SME
Analytical Chemistry · Problem type
Titrating a diprotic acid
Use this for “25.00 mL of 0.0800 M maleic acid (pKₐ₁ = 1.92, pKₐ₂ = 6.27) is titrated with 0.1000 M NaOH. Find both equivalence volumes and the pH at the first.”
Diagram
Each
proton
of
the
acid
takes
CₐVₐ
Cb
=
25
mL
of
NaOH:
equivalence
at
25,
50
mL,
in
the
ratio
1
:
2.
Half-way to each, pH = pKₐ while that step is weak.
At
V₁,
pH
≈
pKₐ₁ + pKₐ₂
2
=
4.00;
the
exact
curve
gives
4.04.
After 25 mL, the pH is 4.04.
This is an example. Type your own values to start a new problem.
Cₐ (M)
×
Vₐ (mL)
=
Cb (M)
×
V₁ (mL)
pH₁
=
(
pKₐ₁
+
pKₐ₂
)
÷
2
=
V₂
Second equivalence volume
calculated · mL
Formulas
V₁
=
Cₐ
×
Vₐ
Cb
25
=
0.1
×
25
0.1
V₂
=
2
×
V₁
50 = 2 × 25
pH₁
=
pKₐ₁ + pKₐ₂
2
4.00
=
2.00 + 6.00
2
Assumptions
•
The acid H₂A gives up its two protons one after the other, so each takes the same volume of NaOH.
•
At the first equivalence point the flask holds HA⁻ alone, so its pH sits halfway between the two pKₐ values.
•
Halfway to each equivalence point the pH equals that step’s pKₐ (the curve’s two buffer regions).
Step-by-step
WE KNOW
pKₐ₁ = 2.00, pKₐ₂ = 6.00, Cₐ = 0.1 M, Vₐ = 25 mL, Cb = 0.1 M
FIND
first equivalence volume (V₁), second equivalence volume (V₂), pH at the first equivalence (pH₁)
Step 1 · Find first equivalence volume (V₁)
Use V₁ = Cₐ × Vₐ ÷ Cb
The first equivalence takes one NaOH per H₂A: the NaOH’s moles match CₐVₐ.
V₁
=
0.1
×
25
0.1
V₁
=
2.5
0.1
V₁
=
25
mL
Step 2 · Find second equivalence volume (V₂)
Use V₂ = 2 × V₁
The
second
proton
takes
as
much
NaOH
again,
so
the
second
equivalence
is
at
twice
V₁
.
V₂
=
2
×
25
V₂
=
50
mL
Step 3 · Find pH at the first equivalence (pH₁)
Use pH₁ = (pKₐ₁ + pKₐ₂) ÷ 2
HA⁻ can both lose and gain a proton; the two balance at the average of the pKₐ values.
pH₁
=
2.00 + 6.00
2
pH₁
=
8
2
pH₁ = 4.00
Check
Put the numbers back in. Both sides should match.
25
=
0.1
×
25
0.1
✓
50 = 2 × 25 ✓
4.00
=
2.00 + 6.00
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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