SME
Organic Chemistry I · Problem type
Which way an acid–base reaction lies
Use this for “Does hydroxide take the proton from phenol (pKₐ 9.95)? Water’s pKₐ is 15.7. Find K.”
Diagram
acid on the left (4.76) gives its proton; acid on the right (15.7) forms.
The arrow points to the weaker acid, the products.
log K = pKₐ(right) − pKₐ(left) = 15.7 − 4.76 = 10.9, K = 8.71 × 10¹⁰.
This is an example. Type your own values to start a new problem.
K
=
10
−
log K
Log of the equilibrium constant
calculated
Formulas
log
K
=
pKₐ(right)
−
pKₐ(left)
10.9 = 15.7 − 4.76
K
=
10
log
K
8.71 ×
10
10
=
10
10.9
Assumptions
•
The acid on the left gives its proton to the base; the acid on the right is what that base becomes, its conjugate acid.
•
The reaction favors the side with the weaker acid, the one with the larger pKₐ, so log K = pKₐ(right) − pKₐ(left).
•
pKₐ values are for water; those far below 0 or above 14 are estimates, so a K from them is a guide.
•
The picture places both acids on a ladder of common acids, with the arrow toward the weaker one.
Step-by-step
WE KNOW
pKₐ(left) = 4.76, pKₐ(right) = 15.7
FIND
log of the equilibrium constant (log K), equilibrium constant (K)
Step 1 · Find log of the equilibrium constant (log K)
Use log K = pKₐ(right) − pKₐ(left)
Take the pKₐ of the acid on the left from the pKₐ of the acid on the right.
log
K
=
15.7
−
4.76
log
K
=
10.94
Step 2 · Find equilibrium constant (K)
Use K = 10ˡᵒᵍᴷ
Undo
the
log:
raise
10
to
log
K
.
K
=
10
10.94
K
=
8.71 ×
10
10
→
K
>
1:
the
acid
on
the
right
is
weaker
(larger
pKₐ),
so
the
products
are
favored.
Check
Put the numbers back in. Both sides should match.
10.94 = 15.7 − 4.76 ✓
8.71 ×
10
10
=
10
10.94
✓
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Rings and π bonds from a formula
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Topic check
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