Volumetric Analysis: Redox & Acid-Base Titrations (Class 12) - Chemca

Volumetric Analysis

Redox & Acid-Base Titrations • Class 12 NCERT Lab Manual

Class 12 Practical Exam / JEE

105 Short Q&A Facts

Part 1: Basic Volumetric Concepts

Q1: What is Volumetric Analysis?
A: A quantitative analysis technique based on measuring the volume of a solution of known concentration required to react completely with a solution of unknown concentration.
Q2: Define Titration.
A: The process of adding the solution from the burette to the solution in the conical flask until the reaction is complete.
Q3: Define Titre.
A: The volume of solution taken from the burette during the titration.
Q4: What is the Endpoint of a titration?
A: The point where the indicator changes color, signaling the completion of the reaction.
Q5: What is the Equivalence Point of a titration?
A: The point where the moles (or equivalents) of the reactant from the burette exactly equal the moles (or equivalents) of the reactant in the flask.
Q6: What is the ideal relationship between the endpoint and the equivalence point?
A: They should be as close as possible.
Q7: Define a Standard Solution.
A: A solution whose concentration is accurately known.
Q8: Define a Primary Standard.
A: A substance that can be obtained in high purity, is stable, and has a known chemical composition (used to make standard solutions directly).
Q9: Give two examples of Primary Standards used in titrations.
A: Oxalic Acid (Acid-Base) and Anhydrous Na₂CO₃ (Acid-Base).
Q10: Define a Secondary Standard.
A: A substance that cannot be obtained in high purity or is unstable, and must be standardized against a primary standard (e.g., NaOH, KMnO₄).

Part 2: Acid-Base Titrations (Acidimetry and Alkalimetry)

Q11: What type of reaction is the basis of an Acid-Base titration?
A: Neutralization Reaction.
Q12: What is the fundamental relationship used at the equivalence point in acid-base titration?
A: N₁V₁ = N₂V₂ (or Moles_A = Moles_B for 1:1 reaction).
Q13: What is the purpose of phenolphthalein indicator?
A: To indicate the endpoint in titrations involving a strong base (pH range 8.3-10.5).
Q14: What color change does phenolphthalein show in a basic solution?
A: Colorless to Pink.
Q15: What color change does phenolphthalein show in an acidic solution?
A: Colorless to Colorless.
Q16: What is the purpose of methyl orange indicator?
A: To indicate the endpoint in titrations involving a strong acid (pH range 3.1-4.4).
Q17: What color change does methyl orange show in an acidic solution?
A: Yellow to Red.
Q18: Which indicator is suitable for the titration of a Strong Acid vs. Strong Base (HCl vs NaOH)?
A: Phenolphthalein or Methyl Orange (endpoint pH ≈ 7, both work).
Q19: Which indicator is suitable for the titration of a Weak Acid vs. Strong Base (Oxalic vs NaOH)?
A: Phenolphthalein (endpoint pH > 7, basic range).
Q20: Which indicator is suitable for the titration of a Strong Acid vs. Weak Base (HCl vs Na₂CO₃)?
A: Methyl Orange (endpoint pH < 7, acidic range).
Q21: Why is NaOH considered a secondary standard?
A: It is hygroscopic (absorbs moisture) and absorbs CO₂ from the air, making its concentration unstable.
Q22: Why is Oxalic Acid considered a primary standard?
A: It is stable, non-hygroscopic, and available in high purity.
Q23: Why is the conical flask rinsed only with water before titration?
A: To avoid changing the moles of the substance being analyzed (water doesn't add moles of solute).
Q24: Why is the burette rinsed with the titrant solution before titration?
A: To ensure the concentration of the liquid dispensed is not diluted by residual water.
Q25: What is the common source of error when reading the burette?
A: Parallax error (not reading at eye level).
Q26: What part of the liquid meniscus should be read in a burette for colorless solutions?
A: The lower meniscus.

Part 3: Redox Titrations (Permanganometry)

Q27: What type of reaction is the basis of a Redox titration?
A: Oxidation-Reduction (electron transfer).
Q28: What is the primary oxidizing agent used in Permanganometry experiments?
A: Potassium Permanganate (KMnO₄).
Q29: Why is KMnO₄ considered a secondary standard?
A: It is difficult to obtain in high purity and may contain MnO₂ impurity.
Q30: Against which primary standard is KMnO₄ usually standardized?
A: Oxalic Acid or Mohr's Salt (Ferrous Ammonium Sulphate).
Q31: Why must KMnO₄ titrations be carried out in an acidic medium?
A: KMnO₄ is the strongest oxidizing agent in acidic solution (MnO₄^- → Mn²⁺).
Q32: What is the reducing agent when KMnO₄ reacts with Oxalic Acid?
A: Oxalate ion (C₂O₄^2-).
Q33: What are the oxidation products of Oxalic Acid in this titration?
A: Carbon Dioxide (CO₂) and Water (C₂O₄^2- → 2CO₂).
Q34: What is the oxidation number change for Mn in acidic medium (KMnO₄)?
A: From +7 to +2 (n-factor = 5).
Q35: What is the oxidation number change for Carbon in Oxalic Acid?
A: From +3 to +4 (n-factor per C₂O₄^2- unit = 2).
Q36: What is the self-indicator in the KMnO₄ vs. Oxalic Acid titration?
A: KMnO₄ itself.
Q37: What is the color change at the endpoint of the KMnO₄ titration?
A: Colorless to Permanent Pale Pink.
Q38: Why is the solution heated to 60-70°C during the KMnO₄ vs Oxalic Acid titration?
A: To increase the reaction rate, as the reaction is slow at room temperature.
Q39: What is the catalyst for the KMnO₄ vs Oxalic Acid titration?
A: Mn²⁺ ions (Autocatalysis).
Q40: What is the reducing agent when KMnO₄ reacts with Mohr's Salt?
A: Ferrous ion (Fe²⁺).
Q41: What is the oxidation product of the Fe²⁺ ion?
A: Ferric ion (Fe³⁺) (Fe²⁺ → Fe³⁺).
Q42: What is the n-factor for Fe²⁺ in this reaction?
A: 1 (Change from +2 to +3).
Q43: Why is FeSO₄ (Ferrous Sulphate) preferred as a secondary standard over FeSO₄ alone?
A: Mohr's Salt (FAS) is more stable and does not readily oxidize in air.
Q44: What is the disadvantage of using KMnO₄ in a basic medium?
A: It reduces to brown/black MnO₂, making the endpoint difficult to detect.

Part 4: Calculation and Equivalence Concepts

Q45: Define Normality (N).
A: Number of gram equivalents of solute per litre of solution.
Q46: Define Equivalent Mass of an acid.
A: Molar Mass / Basicity (number of replaceable H⁺ ions).
Q47: Define Equivalent Mass of a base.
A: Molar Mass / Acidity (number of replaceable OH^- ions).
Q48: Define Equivalent Mass of an oxidizing/reducing agent.
A: Molar Mass / Change in Oxidation Number (n-factor).
Q49: What is the relationship between Normality and Molarity?
A: N = M × n-factor.
Q50: What is the equivalent mass of HCl?
A: 36.5 g/eq (Molar Mass / 1).
Q51: What is the equivalent mass of H₂SO₄ in a neutralization reaction?
A: 49.0 g/eq (Molar Mass / 2).
Q52: What is the equivalent mass of KMnO₄ in an acidic medium?
A: ≈ 31.6 g/eq (Molar Mass / 5).
Q53: What is the equivalent mass of Mohr’s Salt in a redox titration?
A: Molar Mass / 1 (Only Fe²⁺ is oxidized).
Q54: How many grams of NaOH (mol mass 40) are in 1 L of 1 N solution?
A: 40 g (Eq Mass = 40).
Q55: How many grams of H₂SO₄ (mol mass 98) are in 1 L of 1 N solution?
A: 49 g (Eq Mass = 49).

Part 5: Indicators and Endpoint Detection

Q56: What is an Acid-Base Indicator?
A: A weak organic acid or base whose undissociated form has a different color than its ionic form.
Q57: What is the mechanism by which an indicator changes color?
A: The shift in the indicator's dissociation equilibrium due to the change in pH.
Q58: What pH range should an indicator's color change cover for accurate results?
A: The color change should occur within the steep portion of the titration curve.
Q59: Why is the endpoint of CH₃COOH vs NaOH not sharp?
A: The pH change around the equivalence point is less steep (buffering action of the salt).
Q60: Can a common indicator (like phenolphthalein) be used in a redox titration?
A: No, redox indicators are required, or the titrant must be self-indicating (KMnO₄).
Q61: Give an example of an external indicator used in a redox titration.
A: Starch solution (used for I₂ titrations).
Q62: What color complex does Starch form with I₂?
A: Deep blue.
Q63: What is a Concordant Reading?
A: Titre values that agree closely (usually within ±0.1 mL of each other).
Q64: What is the major error when the color change is observed only after a delay of 30 s?
A: Overshooting the endpoint (endpoint should be recorded at the first permanent color change).

Part 6: Practical Precautions and Calculations

Q65: What is the correct way to add the titrant near the endpoint?
A: Drop by drop, with continuous swirling.
Q66: What is the purpose of placing a white tile/paper under the conical flask?
A: To clearly visualize the color change at the endpoint.
Q67: What substance should be avoided near a KMnO₄ titration?
A: Rubber (as KMnO₄ oxidizes rubber), hence glass stopcocks are preferred.
Q68: Why must the burette tip be free of air bubbles before starting?
A: Air bubbles introduce volume errors in the titre reading.
Q69: What is the final step in standardizing a secondary standard solution?
A: Calculate the exact molarity/normality using the titration reading and the known concentration of the primary standard.
Q70: How is the percentage purity of a sample determined using titration data?
A: % Purity = (Mass of pure substance / Mass of impure sample) × 100.
Q71: If V₁ L of M₁ M acid reacts with V₂ L of M₂ M base (1:1 mole ratio), what is the relationship at the endpoint?
A: M₁V₁ = M₂V₂.

Part 7: More on Permanganometry and Oxalic Acid

Q72: What is the full name of Mohr's Salt?
A: Ferrous Ammonium Sulphate (FeSO₄·(NH₄)₂SO₄·6H₂O).
Q73: What is the chemical formula of Oxalic Acid dihydrate?
A: H₂C₂O₄·2H₂O.
Q74: What happens if H₂SO₄ is not added to the conical flask in the KMnO₄ titration?
A: The reaction will take place in a neutral/basic medium, reducing KMnO₄ to MnO₂ (brown ppt), obscuring the endpoint.
Q75: Why is the initial addition of KMnO₄ to hot oxalic acid slow?
A: The initial reaction is slow; Mn²⁺ (autocatalyst) has not yet formed in sufficient concentration.
Q76: Why must the KMnO₄ solution be stored in dark bottles?
A: Light decomposes KMnO₄ into MnO₂, affecting its concentration.
Q77: What precaution is necessary regarding HCl when standardizing KMnO₄?
A: HCl cannot be used for acidification because KMnO₄ will oxidize Cl^- to Cl₂.
Q78: What is the standard concentration used for KMnO₄ solutions in the lab manual?
A: Usually M/20 or N/10.

Part 8: Additional Titration Concepts

Q79: What is a Double Titration?
A: Titrating a mixture of two components (e.g., NaOH and Na₂CO₃) using two different indicators (phenolphthalein and methyl orange).
Q80: In the double titration of NaOH and Na₂CO₃ with HCl, what happens at the phenolphthalein endpoint?
A: NaOH is completely neutralized, and Na₂CO₃ is converted to NaHCO₃.
Q81: What happens between the phenolphthalein endpoint and the methyl orange endpoint?
A: NaHCO₃ is converted to H₂CO₃.
Q82: What is the purpose of adding 1 g of KI in the titration of Cu²⁺?
A: I^- is oxidized by Cu²⁺ to I₂ (Iodometry) which is then titrated.
Q83: Why should the titration of iodine (I₂) be done as soon as it is prepared?
A: I₂ is volatile and its concentration will change.
Q84: Which indicator is used in Iodometric titrations?
A: Starch solution.
Q85: Why is starch indicator added only when the I₂ color is faint yellow?
A: Starch forms a stable complex with high I₂ concentration, making the endpoint inaccurate.
Q86: What is the relationship between the Equivalent Mass of Na₂S₂O₃ and its Molar Mass?
A: 1:1 (Na₂S₂O₃ → Na₂S₄O₆ in the reaction).

Part 9: Error Analysis and Standardisation

Q87: What is the difference between a systematic error and a random error?
A: Systematic errors are constant (e.g., faulty balance); Random errors vary (e.g., reading meniscus).
Q88: What is the main source of random error in titration?
A: Endpoint detection and meniscus reading.
Q89: What is the standard pH for a neutral solution at 25°C?
A: 7.0.
Q90: Why is standardizing secondary solutions crucial?
A: To determine their exact concentration due to their unstable or impure nature.
Q91: What is the primary use of a pipette in titration?
A: To accurately measure a fixed volume of the solution into the conical flask.
Q92: What is the primary use of a volumetric flask?
A: To accurately prepare a solution of a specific volume and known concentration.
Q93: Why should the pipette never be blown out to dispense the final drop?
A: The pipette is calibrated to deliver the volume only when the tip is allowed to drain freely.
Q94: What is the term for the final, permanent color observed?
A: Permanent Endpoint.

Part 10: Final Revision Points

Q95: What is the n-factor of H₃PO₄ in a titration with NaOH?
A: 3 (Tribasic acid).
Q96: What is the n-factor of H₃BO₃ (Boric Acid)?
A: 1 (Monobasic acid, acts as a Lewis acid).
Q97: What is the equivalent mass of Na₂CO₃ in a titration against HCl?
A: Molar Mass / 2 (releases 2 Na⁺ or requires 2 H⁺).
Q98: Why is an acidic medium preferred for KMnO₄ titrations?
A: The reduction is cleanest and fastest, and the endpoint is sharp (colorless to pink).
Q99: What is the chemical that causes the pale pink color at the endpoint of a KMnO₄ titration?
A: Excess MnO₄^- ions (the reducing agent is depleted).
Q100: Why must the conical flask be swirled continuously during titration?
A: To ensure complete and homogeneous mixing of the reactants.
Q101: If the normality is 2 N and the volume is 20 mL, how many equivalents are present?
A: 0.04 equivalents (2 N × 0.02 L).
Q102: What must be done if the initial burette reading is below the 0.0 mL mark?
A: Record the initial reading as is, but ensure the volume is sufficient for the entire titration.
Q103: If the pH changes from 4 to 10 near the equivalence point, what acid/base types are involved?
A: Strong Acid vs. Strong Base (large pH jump).
Q104: What is the term for the method where the analyte is added to the burette?
A: Direct Titration.
Q105: What is the common catalyst added in the standardization of KMnO₄ against Mohr’s Salt?
A: Dilute H₂SO₄ (provides the acidic medium).

Practice Quiz (30 MCQs)

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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

Add Excess Reagent (A): Add a known volume and concentration of Standard Reagent A to the Analyte.
Analyte + A (Excess) $\rightarrow$ Product + A (Remaining)
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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