SME
General Chemistry II · Problem type
pH of a weak base
Use this for “Find the pOH and pH of 0.250 M methylamine, CH₃NH₂ (Kb = 4.4 × 10⁻⁴).”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻¹²·⁰¹ = 9.7367 × 10⁻¹³ mol/L.
Above 7: basic. pOH = 14 − pH = 1.988.
This is an example. Type your own values to start a new problem.
Kb
=
(
[OH⁻] (M)
)
2
÷
(
C (M)
−
[OH⁻] (M)
)
pOH
=
−log(
[OH⁻] (M)
)
=
C is the base’s concentration before any of it reacts, and Kb its base dissociation constant.
pH
pH
calculated
Formulas
Kb
=
[OH⁻]²
÷
(
C
−
[OH⁻])
4.4 ×
10
-4
=
0.0103
2
0.25 − 0.0103
pOH = −log₁₀([OH⁻])
1.99 = −log₁₀(0.0103)
pH = 14.00 − pOH
12.01 = 14.00 − 1.99
Assumptions
•
A weak base B + H₂O ⇌ BH⁺ + OH⁻. If y is the [OH⁻] it makes, then [BH⁺] = y and [B] = C − y at equilibrium.
•
Kb = y² ÷ (C − y) is solved exactly by the quadratic formula. The shortcut y ≈ √(KbC) is fine when it lands within 5%.
•
Water’s own OH⁻ is left out, which is fine while y is above 10⁻⁶ M.
•
The water is at 25 °C, so pH + pOH = 14.00.
Step-by-step
WE KNOW
Kb = 4.4 × 10⁻⁴, C = 0.25 M
FIND
hydroxide ion concentration ([OH⁻]), pOH (pOH), pH (pH)
Step 1 · Find hydroxide ion concentration ([OH⁻])
Use Kb = [OH⁻]² ÷ (C − [OH⁻])
Kb
=
y
2
C
− y
is
y
2
+
Kb·y
−
Kb·
C
=
0.
Its
positive
root
is
−Kb +
√
Kb² + 4KbC
2,
written
here
as
2K
bC
Kb +
√
Kb² + 4KbC
so
a
tiny
y
keeps
its
digits.
[OH⁻]
=
2
×
Kb
×
C
Kb +
√
Kb² + 4 × Kb ×
C
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
÷
(4.4
×
10
-4
+
√(
(4.4 ×
10
-4
)
2
+
4
×
(4.4
×
10
-4
)
×
0.25))
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
÷
((4.4
×
10
-4
)
+
√((1.936
×
10
-7
)
+
4
×
(4.4
×
10
-4
)
×
0.25))
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
÷
((4.4
×
10
-4
)
+
√((1.936
×
10
-7
)
+
0.00176
×
0.25))
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
÷
((4.4
×
10
-4
)
+
√((1.936
×
10
-7
)
+
0.00044))
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
(4.4 ×
10
-4
) +
√
0.0004402
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
(4.4 ×
10
-4
) + 0.02098
[OH⁻]
=
2
×
(4.4
×
10
-4
)
×
0.25
0.02142
[OH⁻]
=
0.00088
×
0.25
0.02142
[OH⁻]
=
0.00022
0.02142
[OH⁻]
=
0.0103
M
→
The
shortcut
√
KbC
gives
0.0105
M,
2.1%
off:
close
enough
to
use
Step 2 · Find pOH (pOH)
Use pOH = −log₁₀([OH⁻])
pOH counts the powers of ten in [OH⁻].
pOH = −log₁₀(0.0103)
pOH = 1.99
Step 3 · Find pH (pH)
Use pH = 14.00 − pOH
In
water
at
25
°
C
,
pH
and
pOH
add
to
14.00.
pH = 14.00 − 1.99
pH = 12.01
Check
Put the numbers back in. Both sides should match.
4.4 ×
10
-4
=
0.0103
2
0.25 − 0.0103
✓
1.99 = −log₁₀(0.0103) ✓
12.01 = 14.00 − 1.99 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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