SME
Grade 10 Science · Problem type
Titrating a strong acid
Use this for “25 mL of 0.1 M HCl is titrated with 0.2 M NaOH. Where is the equivalence point?”
Diagram
The HCl is used up at 12.5 mL of NaOH, where the pH is 7.
After 12.5 mL of NaOH, the pH is 7.
Share of the way to equivalence: 1
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 HCl’s concentration and volume; C₂ is the NaOH’s concentration.
V₂
Base added
example · mL
r
Share of the way to equivalence
calculated
Formulas
Vₑ
=
(
C₁
×
V₁
)/
C₂
12.5 = (0.1 × 25)/0.2
r
=
V₂
/Vₑ
1 = 12.5/12.5
Assumptions
•
Each mole of NaOH uses up one mole of HCl.
•
At the equivalence point a strong acid and strong base leave salt water: pH 7.
Step-by-step
WE KNOW
C₁ = 0.1 mol/L, V₁ = 25 mL, C₂ = 0.2 mol/L, V₂ = 12.5 mL
FIND
base at the equivalence point (Vₑ), share of the way to equivalence (r)
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 × 25)/0.2
Vₑ = 2.5/0.2
Vₑ = 12.5 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
=
12.5/12.5
r
=
1
Check
Put the numbers back in. Both sides should match.
12.5 = (0.1 × 25)/0.2 ✓
1 = 12.5/12.5 ✓
Topic check
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