What amount of solid sodium acetate is needed to prepare a buffer of pH 5.00 from 1L of 0.10 M acetic acid? (pka of acid is 4.75)
Answers
Answer:
This can be contradictory, depending on whether the
0.1 M
is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...
V
A
−
=
9.125 mL
V
H
A
=
15.875 mL
The Henderson-Hasselbalch equation is:
pH
=
pK
a
+
log
[
A
−
]
[
HA
]
We have a
pH 4.5
solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:
[
A
−
]
[
HA
]
=
10
pH
−
pK
a
=
10
4.5
−
4.74
=
0.5754
Now, if the total concentration is
0.10 M
, then:
[
HA
]
+
[
A
−
]
0.5754
[
HA
]
=
0.10 M
⇒
[
HA
]
=
0.10 M
1.0000
+
0.5754
=
0.0635 M
−−−−−−−−
⇒
[
A
−
]
=
0.0365 M
−−−−−−−−
and these concentrations are AFTER mixing. Since the total volume is
50 mL
, or
0.050 L
, the mols of each component (which are constant!) are:
n
A
−
=
0.0365 mol
L
×
0.050
L
=
0.001825 mols
−−−−−−−−−−−−
n
H
A
=
0.0635 mol
L
×
0.050
L
=
0.003175 mols
−−−−−−−−−−−−
So, if both of the starting concentrations were
0.20 M
, we can find the volume they each start with:
V
A
−
=
1 L
0.20
mols A
−
×
0.001825
mols A
−
=
0.009125 L
=
9.125 mL
−−−−−−−−
V
H
A
=
1 L
0.20
mols HA
×
0.003175
mols HA
=
0.015875 L
=
15.875 mL
−−−−−−−−−
And this should make sense, because the total starting volume is
25.000 mL
, the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.
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