Titration Formula
Quantitative analysis technique to determine unknown concentration using neutralization reaction
Titration Formulas
For 1:1 Reactions (e.g., HCl + NaOH)
Mâ‚Vâ‚ = Mâ‚‚Vâ‚‚
General Stoichiometric Formula
nâ‚Mâ‚Vâ‚ = nâ‚‚Mâ‚‚Vâ‚‚
where n is the stoichiometric coefficient
Moles at Equivalence Point
moles acid = moles base
Variables Explained
Mâ‚ (Molarity of acid)
Concentration of analyte (unknown) in mol/L
Vâ‚ (Volume of acid)
Volume of analyte solution in mL or L
Mâ‚‚ (Molarity of base)
Concentration of titrant (known) in mol/L
Vâ‚‚ (Volume of base)
Volume of titrant added at equivalence point
Worked Examples
Example 1: Simple 1:1 Titration
Problem: 25.0 mL of HCl is titrated with 0.100 M NaOH. It takes 30.5 mL of NaOH to reach equivalence. Find [HCl].
Reaction: HCl + NaOH → NaCl + H₂O (1:1 ratio)
Given:
Vâ‚ (HCl) = 25.0 mL
Mâ‚‚ (NaOH) = 0.100 M
Vâ‚‚ (NaOH) = 30.5 mL
Solution:
Mâ‚Vâ‚ = Mâ‚‚Vâ‚‚
Mâ‚(25.0) = (0.100)(30.5)
Mâ‚ = 3.05 / 25.0
Mâ‚ = 0.122 M HCl
Example 2: 1:2 Stoichiometry
Problem: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
20.0 mL of Hâ‚‚SOâ‚„ requires 35.0 mL of 0.150 M NaOH. Find [Hâ‚‚SOâ‚„].
Solution using nâ‚Mâ‚Vâ‚ = nâ‚‚Mâ‚‚Vâ‚‚:
nâ‚ = 1 (Hâ‚‚SOâ‚„), nâ‚‚ = 2 (NaOH)
(1)Mâ‚(20.0) = (2)(0.150)(35.0)
Mâ‚(20.0) = 10.5
Mâ‚ = 0.525 M Hâ‚‚SOâ‚„
Example 3: Finding Volume Required
Problem: What volume of 0.200 M HCl neutralizes 40.0 mL of 0.150 M KOH?
Solution:
Mâ‚Vâ‚ = Mâ‚‚Vâ‚‚
(0.200)Vâ‚ = (0.150)(40.0)
Vâ‚ = 6.00 / 0.200
Vâ‚ = 30.0 mL HCl needed
Common Mistakes
Ignoring Stoichiometry
Must use nâ‚Mâ‚Vâ‚ = nâ‚‚Mâ‚‚Vâ‚‚ when mole ratio ≠1:1
Unit Inconsistency
Both volumes must be in same units (both mL or both L)
Wrong Equivalence Point Volume
Use volume at equivalence point, not endpoint (indicator may change slightly before/after)
Polyprotic Acids
H₂SO₄, H₃PO₄ have multiple H⺠- check which H⺠is being neutralized