12. กำรแตกตัวของกรดอ่อน
สมกำรรูปเต็ม สมกำรรูปง่ำย
[ ][ ]
HA + H O = HCN +
OH
+ -
K =
H O A
[ ]
[ H O
+
]
C -
[ H O
]
HA 3
[ H O
+
] [ ] 3 HA a
HA
2
3
2
3
3
a
2
+
H O C K
C
HA
=
+
» Þ =
-
17. พิสูจน์
HA + H O = H O +
A K
2 3 a
- -
A + H O = HA +
OH K
[ ][ ]
[ ]
K H O A
3
[ ][ ]
[ ]
[ ]
K
[ ] b
+ - -
OH
H O A
3 w
a
2 b
K
OH
HA
HA
´ =
=
-
+ -
+ -
18.
19. หำควำมเข้มข้น HA H3O+ A- ใน
สำรละลำย HA
0.20 M Ka = 1.0x10-5
initial 0.20M 0M 0M
change - x + xM +
xM
eq (0.2 x )M xM xM
[ ][ ]
[ ]
2 3
H O A
3 5
5
1.0 10
( x )( x )
(0.20 x )
1.0 10
HA
HA H O H O A
-
-
+ -
+ -
= ´
-
= ´
-
+ = +
20. 1.0 10
x
0.20
= ´
2 6
x = 2.0 ´
10
x = 1.4 ´
10
-
-
-
x = [ H O + ] = [ A - ]
= 1.4 ´
10
-
3
3
[ HA ]
(0.20 x )
0.20M 0.0014M 0.20M
3
5
2
= -
= - »
21. หำร้อยละในกำรแตกตัวของ
HNO2 0.20 M
Ka = 4.5x10-4
HNO H O H O NO
start 0.20M 0M 0M
c - x + x +
x
eq (0.20 x ) x x
[ ][ ]
[ ]
4
4
H O NO
3 2
2
2 2 3 2
4.5 10
( x )( x )
0.20 x
4.5 10
HNO
-
-
+ -
+ -
= ´
-
= ´
-
+ = +
22. 2 - 4 -
4
x = 0.90 ´ 10 - 4.5 ´
10 x
2 - 4 -
4
x + 4.5x10 x - 0.90 ´ 10 =
0
x b b 4ac
( )
2
= - ± -
2a
4 4 2 4
- - -
4.5 10 4.5 10 4(1)( 0.90 10 )
- ´ ± ´ - - ´
=
2
- [ + ] [ - ]
= ´ = =
3 2
3
x 9.3 10 M H O NO
23. [ HNO ] [ H O
+
]
[ ]
[ ]
[ ]
[ ]
HNO
2 ionized
=
= ´
+
= ´
100% 4.6% ionized
H O
0.0093M
0.20M
100
HNO
100
HNO
% ionization
2 total
3
2 total
2 ionized 3
= ´ =
24. คำำนวณร้อยละในกำรแตกตัว
ของสำรละลำยอะนิลีน 0.30 M Kb =
4.2 x 10-10
C H NH H O C H NH OH
6 5 2 2 6 5 3
ini 0.30M 0M 0M
c - x + x +
x
eq (0.30 x ) x x
[ ][ ]
[ ]
4.2 10 -
x
+ -
+ -
0.30 x
C H NH OH
6 5 3
C H NH
- [ + ] [ - ]
= ´ = =
-
= ´ =
-
+ = +
x 1.1 10 M C H NH OH
6 5 3
5
2
10
6 5 2
25. [ ]
[ ]
100 0.0037%
1.1 10
0.30
100%
C H NH
6 5 2 ionized
= ´
C H NH
% ionization
5
6 5 2 total
-
= ´ ´ =
26. หำ pH และควำมเข้มข้นของทุก
สปีชีส์ใน 0.25 M NaCN KaHCN =
4.0x10-10
- -
CN + H O = HCN +
OH
init 0.25M 0M 0M
c - xM + xM +
xM
-
eq(0.2 5 -
x ) xM xM
[ ][ ]
[ ]
K
K
K HCN OH
-
= =
2.5 10
CN
14
-
= ´
2.5 10
1.0 10
= ´
4.0 10
x
´
0.25 x
w
-
-
= ´
-
-
-
x 2.5 10 3
[OH ] [HCN]
5
2
5
10
a
b
2
- -
= ´ = =
27. [ ]
[ ]
CN = (0.25 -
x )
(0.25 0.0025 ) 0.25
= - »
H O = 1.0 ´
10
-
2.5 10
12
3
4.0 10 M
14
3
= ´
pH =
11.40 very basic
´
-
-
+
-
28. หำ pH และร้อยละของกำรแตกตัว
(ไฮโดรลิซิส) ของ 0.15 M
trimethylammonium nitrate
(CH3)3NHNO3 Kb trimethylamine =
7.4x10-5
( ) ( )
+ +
CH NH H O CH N H O
(0.15 x ) x x
[( ) ][ ]
14
[( 10
) ] 5
= = = ´
w
b
K CH N H O
3 3
3 3
a
3 3 2 3 3
1.4 10
1.0 10
-
7.4 10
K
K
CH NH
-
-
+
+
= ´
´
-
+ = +
29. -
- [ ] [( ) ]
x
x = 4.6 ´ 10 = H O =
CH N
3 3 3
[ ] [( ) ]
x
x = 4.6 ´ 10 = H O =
CH N
3 3 3
%hydrolysis 4.6 10
100 0.0031%
%hydrolysis 4.6 10
0.15
pH 5.34
1.4 10
0.15 x
6
6
10
2
= ´ ´ =
=
= ´
-
-
- +
100 0.0031%
0.15
pH 5.34
1.4 10
0.15 x
6
6
10
2
= ´ ´ =
=
= ´
-
-
- +
30. สำรละลำยบัฟเฟอร์
สมกำรรูปเต็ม สมกำรรูปง่ำย
+ -
+ = +
2 3 a
A - + H O = HA + OH -
K =
K
[ + ][ -
]
[ ]
[ ]( [ ])
( [ ])
K H O A
HA
+ +
H O c +
H O
3 A -
3
c H O
w
+
[ ] [ ]
+ + -
H O molA
H O c
3 A
-
=
=
-
-
á = =
Handerson Hasselbach
c
c
c
K
= -
pH pK log
c
pK pH log
molHA
c
K 10 K
K
HA H O H O A K
A
A
-
-
HA
a
HA
a
3
HA
a
3
a
HA 3
a
3
a
a
2 b
= + -
36. คำานวณความเข้มข้นของทุกสปีชี
ส์ใน 0.20 M H3PO4 K1=7.5x10-3, K2 =
6.2 x 10-8 และ K3 =3.6x10-13
+ -
H PO + H O = H O +
H PO
3 4 2 3 2
(0.20 -
x ) x x
[ ][ ]
[ ]
( )( )
7.5 10 x x
(0.20 x )
+ -
H O H PO
-
[ ] [ ]
3 2
H PO
K
x = 3.2 ´ 10 M = H O =
H PO
[H PO ] 0.20 x 0.165M 0.16M
3 4
3 2
2
3
3 4
1
4
4
4
= - = =
-
= = ´ =
- + -
37. - + -
H PO + H O = H O +
HPO
2 4 2 3 4
3.5 ´ 10 -
y y y
[ + ][ -
]
[ ]
( )
6.2 10
K H O HPO
= = ´
6.2 10
= ´
3 4
H PO
( 3.5 ´ 10 +
y )( y )
3.5 ´ 10 -
y
- [ - ]
-
-
-
-
-
-
= ´ = =
2 4
8
8
2
2
8
2 4
2
2
y 6.2 10 K HPO
38. - + -
HPO + H O = H O +
PO
6.2 10 z z z
[ + ][ -
]
[ ]
K H O PO
= = ´
( 3.5 ´ 10 - 2 + 6.2 ´ 10 -
8
+
z )( z
)
( 3.5 10 )( z
)
( )
= 3.6 ´
10
-
- [ - ]
-
-
-
-
-
-
-
´
6.2 ´
10
= ´ =
= ´
´ -
´ -
3
4
19
13
8
2
13
8
13
2
4
3
3 4
3
8
3
2 3 4
2
4
z 6.4 10 M PO
3.6 10
6.2 10 z
3.6 10
HPO
39. Species 0.10MH PO 0.20MH PO
3 4 3 4
H PO 0.076M 0.16M
2 2
3 4
+ - -
H O 2.4 ´ 10 M 3.5 ´
10 M
2 2
3
- - -
H PO 2.4 10 M 3.5 10 M
2 4
´ ´
2 - - 8 -
8
4
HPO 6.2 ´ 10 M 6.2 ´
10 M
3 - - 19 -
19
4
PO 9.3 ´ 10 M 6.4 ´
10 M
13 13
- - -
OH 4.2 ´ 10 M 2.9 ´
10 M
%dis(1) 31% 21%
40. บัฟเฟอร์โพลีโปรติก คิดแบบ
บัฟเฟอร์ธรรมดา โดยพิจารณา
สมดุลหลัก
หาความเข้มข้นของไฮโดรเนียมไอออนใน
สารละลายบัฟเฟอร์ที่มี 2.00 M กรดฟอสฟอริคและ
1.50M โปตัสเซียมไดไฮโดรเจนฟอสเฟต
สมดุลหลัก K1 [ ][ ]
K H O H PO
3 2 4
[ ]
=
[ ]
K c
= 1 H PO
= ´ ´
3
3
3 4
H PO
3
3 4
1
9.48 10
7.11 10 2.00
1.50
c
H O
H PO
2 4
-
-
+
+ -
= ´
-
53. เติม HCl 30.00 mL
= + [ ]
H O 30.00 0.1000 2.5
[ H O + ] [ H O + ] [ H O
+
]
[ ] [ ]3 tot
3 tot 3 acid 3 water
H O C K
w
3 tot HX
H O
+
+
= +
= ´
6.25 10
pH 2.20
80.00
3
3
=
= ´ -
-
+
89. สมดุลของเกลือที่ละลายนำ้า
ได้น้อย
ค่า Ksp
ความสัมพันธ์ระหว่าง Ksp และ s
Salt type relationship
d n
for sat sol
AB s K
K
K
K
K
=
=
=
4 sp
3
4 sp
2 2
3 sp
2
3 sp
2
sp
27
AB s
16
A B s
4
AB s
4
A B s
=
=
123. Goal and Objective
State the desired goal
State the desired objective
Use multiple points if necessary
124. Today’s Situation
Summary of the current situation
Use brief bullets, discuss details verbally
125. How Did We Get Here?
Any relevant historical information
Original assumptions that are no longer valid
126. Available Options
State the alternative strategies
List advantages & disadvantages of each
State cost of each option
127. Recommendation
Recommend one or more of the strategies
Summarize the results if things go as proposed
What to do next
Identify action items