SME
Grade 10 Science · Problem type
Titrating a weak acid
Use this for “20 mL of 0.1 M acetic acid (Kₐ = 1.8 × 10⁻⁵) is titrated with 0.1 M NaOH.”
Diagram
The acetic acid is used up at 20 mL of NaOH, where the pH is 8.722: above 7, since the acid’s partner base is left.
After 10 mL of NaOH, the pH is 4.745.
Share of the way to equivalence: 0.5
pKₐ: 4.745
Base at the half-way point: 10 mL
This is an example. Type your own values to start a new problem.
C₁ (mol/L)
×
V₁ (mL)
=
C₂ (mol/L)
×
Vₑ (mL)
C₁ and V₁ are the acetic acid’s concentration and volume; C₂ is the NaOH’s concentration.
V₂
Base added
example · mL
r
Share of the way to equivalence
calculated
Kₐ
Acid dissociation constant
example
pKₐ
pKₐ
calculated
V½
Base at the half-way point
calculated · mL
Formulas
Vₑ
=
(
C₁
×
V₁
)/
C₂
20 = (0.1 × 20)/0.1
r
=
V₂
/Vₑ
0.5
=
10
20
pKₐ = −log₁₀(Kₐ)
4.745
=
−log₁₀(1.8
×
10
-5
)
V
1
2
=
Vₑ/2
10
=
20
2
Assumptions
•
Halfway to equivalence, half the acid is turned to its partner base, so pH = pKₐ.
•
At equivalence only the partner base is left, so the pH is above 7.
Step-by-step
WE KNOW
C₁ = 0.1 mol/L, V₁ = 20 mL, C₂ = 0.1 mol/L, V₂ = 10 mL, Kₐ = 1.8 × 10⁻⁵
FIND
base at the equivalence point (Vₑ), share of the way to equivalence (r), pKₐ (pKₐ), base at the half-way point (V½)
Step 1 · Find base at the equivalence point (Vₑ)
Use Vₑ = (C₁ × V₁)/C₂
At
equivalence
the
moles
of
base
equal
the
moles
of
acid:
C₂Vₑ
=
C₁
V₁
.
Vₑ = (0.1 × 20)/0.1
Vₑ = 2/0.1
Vₑ = 20 mL
Step 2 · Find share of the way to equivalence (r)
Use r = V₂/Vₑ
Compare the base added with the base equivalence takes.
r
=
10
20
r
=
0.5
Step 3 · Find pKₐ (pKₐ)
Use pKₐ = −log₁₀(Kₐ)
Each step of 1 on the pKₐ scale is ten times the Kₐ: the log counts the powers of ten.
pKₐ
=
−log₁₀(1.8
×
10
-5
)
pKₐ = 4.745
Step 4 · Find base at the half-way point (V½)
Use V½ = Vₑ/2
Half the base turns half the acid into its partner base: there pH = pKₐ.
V
1
2
=
20
2
V
1
2
=
10
mL
Check
Put the numbers back in. Both sides should match.
20 = (0.1 × 20)/0.1 ✓
0.5
=
10
20
✓
4.745
≈
−log₁₀(1.8
×
10
-5
)
✓
10
=
20
2
✓
Topic check
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