How to Find Valence Factor?

The valence factor (also known as the n-factor) depends on the type of substance you're dealing with: acid, base, salt, or oxidizing/reducing agent. Here's how to determine it for each:

1. For Acids:

• Definition: The n-factor of an acid is the number of replaceable hydrogen (H⁺) ions per molecule of the acid. This is also known as its basicity.

• How to Find It: Count the number of H⁺ ions that can be donated by one molecule of the acid in a reaction.

• Examples:

  • Hydrochloric acid (HCl): n-factor = 1 (one replaceable H⁺ ion)
  • Sulfuric acid (H₂SO₄): n-factor = 2 (two replaceable H⁺ ions)
  • Phosphoric acid (H₃PO₄): n-factor = 3 (three replaceable H⁺ ions)
  • Acetic acid (CH₃COOH): n-factor = 1 (only the H in the COOH group is replaceable)

2. For Bases:

• Definition: The n-factor of a base is the number of replaceable hydroxide (OH^-) ions per molecule of the base. This is also known as its acidity.

• How to Find It: Count the number of OH^- ions that can be donated or the number of H⁺ ions it can accept in a reaction.

• Examples:

  • Sodium hydroxide (NaOH): n-factor = 1 (one replaceable OH^- ion)
  • Calcium hydroxide (Ca(OH)₂): n-factor = 2 (two replaceable OH^- ions)
  • Aluminum hydroxide (Al(OH)₃): n-factor = 3 (three replaceable OH^- ions)

3. For Salts:

• Definition: The n-factor of a salt is the total positive or negative charge on the cation or anion of the salt.

• How to Find It: Determine the total positive charge on the cation or the total negative charge on the anion. These should be equal in magnitude.

• Examples:

  • Sodium chloride (NaCl): n-factor = 1 (Na⁺ has a charge of +1, Cl^- has a charge of -1)
  • Calcium chloride (CaCl₂): n-factor = 2 (Ca²⁺ has a charge of +2, and there are two Cl^- ions each with a charge of -1)
  • Aluminum sulfate (Al₂(SO₄)₃): n-factor = 6 (Al³⁺ has a charge of +3, and there are two Al³⁺ ions, so total positive charge is +6. Alternatively, SO₄^2- has a charge of -2, and there are three SO₄^2- ions so the total negative charge is -6.)

4. For Oxidizing and Reducing Agents (Redox Reactions):

• Definition: The n-factor is the number of moles of electrons lost or gained by one mole of the substance during a redox reaction.

• How to Find It: Determine the change in oxidation state of the element that is being oxidized or reduced. Multiply this change by the number of atoms of that element present in one molecule of the substance.

• Examples:

  • Potassium permanganate (KMnO₄) in acidic medium: Mn⁷⁺ + 5e^- → Mn²⁺. The oxidation state of Mn changes from +7 to +2, so the n-factor = 5.
  • Potassium dichromate (K₂Cr₂O₇) in acidic medium: Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O. The oxidation state of Cr changes from +6 to +3. Since there are two Cr atoms, the n-factor = 2 * (6 - 3) = 6.
  • Ferrous sulfate (FeSO₄) in a reaction where iron is oxidized to ferric ion: Fe²⁺ → Fe³⁺ + e^-. The oxidation state of Fe changes from +2 to +3, so the n-factor = 1.

In summary, the valence factor or n-factor is a crucial concept in stoichiometry that helps determine the number of equivalents of a substance involved in a reaction. Its value depends on the specific type of compound and the reaction it undergoes.