Normality ($N$) & Equivalent Concept
Concentration Terms | Stoichiometry | Titrations
1. Definition & Formula
Normality ($N$) is defined as the number of gram equivalents of solute present per liter of the solution.
$$ N = \frac{\text{Gram Equivalents of Solute}}{\text{Volume of Solution (in Liters)}} $$
$$ \text{Gram Eq} = \frac{\text{Mass (w)}}{\text{Equivalent Mass (E)}} $$
$$ \text{Gram Eq} = \frac{\text{Mass (w)}}{\text{Equivalent Mass (E)}} $$
Combined Formula:
$$ N = \frac{w}{E \times V(L)} $$Note: Like Molarity, Normality changes with temperature because it involves volume.
2. Equivalent Mass ($E$) and $n$-factor
The Equivalent Mass is determined by the Molar Mass ($M$) divided by a valency factor ($n$-factor).
$$ E = \frac{M}{n} $$Calculation of $n$-factor:
| Substance | n-factor Definition | Example |
|---|---|---|
| Acids | Basicity (No. of displaceable $H^+$) | $HCl (1)$, $H_2SO_4 (2)$, $H_3PO_4 (3)$ |
| Bases | Acidity (No. of displaceable $OH^-$) | $NaOH (1)$, $Ca(OH)_2 (2)$, $Al(OH)_3 (3)$ |
| Salts | Total magnitude of +ve or -ve charge | $Na_2CO_3 (2)$, $AlCl_3 (3)$, $Al_2(SO_4)_3 (6)$ |
| Oxidizing/Reducing Agents | Change in Oxidation Number per molecule | $KMnO_4$ (Acidic): $Mn^{+7} \to Mn^{+2}$ ($n=5$) |
3. Relation between Normality and Molarity
Since $N = \frac{w}{(M/n) \times V}$ and $Molarity (M) = \frac{w}{M \times V}$:
$$ N = M \times n $$
Since $n \ge 1$, Normality is always $\ge$ Molarity.
4. Laws of Dilution and Mixing
Dilution Law
Adding water does not change the number of equivalents of solute.
$$ N_1 V_1 = N_2 V_2 $$Mixing Law (Same Solute)
$$ N_{mix} = \frac{N_1 V_1 + N_2 V_2}{V_1 + V_2} $$Titration / Neutralization
At the equivalence point:
$$ \text{Gram Eq of Acid} = \text{Gram Eq of Base} $$ $$ N_1 V_1 = N_2 V_2 $$Practice Quiz
Test your grasp on Normality and n-factors.
No comments:
Post a Comment