SME
Organic Chemistry II · Problem type
Which layer an amine is in
Use this for “An amine’s conjugate acid has pKₐ 9.8 and KD is 15. Does it move into ether from water at pH 12.4?”
This is an example. Type your own values to start a new problem.
pKₐH
pKₐ of the conjugate acid
example
pH
pH of the water layer
example
f
Fraction neutral
calculated
KD
Partition coefficient
example
D
Distribution ratio
calculated
Formulas
f
=
1
1 +
10
pKₐH − pH
2.5119 ×
10
-9
=
1
1 +
10
10.6 − 2
D
=
KD
×
f
5.0238 ×
10
-8
=
20
×
(2.5119
×
10
-9
)
Assumptions
•
Only the neutral amine dissolves in the organic layer; its protonated form, R₃NH⁺, stays in the water.
•
pKₐH is the pKₐ of the amine’s conjugate acid, and the water layer’s pH is held by excess acid, base or a buffer.
•
KD is the neutral amine’s concentration in the organic layer ÷ its concentration in the water, at equilibrium.
•
D is all of the amine in the organic layer ÷ all of it in the water, both forms counted.
Step-by-step
WE KNOW
pKₐH = 10.6, pH = 2, KD = 20
FIND
fraction neutral (f), distribution ratio (D)
Step 1 · Find fraction neutral (f)
Use f = 1 ÷ (1 + 10pKₐH − pH)
R₃NH⁺ ÷ R₃N is 10 raised to pKₐH − pH, so the neutral share is 1 over 1 plus that ratio.
f
=
1
1 +
10
10.6 − 2
f
=
1
1 +
10
8.6
f
=
1
1 + 398,100,000
f
=
1
398,100,001
f
=
2.5119 ×
10
-9
Step 2 · Find distribution ratio (D)
Use D = KD × f
Only the neutral share moves into the organic layer, so scale KD by it.
D
=
20
×
(2.5119
×
10
-9
)
D
=
5.0238 ×
10
-8
→
D
<
1:
with
equal
volumes,
most
of
the
amine
stays
in
the
water
layer,
as
R₃NH⁺.
Check
Put the numbers back in. Both sides should match.
2.5119 ×
10
-9
=
1
1 +
10
10.6 − 2
✓
5.0238 ×
10
-8
=
20
×
(2.5119
×
10
-9
)
✓
Topic check
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