Hcl note you may ignore the slight change in volume

HCl? Note: You may ignore the slight change
in volume caused by the HCl addition.
n ofNa HPO mol
2 4
0072 01 00072 = ´ =
( )
. . .
(2.32)
n ofNaH PO mol
2 4
0028 01 00028 = ´ =
( )
. . . (2.33)
n M v ofHCl mol
mol
( )
= ´ = ´
=
( )
2 0001
0002
.
. (2.34)
n ofNa HPO afterHCladdition mol
2 4
00072 0002 00052
( )
- =
:
. . .
(2.35)
n ofNaH PO afterHCladdition mol
2 4
00028 0002 00048
( )
+ =
:
. . .
(2.36)
pH of buffer after HCl addition and conversion
of n into M through Eq. 2.1 :
pH = +
[ ]
[ ]
= 671
0052
0048
674 . log
.
.
.
(2.37)
Note: The ratio of A
−
to AH stays the same
whether concentrations or amounts are inserted,
since units are canceled, so technically n does not
need to be converted into M to obtain the correct
result.
Example C3 Prepare 250 mL of 0.1 M acetate
buffer with pH 5. The pK
a
of acetic acid is 4.76
(see Table 2.2 ). The molecular weights of acetic
acid and sodium acetate are 60.06 and 82.03 g/
mol, respectively.
Solution This example matches the tasks at
hand in a lab better than Example C1 . One needs
a buffer to work at a certain pH, looks up the
pK
a
value, and decides on the molarity and vol-
ume needed. The molarity of a buffer equals
[A
−
] + [AH]. To solve Example C3 , one of those
concentrations needs to be expressed in terms of
the other, so that the equation only contains one
unknown quantity. For this example, it will be
[A
−
], but the results would be the same if [AH]
had been chosen. Together with the target pH
(5) and the pK
a
(4.76), the values are inserted
into Eq. 2.25 :
Molarityofbuffer
mol
L
AH A
æ
è
ç
ö
ø
÷
=
[ ]
+ - [ ] (2.38)
01 . [ ] = +
[ ]
A AH
-
(2.39)
[A AH – ] . – =
[ ]
01
(2.40)
C. Tyl and B.P. Ismail

29
There are three ways to prepare such a buffer:
1. Prepare 0.1 M acetic acid and 0.1 M sodium ace-
tate solutions. For our example, 1 L will be pre-
pared. Sodium acetate is a solid and can be
weighed out directly using Eq. 2.5 . For acetic
acid, it is easier to pipet the necessary amount.
Rearrange Eq. 2.8 to calculate the volume of
concentrated acetic acid (density = 1.05) needed
to prepare a volume of 1 L. Then express the
mass through Eq. 2.5 .
m M v ofsodiumacetate g MW
( )
= ´ ´ (2.5)
m ofsodiumacetate g
mol
L
L
g
mol
g
( )
=
æ
è
ç
ö
ø
÷
´
( )
´
æ
è
ç
ö
ø
÷
=
( )
01 1 82 82 . . (2.50)
v ofaceticacid mL
m g
d
g
mL
( )
=
( )
æ
è
ç
ö
ø
÷
(2.51)
v
M v
ofaceticacid mL
mol
L
L MW
g
mol
d
g
mL
( )
=
æ
è
ç
ö
ø
÷
´
( )
´
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
(2.52)
v ofaceticacid mL
g
mol
mol
L
L
g
m
( )
=
æ
è
ç
ö
ø
÷
´
æ
è
ç
ö
ø
÷
´
( )
6006 01 1
105
. .
.
L
mL
æ
è
ç
ö
ø
÷
=
( )
572 . (2.53)
Dissolving each of these amounts of sodium
acetate and acetic acid in 1 L of water gives two
1  L stock solutions with a concentration of
0.1 M . To calculate how to mix the stock solu-
tions, use their concentrations in the buffer, i.e.,
0.0365 as obtained from Eqs. 2.48 and 2.49 –
0.0365
mol
L
æ
è
ç
ö
ø
÷
for acetic acid and 0.0635
mol
L
æ
è
ç
ö
ø
÷
for sodium acetate and Eq. 2.15 :
v
M
ofaceticacidstocksolution L
ofaceticacidinbuffer
mol
L
( )
=
æ
è
ç
ö
ø
÷
´
( )
æ
è
ç
ö
ø
÷
v
M
ofbuffer L
ofstocksolution
mol
L
(2.54)
v ofaceticacidstocksolution L
L
( )
=
´
=
( )
00365 025
01
0091
. .
.
.
(2.55)
v ofsodiumacetatestocksolution L
L
( )
=
´
=
( )
00635 025
01
0159
. .
.
.
(2.56)
Combining the 0.091 L of acetic acid stock and
0.159  L of sodium acetate stock solution gives
0.25 L of buffer with the correct pH and molarity.
2. Directly dissolve appropriate amounts of both
components in the same container. Equations 2.48
and 2.49 yield the molarities of acetic acid and
sodium acetate in a buffer, i.e., the moles per 1