-4.6 )I�H� -�'(�) Y�
Congo red 3.0-5.0 �'(�) Y�-��
66
Congo red 3.0-5.0 �'(�) Y�-��
Methyl orange 3.1-4.4 �� -)I�H�
Bromocresol green 3.8-5.4 )I�H� -�'(�) Y�
Indicator pH �7678)O�78��
Methyl red 4.2-6.3 �� -)I�H�
Azolitmin (litmus) 5.0-8.0 �� -�'(�) Y�
������� �Y��Y)=)��#%
67
Bromocresol purple 5.2-6.8 )I�H� -���
Bromthymol blue 6.0-7.6 )I�H� -�'(�) Y�
Phenol red 6.8-8.4 )I�H� - ��
Phenolphthalein 8.3-10.0 ����7�7-*�]"
)OP���#�Y)=#�9I%I�O#Y��$��#I#H�=���)K��K��K� ��#�9��� <����#�(���#�9���678��� ��#�Y)=#�9I%��6(�OkY�Y#Y����!��#6786#�!=��� )K��K��678�������
��#�6)6#� (Titration)
68
��#6786#�!=���)K��K��678������ )#7�������#�9������#��� (Standard solution)
+&�=!(!�)�9�7/& +�--(01
•••• )�H8��#�-)!�6(�OkY�Y#Y��������"�]��7 )#7��>?,+,�,'&19(!,3�9�7/& +�--(01
•••• M��678��# 2 *�Y�6(�OkY�Y#Y�����]��7 : @*-1%%A8
•••• M��678��#�9���)O�78���7 ()�Y�M�� indicator) : @*-'*),
69
•••• M��678��#�9���)O�78���7 ()�Y�M�� indicator) : @*-'*),
•••• I� pH K� ��#�9���#9I��� ��#�6)6#�M��- ��#������� pH meter- ��#=(���$- Titration curve #9I��� pH ��! O#Y���#
+&�=!(!�)�9�7/& +�-.+/-(01.+/
(1.0M HCl 25 mL+ 1.0 M NaOH) $ M�����"� pH =7
pH range 5-9G Phenol red6.8-8.4
70
6.8-8.4 (yellow- red)G Bromthylmol blue 6.0-7.6
(yellow y blue)
+&�=!(!�)�9�7/& +�-�/�3-(01.+/
0.1 M CH3COOH 25 mL + 0.1 M NaOH $ M�����"�pH = 9
pH range 8-10
71
pH range 8-10Phenolphthalein8.3-10.0(no color y pink)
1.0 M NH3 40 mL + 1.0 M HCl $ M�����"� pH ≈≈≈≈ 5
��#�6)6#�#9I��� �#����-)!�����
pH range 4-7G Methyl red4.2-6.3
72
4.2-6.3 red-yellow
G Bromocresolgreen 3.8-5.4yellow-blue
�#����������������� 1 =#�' )*�� H2A + NaOH
M�����"�678 1 pH = 6.0
+&�=!(!�)+�-BC8,B>�),+-(01.+/
73
pH = 6.0
M�����"�678 2 pH = 11.9
������� Calculate the pH at 0, 10, 90, 100 and 110 % titrant for the titration of 50.0 mL of 0.100 M HCl with 0.100 M NaOH .
At 0 %pH = - log 0.100 =1.00
74
AT 10 %5.0 mL NaOH is added.
We start with 0.100 M x 50.0 mL = 5.00 mmol H+
Calculate the concentration of H+ after adding the NaOH :mmol H+ at start = 5.00 mmol H+
mmol OH-added = 0.100 M x 5.0 ml= 0.500 mmol OH-
75
= 0.500 mmol OH-
mmol H+ left = 4.50 mmol H+ in 55.0 mL.[H+] = 4.50 mmol / 55.0 mL = 0.0818 MpH = -log 0.0818 = 1.09
At 90 %mmol H+ at start = 5.00 mmol H+
mmol OH- added = 0.100 M x 45.0 mL = 4.50 mmol OH-
mmol H+ left = 0.50 mmol H+ in 95.0 mL.[H+] = 0.00526 M
76
[H+] = 0.00526 MpH = -log 0.00526 = 2.28
At 100 %All the H+ has been reacted with OH-, and we have a 0.0500 M solution of NaCl. Therefore, the pH is 7.00.
At 110 %We now have a solution consisting of NaCl and excess added NaOH.mmol OH- = 0.100 M x 5.00 mL
=0.50 mmol OH- in 105 mL[OH-] = 0.00476 M
77
[OH-] = 0.00476 MpOH = -log 0.00476 = 2.32pH = 11.68
'��5�6-7���8-9/��(� )#$% :�;#;-)
)*�� AgCl, BaSO4, Ag2SO4
AgCl �9���&��'(�AgCl(s) Ag+(aq) + Cl-(aq)
][][ ClAg−+
78
Ksp= =��= 678l�="$��#�9������[Ag+] [Cl- ] = l�="$����� (ion product)
][][
][][
][][
][
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ClAgK
ClAg
ClAg
sp
AgClK
AgClK
−+
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−+
=
=
=
ion product <<<< Ksp ����#N)�Y���#�9�������7�Ion product = Ksp ����� (��#�9����Y8����)ion product >>>> Ksp )�Y��9���KL'�&���#�9���
���Y ;�Z#�) K =;�=����� ���������$�����
1%-*8D� (+8��!"#898&'3EF&=-G3G�'
79
���Y ;�Z#�) Ksp =;�=����� ���������$�����
��#�7=�� Ksp �8(� M9���9������ ���
��#�7=�� Ksp �" M9�9��������� I#H����9���������
CaSO4 ���� �����"� > BaSO4
Ksp BaSO4 = 1.1x10-10
Ksp CaSO4 = 1.1x10-5
����������� ���������$�����
80
CaSO4 ���� �����"� > BaSO4
��8�=H� N��&���#�9����7 [Ba2+ ] = [Ca2+ ] )�H8�)�Y�SO4
2- M9)�Y��9���K� BaSO4 ���� ��9N��&*� [SO42-]
678)I��9��M9��� BaSO4 ���I��
������� &���#�9����7 Cu2+ 1.0 ×××× 10-4 M l����! Pb2+ 2.0 ×××× 10 y3 M N��)�Y� I- M9)�Y��9��� CuI2 I#H� PbI2 ���� N�� Ksp K� CuI2 = 1.0 ×××× 10-12
K K� PbI = 8.0 ×××× 10-9
81
Ksp K� PbI2 = 8.0 ×××× 10-9
��9��� &*�=���)K��K��K� I- )6���# ML M96(�&I�)�Y��9���]��7
H8D� =���3�/7% (Commom ion effect)
=H���#)O�78���O� 678)�Y�KL'�)�H8�)�Y������678�7��"�&�#9!! )*�� )�Y� NaCl &���#�9����Y8���� AgCl
AgCl(s) Ag+(aq) + Cl- (aq)NaCl(s) Na+(aq) + Cl- (aq)
82
[ Cl- ] )]Y8�KL'� [ Ag+ ] [ Cl- ] >>>> Ksp 6(�&I������)�H8��6� >���
[ Ag+ ] ��� M� [ Ag+ ] [ Cl- ] = Ksp
NaCl(s) Na+(aq) + Cl- (aq)������(��
1%-*8D� =���3(L, MG�3(Complex ion equilibrium)
��#)�Y������)*Y >���*���&I���#678�9����'(������������#N�9�������7KL'� )*�� )�Y� NH3 &� ��#�9����Y8���� AgCl
AgCl(s) Ag+(aq) + Cl - (aq)
83
AgCl(s) Ag+(aq) + Cl - (aq)
Ag+(aq) + NH3 [ Ag (NH3 )2 ]+
Complex ion[ Ag+ ] ��� �����)�H8��6� K��AgCl(s) �9������KL'�
������� �����)*Y >���
Ag+(aq.) + 2CN
- [ Ag (CN )2 ] -
Zn 2+ (aq.) + 4OH- [ Zn (OH )4 ] 2-
84
Zn 2+ (aq.) + 4OH- [ Zn (OH )4 ] 2-
[ Co(H2O)6 ] 2+
(aq.) + 4Cl- [CoCl4 ]
2-(aq.) + 6H2O