Weak Acids/Bases (part I)
, H is the activity of Itt
pH log a AH
or effective concentration
1 gamma
C
activity
pH meter a Y a
OE TE l coefficient
see a dif reality 1M the amount you
than what we
calculate weighed out in a
certain volume
borosilicate glass
membrane
1
-1/+1
HA Ht VA
Y x -2/+2
0 p -3/+3
At buffer concentrations of conc M
0.1 M or higher p's 1
pH log Ht
, Acid HCl if you don't see Her it's weak
Strong other strong Oz metal analysis
NaOH
2504 forms appt
Strong base weak base
KOH j move the buffer at the wanted pH
weak
pKa tog ka
HA + H2O H3O+ + A-
acid
55.6M
HA H + + A-
y -x x x
pH
Ka = [H+] [A-] / [HA] equilibrium constant
quadratic is for dilute
0 Ka = x2/y-x solutions
mol OH- p
y Ka y
X ME Ht pH log Ht
, weak A- + H2O HA + OH-
base
A- HA + OH- 14
ka Kb 1 10
y- x x x
pKa 1 pKb 14
Kb = [HA] [OH-] / [A-]
v
Kb = x2/y-x
40
OH
X OH pOH log
pH
pHtpOH 14
_14 POH
pH
0
mol OH-
, weak acid
HA H+ + A-
y- x x x
pH Ka = [H+] [A-] / [HA]
logKatiogatt log
0 E T
mol OH-
log Ht tog ka t log
mol H+ 4
base
buffer 1 unit pH = pKa +
Tweak
log {[b]/[a]}
EX pH 7 pH 186 weak acid
Henderson Hasselbach eqh
pH pKa t log toady all
defined PH pka
If [b] = [a], then ___________.
log I o
be percentages molar etc Same
can O
value in same unit
, 25 HCl
NaOH mtotalbuffer concentration
a 25 tf Prepare 100 ml of 50 mM acetate buffer, pH 5, using 1.00 M
CH3COOH, 1.00 M NaOH, and/or 1.00 M HCl.
5050 go up curve down curve
pH 63,5l go
4.76 a b
to (0.100 l) (0.050 M acetate) = 0.0050 mol CH3COOH/CH3COO-
1001
0 0.0050 mol acetate = (x l) (1.00 M CH3COOH)
mol OH-
x = 0.005 liter = 5 ml
what is the 1 bat pH 5 0 635 0.0050MolCHz
COOH
oH pKa for Mme
log gab
eqn o 003175 moi OH reqir
0.003175MOIOH
5 b 19 00MNaOH
4.76 1 log x
III
0.24 109519 X O
00317 .7378
1.7378 b
bla a 1 3.17 MI
z
2.7378 i 5mL 1.00MCH3COOH
b 1.7378 NaOH
A 110 1001 3.17mi 1.00 M
atb I O 635
water to 100 Ml
at b 50mm
, H is the activity of Itt
pH log a AH
or effective concentration
1 gamma
C
activity
pH meter a Y a
OE TE l coefficient
see a dif reality 1M the amount you
than what we
calculate weighed out in a
certain volume
borosilicate glass
membrane
1
-1/+1
HA Ht VA
Y x -2/+2
0 p -3/+3
At buffer concentrations of conc M
0.1 M or higher p's 1
pH log Ht
, Acid HCl if you don't see Her it's weak
Strong other strong Oz metal analysis
NaOH
2504 forms appt
Strong base weak base
KOH j move the buffer at the wanted pH
weak
pKa tog ka
HA + H2O H3O+ + A-
acid
55.6M
HA H + + A-
y -x x x
pH
Ka = [H+] [A-] / [HA] equilibrium constant
quadratic is for dilute
0 Ka = x2/y-x solutions
mol OH- p
y Ka y
X ME Ht pH log Ht
, weak A- + H2O HA + OH-
base
A- HA + OH- 14
ka Kb 1 10
y- x x x
pKa 1 pKb 14
Kb = [HA] [OH-] / [A-]
v
Kb = x2/y-x
40
OH
X OH pOH log
pH
pHtpOH 14
_14 POH
pH
0
mol OH-
, weak acid
HA H+ + A-
y- x x x
pH Ka = [H+] [A-] / [HA]
logKatiogatt log
0 E T
mol OH-
log Ht tog ka t log
mol H+ 4
base
buffer 1 unit pH = pKa +
Tweak
log {[b]/[a]}
EX pH 7 pH 186 weak acid
Henderson Hasselbach eqh
pH pKa t log toady all
defined PH pka
If [b] = [a], then ___________.
log I o
be percentages molar etc Same
can O
value in same unit
, 25 HCl
NaOH mtotalbuffer concentration
a 25 tf Prepare 100 ml of 50 mM acetate buffer, pH 5, using 1.00 M
CH3COOH, 1.00 M NaOH, and/or 1.00 M HCl.
5050 go up curve down curve
pH 63,5l go
4.76 a b
to (0.100 l) (0.050 M acetate) = 0.0050 mol CH3COOH/CH3COO-
1001
0 0.0050 mol acetate = (x l) (1.00 M CH3COOH)
mol OH-
x = 0.005 liter = 5 ml
what is the 1 bat pH 5 0 635 0.0050MolCHz
COOH
oH pKa for Mme
log gab
eqn o 003175 moi OH reqir
0.003175MOIOH
5 b 19 00MNaOH
4.76 1 log x
III
0.24 109519 X O
00317 .7378
1.7378 b
bla a 1 3.17 MI
z
2.7378 i 5mL 1.00MCH3COOH
b 1.7378 NaOH
A 110 1001 3.17mi 1.00 M
atb I O 635
water to 100 Ml
at b 50mm
