SME
General Chemistry I · Problem type
Solution stoichiometry: the volume that reacts
Use this for “What volume of 0.100 M NaOH reacts completely with 25.00 mL of 0.150 M H₂SO₄?”
Diagram
n(H₂SO₄) = 0.15 M × 0.025 L
n(H₂SO₄) = 0.00375 mol
Mole ratio: n(NaOH) = 0.00375 × 2 = 0.0075 mol.
V
=
0.0075
mol
0.1
M
=
0.075
L
=
75
mL.
Every path goes through moles: the filled box is the value given.
This is an example. Type your own values to start a new problem.
r
×
C₁ (M)
×
V₁ (mL)
=
C₂ (M)
×
V₂ (mL)
The picture’s n = C × V counts moles: the base’s n₂ = C₂ × V₂ is r times the acid’s n₁ = C₁ × V₁, so r × C₁ × V₁ = C₂ × V₂.
n₁
Moles of acid
calculated
n₂
Moles of base
calculated
Formulas
n₁
=
C₁
×
V₁
1,000
0.00375
=
0.15
×
25
1,000
n₂
=
r
×
n₁
0.0075 = 2 × 0.00375
n₂
=
C₂
×
V₂
1,000
0.0075
=
0.1
×
75
1,000
Assumptions
•
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so the mole ratio r is 2 mol of base for each mol of acid. Read r from the coefficients of your own balanced equation.
•
Moles of solute = molarity × volume in liters; a volume in mL is divided by 1000.
•
The reaction goes to completion: the base added is exactly what the acid needs (the equivalence point).
Step-by-step
WE KNOW
C₁ = 0.15 M, V₁ = 25 mL, r = 2, C₂ = 0.1 M
FIND
moles of acid (n₁), moles of base (n₂), base volume (V₂)
Step 1 · Find moles of acid (n₁)
Use n₁ = C₁ × V₁ ÷ 1,000
Molarity
times
the
volume
in
liters
(
mL
1000
)
counts
the
moles
of
acid.
n₁
=
0.15
×
25
1,000
n₁
=
3.75
1,000
n₁
=
0.00375
mol
Step 2 · Find moles of base (n₂)
Use n₂ = r × n₁
The mole ratio from the balanced equation carries the moles from acid to base.
n₂
=
2
×
0.00375
n₂
=
0.0075
mol
Step 3 · Find base volume (V₂)
Use n₂ = C₂ × V₂ ÷ 1,000
Moles of base over its molarity gives liters; × 1000 turns them into mL.
V₂
=
1,000
×
n₂
C₂
V₂
=
1,000
×
0.0075
0.1
V₂
=
7.5
0.1
V₂
=
75
mL
Check
Put the numbers back in. Both sides should match.
0.00375
=
0.15
×
25
1,000
✓
0.0075 = 2 × 0.00375 ✓
0.0075
=
0.1
×
75
1,000
✓
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The molecular formula from the empirical formula
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Water of hydration from a heating experiment
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