Equivalents, Equivalent Weight, and Normality in GOB Chemistry

Equivalents and Equivalent Weight

Definition of Equivalents

In chemistry, equivalents are used to measure the number of H+ ions or OH- ions in acids and bases, respectively. The concept of equivalents allows chemists to compare the reactive capacities of different acids and bases.

• Equivalent (Eq) of Acid: The amount of acid that contributes 1 mole of H+ ions.

• Equivalent (Eq) of Base: The amount of base that contributes 1 mole of OH- ions.

Example:

• 1 mol HBr = 1 Eq HBr

• 1 mol Ca(OH)2 = 2 Eq Ca(OH)2

Calculating Equivalents

To calculate the number of equivalents of acid or base, multiply the number of moles by the number of H+ or OH- ions each molecule can donate or accept.

Type	Formula
Acid Equivalent (Eq)	Eq = n × moles of acid
Base Equivalent (Eq)	Eq = n × moles of base

Where n = number of H+ or OH- ions per molecule.

Example: Calculating Equivalents

• a) 1 mol of H3PO4: H3PO4 can donate 3 H+ ions, so n = 3. Eq = 3 × 1 mol = 3 Eq H3PO4

• b) 2.7 g of ROH: Assume ROH is a monoprotic base (n = 1). Moles = 2.7 g / (molar mass of ROH) Eq = n × moles = 1 × moles of ROH

Equivalent Weight

The equivalent weight represents the mass (in grams) of an acid or base that supplies or reacts with 1 mole of H+ or OH- ions.

Formula
Eq Weight = $\frac{\text{molar mass (M)}}{n}$

Example: Calculate the equivalent weight of H2SO4 (n = 2, since it can donate 2 H+ ions):

• Molar mass of H2SO4 = 98.086 g/mol

• Eq weight = $\frac{98.086}{2} = 49.043$ g/mol

Normality (N)

Definition of Normality

Normality (N) is a measure of concentration that expresses the number of equivalents of solute per liter of solution. It is commonly used for acid-base and redox reactions.

• Normality (N) = $\frac{\text{Equivalent}}{\text{L solution}}$

• Alternatively, Normality (N) = n × Molarity (M)

Formula	Explanation
Normality (N) = $\frac{\text{Equivalent}}{\text{L solution}}$	Number of equivalents per liter
Normality (N) = n × M	n = number of H+ or OH- ions per molecule, M = molarity

Example: Calculating Normality

• a) 4.6 × 10-1 M NaOH: NaOH provides 1 OH- per molecule (n = 1). N = n × M = 1 × 0.46 = 0.46 N

• b) 0.35 g of H3PO4 in 1 L: Moles = 0.35 g / 98 g/mol = 0.00357 mol n = 3 (H3PO4 can donate 3 H+) Eq = n × moles = 3 × 0.00357 = 0.0107 Eq N = Eq / 1 L = 0.0107 N

Practice Problems and Applications

• Calculate mass (grams) needed for a given base equivalent: Example: 0.18 mEq of Mg(OH)2 (1.6 g needed)

• Identify the acid with a given equivalent weight: Example: Which acid has an equivalent weight of 63 g? (Answer: H2SO4)

• Determine volume (L) needed to prepare a solution of given normality: Example: 0.73 g of Ca(OH)2 with 1.25 N (0.9125 L needed)

Summary Table: Key Formulas

Concept	Formula
Acid/Base Equivalent	Eq = n × moles
Equivalent Weight	Eq Weight = $\frac{\text{molar mass}}{n}$
Normality (N)	N = $\frac{\text{Eq}}{\text{L solution}}$ or N = n × M

Additional info: The notes provide foundational concepts for acid-base titrations and solution preparation, which are essential for laboratory calculations in GOB Chemistry.