Back Titration | Volumetric Analysis Class 11

Back Titration

Indirect Volumetric Analysis | Stoichiometry

1. What is Back Titration?

Definition: A method where the amount of a substance (Analyte) is determined by reacting it with a known excess amount of a reagent, and then titrating the remaining (unreacted) excess reagent with another standard solution.

The Analogy:

Imagine buying an item worth \$x (Analyte). You give the shopkeeper a \$100 bill (Excess Reagent). The shopkeeper returns \$30 (Titration of excess).
Cost of Item = Total Amount - Amount Returned
Cost = \$100 - \$30 = \$70.

2. When is Back Titration Used?

Direct titration is preferred, but Back Titration is used when:

• The analyte is volatile (e.g., Ammonia $NH_3$) and might escape during direct titration.
• The analyte is an insoluble salt (e.g., $CaCO_3$) that needs an excess of acid to dissolve.
• The reaction is too slow for a direct endpoint determination.
• No suitable indicator is available for the direct reaction.

3. General Procedure & Formula

1. Add Excess Reagent (A): Add a known volume and concentration of Standard Reagent A to the Analyte.
   Analyte + A (Excess) $\rightarrow$ Product + A (Remaining)
2. Back Titration: Titrate the remaining Reagent A with Standard Reagent B.
   A (Remaining) + B $\rightarrow$ Product

$$ \text{Eq. of Analyte} = \text{Total Eq. of A added} - \text{Eq. of A remaining} $$
$$ \text{Eq. of Analyte} = (N_A V_A)_{total} - (N_B V_B)_{titration} $$

Note: We use the Law of Equivalence ($N_1V_1 = N_2V_2$) because it handles stoichiometry automatically via n-factors.

4. Solved Example

Problem: 4.0 g of impure $CaCO_3$ is treated with 100 mL of 1 N $HCl$. After the reaction, the excess acid required 20 mL of 1 N $NaOH$ for neutralization. Calculate the % purity of $CaCO_3$.

Solution:

• Step 1: Total Acid Added
  $\text{Eq of } HCl = N \times V(L) = 1 \times 0.1 = 0.1 \text{ eq}$.
• Step 2: Acid Left (Back Titration)
  $\text{Eq of } HCl \text{ left} = \text{Eq of } NaOH = 1 \times 0.02 = 0.02 \text{ eq}$.
• Step 3: Acid Reacted with $CaCO_3$
  $\text{Reacted} = \text{Total} - \text{Left} = 0.1 - 0.02 = 0.08 \text{ eq}$.
• Step 4: Mass of $CaCO_3$
  $\text{Eq wt of } CaCO_3 = \text{Molar Mass} / n = 100 / 2 = 50$.
  $\text{Mass} = \text{Eq} \times \text{Eq wt} = 0.08 \times 50 = 4.0 \text{ g}$.
• Step 5: Purity
  Since calculated mass (4g) equals sample mass (4g), purity is 100%. (Wait, calculation check: 0.08 * 50 = 4g. If sample was 5g impure, % would be 4/5 * 100 = 80%).

Practice Quiz

Test your understanding of Back Titration logic.

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