SME
General Chemistry II · Problem type
pH at the equivalence point
Use this for “20.00 mL of 0.120 M formic acid (Kₐ = 1.8 × 10⁻⁴) is titrated with 0.150 M NaOH. Find the pH at the equivalence point.”
Diagram
The HA is used up at 16 mL of NaOH, where the pH is 8.285: above 7, since the acid’s partner base is left.
After 16 mL of NaOH, the pH is 8.285.
This is an example. Type your own values to start a new problem.
Cₐ
Acid concentration
example · M
Vₐ
Acid volume
example · mL
Cb
NaOH concentration
example · M
Vₑ
Equivalence volume
calculated · mL
[A⁻]
Conjugate base at equivalence
calculated · M
Kₐ
Weak acid’s Kₐ
example
Kb
Conjugate base’s Kb
calculated
[OH⁻]
Hydroxide ion concentration
calculated · M
pOH
pOH
calculated
pH
pH
calculated
Formulas
Vₑ
=
Cₐ
×
Vₐ
Cb
16
=
0.12
×
20
0.15
[A⁻]
=
Cₐ
×
Vₐ
Vₐ + Vₑ
0.0667
=
0.12
×
20
20 + 16
Kb
=
1 ×
10
-14
Kₐ
5.56 ×
10
-11
=
1 ×
10
-14
1.8 ×
10
-4
Kb = [OH⁻]² ÷ ([A⁻] − [OH⁻])
5.56 ×
10
-11
=
(1.92 ×
10
-6
)
2
÷
(0.0667
−
(1.92
×
10
-6
))
pOH = −log₁₀([OH⁻])
5.72
=
−log₁₀(1.92
×
10
-6
)
pH = 14.00 − pOH
8.28 = 14.00 − 5.72
Assumptions
•
At the equivalence point the NaOH added has turned all the weak acid HA into its conjugate base A⁻, so only A⁻ sets the pH.
•
The moles of A⁻ equal the moles of acid at the start, now spread through the acid and base volumes together.
•
A⁻ + H₂O ⇌ HA + OH⁻ with Kb = Kw ÷ Kₐ and Kw = 1 × 10⁻¹⁴ at 25 °C, so the pH is above 7.
•
Water’s own OH⁻ is left out, which is fine while [OH⁻] is above 10⁻⁶ M.
Step-by-step
WE KNOW
Cₐ = 0.12 M, Vₐ = 20 mL, Cb = 0.15 M, Kₐ = 1.8 × 10⁻⁴
FIND
equivalence volume (Vₑ), conjugate base at equivalence ([A⁻]), conjugate base’s Kb (Kb), hydroxide ion concentration ([OH⁻]), pOH (pOH), pH (pH)
Step 1 · Find equivalence volume (Vₑ)
Use Vₑ = Cₐ × Vₐ ÷ Cb
At equivalence the moles of NaOH match the moles of acid: Cb × Vₑ = Cₐ × Vₐ. Both volumes stay in mL.
Vₑ
=
0.12
×
20
0.15
Vₑ
=
2.4
0.15
Vₑ = 16 mL
Step 2 · Find conjugate base at equivalence ([A⁻])
Use [A⁻] = Cₐ × Vₐ ÷ (Vₐ + Vₑ)
Every mole of acid became a mole of A⁻, now in the acid and the NaOH together, so divide by the total volume.
[A⁻]
=
0.12
×
20
20 + 16
[A⁻]
=
0.12
×
20
36
[A⁻]
=
2.4
36
[A⁻] = 0.0667 M
Step 3 · Find conjugate base’s Kb (Kb)
Use Kb = (1 × 10⁻¹⁴) ÷ Kₐ
For an acid and its conjugate base, Kₐ × Kb = Kw. The weaker the acid, the stronger A⁻ is as a base.
Kb
=
10
-14
1.8 ×
10
-4
Kb
=
5.56 ×
10
-11
Step 4 · Find hydroxide ion concentration ([OH⁻])
Use Kb = [OH⁻]² ÷ ([A⁻] − [OH⁻])
Kb
=
y
2
[A⁻] − y
is
y
2
+
Kb·y
−
Kb[A⁻]
=
0.
Its
positive
root
is
written
as
2Kb[A⁻]
÷
(Kb
+
√
Kb² + 4Kb[A⁻]
)
so
a
tiny
y
keeps
its
digits.
[OH⁻]
=
2
×
Kb
×
[A⁻]
÷
(Kb
+
√
Kb² + 4 × Kb × [A⁻]
)
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
÷
(5.56
×
10
-11
+
√(
(5.56 ×
10
-11
)
2
+
4
×
(5.56
×
10
-11
)
×
0.0667))
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
÷
((5.56
×
10
-11
)
+
√((3.0914
×
10
-21
)
+
4
×
(5.56
×
10
-11
)
×
0.0667))
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
÷
((5.56
×
10
-11
)
+
√((3.0914
×
10
-21
)
+
(2.224
×
10
-10
)
×
0.0667))
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
÷
((5.56
×
10
-11
)
+
√((3.0914
×
10
-21
)
+
(1.4834
×
10
-11
)))
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
(5.56 ×
10
-11
) +
√
1.4834 ×
10
-11
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
(5.56 ×
10
-11
) + (3.852 ×
10
-6
)
[OH⁻]
=
2
×
(5.56
×
10
-11
)
×
0.0667
3.852 ×
10
-6
[OH⁻]
=
(1.112
×
10
-10
)
×
0.0667
3.852 ×
10
-6
[OH⁻]
=
7.417 ×
10
-12
3.852 ×
10
-6
[OH⁻]
=
1.92 ×
10
-6
M
→
The
shortcut
√
Kb[A⁻]
gives
1.92 ×
10
-6
M,
under
0.1%
off:
close
enough
to
use
Step 5 · Find pOH (pOH)
Use pOH = −log₁₀([OH⁻])
pOH counts the powers of ten in [OH⁻].
pOH
=
−log₁₀(1.92
×
10
-6
)
pOH = 5.72
Step 6 · Find pH (pH)
Use pH = 14.00 − pOH
In water at 25 °C, pH and pOH add to 14.00.
pH = 14.00 − 5.72
pH = 8.28
Check
Put the numbers back in. Both sides should match.
16
=
0.12
×
20
0.15
✓
0.0667
=
0.12
×
20
20 + 16
✓
5.56 ×
10
-11
=
10
-14
1.8 ×
10
-4
✓
5.56 ×
10
-11
=
(1.92 ×
10
-6
)
2
÷
(0.0667
−
(1.92
×
10
-6
))
✓
5.72
=
−log₁₀(1.92
×
10
-6
)
✓
8.28 = 14.00 − 5.72 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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