Transcript Slide 1

Buffers
solutions that resist pH changes addition of acid
base
contain acidic component HA + OH-  H2O + Abasic component A- + H+  HA
conjugate pair weak acid + conjugate base
weak base + conjugate acid
Henderson-Hasselbalch Equation
pH = pKa + log [A-]
best buffering
[A-]  [HA]
[HA]
[A-] =[HA] pH = pKa
-]
[A
adjust pH by changing
[HA]
10:1 maximum
pH = pKa  1
HC  H+ + C-
If you want a buffer at pH = 8.60
10-8.6 = 2.5 x 10-9
8.60 = 8.59 + log [C-]
a) HA Ka = 2.7 x 10-3
[HC]
b) HB Ka = 4.4 x 10-6
[C-]
= 1.02
-9
8.59 [HC]
c) HC Ka = 2.6 x 10
Calculate the pH of a buffer 1.0 M CHOOH HA
1.0
M
KCHOO
A
-4
K = 1.8 x 10
a
-]
[A
(0.9) ==3.74
pH = pKa + log
= 3.74 + log
log [1.00]
3.65
[HA]
[1.00]
(1.1)
What is the pH after addition 0.1 mole of HCl to 1.0 L
HCl  H+ + ClH+ acid reacts with CHOO- base
CHOO- + H+  CHOOH
[CHOOH] [CHOO-] [H+]
in H2O
I
1.1
0.9
0.0
pH = 1
-x
+x
+x
C
E 1.1 - x
0.9 + x
x
Preparation of a buffer of specific pH
-]
[A
pH = pKa + log
“phosphate buffer” pH = 7.4
[HA]
10-7.4 = 3.98 x 10-8
H3PO4  H2PO4- + H+ Ka1 = 7.5 x 10-3 pK = 7.21
a
H2PO4-  HPO42- + H+ Ka2 = 6.2 x 10-8
HPO42-  PO43- + H+ Ka3 = 4.8 x 10-13
dissolve NaH2PO4 and Na2HPO4 in water
7.4 = -log (6.2 x 10-8) + log [HPO42-]
-]
[H
PO
22
4
0.19 = log [HPO ]
4
[H2PO4
-]
2-]
[HPO
4
1.55 =
[H2PO4-]
Buffers
1. Mix weak acid + salt of conjugate base
Mix weak base + salt of conjugate acid
2. Partial neutralization of weak acid with strong base
weak base with strong acid
there must be an excess of weak acid (weak base)
assume that all of the strong base reacts
forming stoichiometric amount of Aleaving unreacted HA
Calculate the pH of a buffer prepared by mixing:
40.0 mL of 1.0 M C2H5OOH
Ka = 1.3 x 10-5
60.0 mL of 0.1 M NaOH
mol C2H5OOH = 0.04 L x 1.0 mol = 0.04 mol
L
mol OH- = 0.06 L x 0.1 mol = 0.006 mol
L
C2H5OOH + OH-  C2H5OO- + H2O
mol C2H5OOH = 0.040 - 0.006 = 0.034
mol C2H5OO- = 0.006 volume = 0.100 L
[C2H5OOH] = 0.34 M
[C2H5OO-] = 0.06 M
Calculate the pH of a buffer prepared by mixing:
Ka = 1.3 x 10-5
40.0 mL of 1.0 M C2H5OOH
60.0 mL of 0.1 M NaOH
pKa = - log (1.3 x 10-5)
pH = pKa + log [C2H5COO-]
[C2H5OOH]
pH = 4.89 + log 0.06 = 4.14
0.34
0.16 (0.1 – 10)
[C2H5OOH] = 0.34 M
3.89 <
< 5.89
[C2H5OO-] = 0.06 M