chem 123
Semester 2
January
2025
,week 1
Acid & Base
↑ Ka =
↑ dissociation ↑ kb = ↑ dissociation
( + 30
+
](A-]
ka =
[HA]
pka
=
-log(ka) PKD =
-log(kb)
(OH-](HA]
ko =
Kakb =
kw
(A -]
Kw = 1 .
0x10-14 at 25 water constant
-log (Hz0 ]
+
PH =
-
(OH] (OH -] [H30 + ]
-log kw
=
pot =
(-log (OH ])( log (H30 3)
=
-
+
pkw
-
DOt PH log rules
Pkw +
pH =
14
=
pOH
What happens to pH/pott of water when heated ?
Because rux rate 4 , Ch30 + ] & [OH-] ↑ ,
which
8 ↓
makes PH
.
pot but STAYS NEUTRAL
Conjugate Acid & base
acds + donate proton
HA * Hz0t
-
+
H20 A + base -
accepts proton
acid base base add
conjugate base
- +
dissociates will be
strong WEAK
T
acid HCI - H
eg
.
strong
acid
strong base dissociates -
conjugate add weak
base
eg. Calculate the pH if 0.002 moles of benzoic acid (C6H5COOH) is added to 100.0 mL of water at 298 K. The pKa of benzoic acid is 4.20.
1 RICE
Chart
R CHSCOOH H20 C6HsCOC HzOt
-
+ +
0 . 002
=
0 02M O C
L
.
I
-
X + X + X
C
0 02 x X
E -
0 02 X
.
.
3
2 pka- Ka ka egn
(H30 ] (C6Hs200]
+
ka =
4
[C6HsCOOH]
pka
= .
20
=
5
4 20
3x105
-
- .
ka =
10
=
6 30X10
.
6 .
-
3
x = 1 .
12x10
↳
PH
]
+
pH log [Hz0
= -
-
(1 12x10-3]
log
=
.
9
=
2 .
:
Summary
-dissociation egh can tell us [H30 ]/COH]+
+
p0H 14
pH
-
=
-
kW = 1 X 18- kW =
Kake
, Buffers
&
Henderson-Hasselback Egn Buffers -
neutralizes OH-1Hz0
+
eauces
change
a
CH -]
>
-
formed by weak add -
conjugate base in appreciable amount
PH pka +
log
=
(HA]
uses moles or eg. Calculate the pH if 10.0 mL of 0.020 M benzoic acid is added to 40.0 mL of 0.004 M sodium benzoate. The pKa of benzoic acid is 4.20.
molarity ·
Calculate moles of CHA] & (A)
moles of benzoic acid =
10 020M) . /0 .
014 = 0 . 0002 moles
moles of benzoate
=
10 .
004M) (0 .
041) =
0 .
00016 moles
H H
eqn
-
2
[Conj base)
+
PH
=
pka log Lacid]
=
4 . 20 +
log (000 ]
=
4 .
103
eg. Calculate the pH if 10.0 mL of 0.10 M NaOH is added to 100.0 mL of 0.020 M benzoic acid. The pKa of benzoic acid is 4.20.
1 moles of each CHAT / [A]
of moles
-
moles OH 0 .
001
moles of HA =
0 . 002 moles
a
use up supply
HA-OH- =
0 .
001 moles HA left+ this is also amount of A made
3
PH =
pka
+
log
-
4 . 20
:
summary
H-H used for Buffers
egh only
Semester 2
January
2025
,week 1
Acid & Base
↑ Ka =
↑ dissociation ↑ kb = ↑ dissociation
( + 30
+
](A-]
ka =
[HA]
pka
=
-log(ka) PKD =
-log(kb)
(OH-](HA]
ko =
Kakb =
kw
(A -]
Kw = 1 .
0x10-14 at 25 water constant
-log (Hz0 ]
+
PH =
-
(OH] (OH -] [H30 + ]
-log kw
=
pot =
(-log (OH ])( log (H30 3)
=
-
+
pkw
-
DOt PH log rules
Pkw +
pH =
14
=
pOH
What happens to pH/pott of water when heated ?
Because rux rate 4 , Ch30 + ] & [OH-] ↑ ,
which
8 ↓
makes PH
.
pot but STAYS NEUTRAL
Conjugate Acid & base
acds + donate proton
HA * Hz0t
-
+
H20 A + base -
accepts proton
acid base base add
conjugate base
- +
dissociates will be
strong WEAK
T
acid HCI - H
eg
.
strong
acid
strong base dissociates -
conjugate add weak
base
eg. Calculate the pH if 0.002 moles of benzoic acid (C6H5COOH) is added to 100.0 mL of water at 298 K. The pKa of benzoic acid is 4.20.
1 RICE
Chart
R CHSCOOH H20 C6HsCOC HzOt
-
+ +
0 . 002
=
0 02M O C
L
.
I
-
X + X + X
C
0 02 x X
E -
0 02 X
.
.
3
2 pka- Ka ka egn
(H30 ] (C6Hs200]
+
ka =
4
[C6HsCOOH]
pka
= .
20
=
5
4 20
3x105
-
- .
ka =
10
=
6 30X10
.
6 .
-
3
x = 1 .
12x10
↳
PH
]
+
pH log [Hz0
= -
-
(1 12x10-3]
log
=
.
9
=
2 .
:
Summary
-dissociation egh can tell us [H30 ]/COH]+
+
p0H 14
pH
-
=
-
kW = 1 X 18- kW =
Kake
, Buffers
&
Henderson-Hasselback Egn Buffers -
neutralizes OH-1Hz0
+
eauces
change
a
CH -]
>
-
formed by weak add -
conjugate base in appreciable amount
PH pka +
log
=
(HA]
uses moles or eg. Calculate the pH if 10.0 mL of 0.020 M benzoic acid is added to 40.0 mL of 0.004 M sodium benzoate. The pKa of benzoic acid is 4.20.
molarity ·
Calculate moles of CHA] & (A)
moles of benzoic acid =
10 020M) . /0 .
014 = 0 . 0002 moles
moles of benzoate
=
10 .
004M) (0 .
041) =
0 .
00016 moles
H H
eqn
-
2
[Conj base)
+
PH
=
pka log Lacid]
=
4 . 20 +
log (000 ]
=
4 .
103
eg. Calculate the pH if 10.0 mL of 0.10 M NaOH is added to 100.0 mL of 0.020 M benzoic acid. The pKa of benzoic acid is 4.20.
1 moles of each CHAT / [A]
of moles
-
moles OH 0 .
001
moles of HA =
0 . 002 moles
a
use up supply
HA-OH- =
0 .
001 moles HA left+ this is also amount of A made
3
PH =
pka
+
log
-
4 . 20
:
summary
H-H used for Buffers
egh only
