Problem 38
pH = pK
a2
+
log
[HPO
4
=
]
[H
2
PO
4

]
Since the buffer will be at pH 6.9, the Bronsted base will be HPO
4
=
and the
Bronsted acid will be H
2
PO
4

.
Use the Henderson Hasselbalch equation as follows
to determine the ratio of [Bronsted base] to [Bronsted acid].:
6.9 = 7.21 + log
[HPO
4
=
]
[H
2
PO
4

]
.31 = log
[HPO
4
=
]
[H
2
PO
4

]
If x =
2 x =
(Note:
See problem136 in Segel Book, Page 57 for a similar problem.)
or
.31 =  log
[H
2
PO
4

]
[HPO
4
=
]
(Note:
I simply rearranged the
equation by multiplying each
side by 1 to make it easier to
deal with and avoid getting a
negative antilog.)
2 = log
[H
2
PO
4

]
[HPO
4
=
]
[HPO
4
=
]
[H
2
PO
4

]
40 liters x .02 moles/liter = .8 moles of phosphate is the total amount needed.
To prepare 40 liters of .02 M phosphate buffer, you will need::
You will need .8 moles of base to convert H
3
PO
4
to H
2
PO
4
You will then have to add additional base to convert some of the
H
2
PO
4

to
HPO
4
=
To calculate this:
x + 2x = .8
x = .2668 mole = amount of base to add
2x = .533 mole = amount of acid left
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 Winter '08
 Hilt
 pH, Phosphoric acid, Phosphate

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