Available Chlorine in Bleaching Powder: Formula & Examples
Bleaching powder is an essential topic in school chemistry and competitive tests like JEE Main, NEET, and CUET. In chapters on redox titrations and halogen chemistry, numerical problems frequently ask for the percentage of available chlorine in commercial bleaching powder. Understanding the underlying reactions during iodometric titration and setting up exact stoichiometric ratios ensures high accuracy in examinations.
Exam details
- Source: Practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, CBSE Class 12, NEET, JEE Main, CUET
- Topic: Available Chlorine in Bleaching Powder
- Difficulty: Moderate
What Is Available Chlorine?
Commercial bleaching powder is prepared by passing chlorine gas over dry slaked lime, Ca(OH)2. Textbooks represent bleaching powder with the formula CaOCl2 or Ca(OCl)Cl (calcium hypochlorite chloride). Real samples contain a mixture of calcium hypochlorite [Ca(OCl)2], basic calcium chloride, and residual Ca(OH)2.
When bleaching powder reacts with dilute acids, chlorine gas is evolved:
CaOCl2 + 2 CH3COOH → (CH3COO)2Ca + H2O + Cl2
CaOCl2 + H2SO4 → CaSO4 + H2O + Cl2
Available chlorine is the percentage by mass of chlorine gas liberated when bleaching powder reacts with dilute acid or excess potassium iodide. Although the solid contains chlorine in both +1 and −1 oxidation states, its total oxidising power is reported as equivalent elemental chlorine, Cl2.
Using IUPAC atomic masses (Ca = 40.08 g/mol, O = 16.00 g/mol, Cl = 35.45 g/mol; or textbook values Ca = 40, O = 16, Cl = 35.5), pure CaOCl2 has molar mass 126.98 g/mol (or 127.0 g/mol). One mole of CaOCl2 yields one mole of Cl2 (70.90 g, or 71.0 g), giving a theoretical maximum of 55.84% (or 55.91%) available chlorine. Commercial samples typically show 30% to 38% available chlorine due to inert constituents.
Iodometric Titration Method
In the laboratory, available chlorine is determined by iodometric titration. A suspension of bleaching powder in water is acidified with dilute acetic acid and treated with excess potassium iodide (KI):
CaOCl2 + 2 CH3COOH + 2 KI → (CH3COO)2Ca + 2 CH3COOK + H2O + I2
The liberated iodine is titrated against standard sodium thiosulphate (Na2S2O3) using starch indicator:
I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6
Starch indicator is added once the solution turns pale yellow. The dark blue colour disappears sharply at the end point.
Key Formula and Quantitative Relations
Stoichiometrically, 1 mole of Cl2 liberates 1 mole of I2, which reacts with 2 moles of Na2S2O3:
1 mol Cl2 ≡ 1 mol I2 ≡ 2 mol Na2S2O3 ≡ 2 equivalents of Na2S2O3
Milliequivalents:
meq of Cl = Normality (N) × Volume of Na2S2O3 (mL)
Mass of available Cl (g) = (N × V × 35.45) / 1000
Percentage of Available Chlorine:
% Available Chlorine = [ Mass of available Cl (g) / Mass of sample (g) ] × 100
Aliquot formula:
% Available Chlorine = [ (N × V × 0.03545 × V_total) / (V_aliquot × W_sample) ] × 100
Variables: N = normality of Na2S2O3; V = volume of Na2S2O3 (mL); V_total = total suspension volume (mL); V_aliquot = titrated volume (mL); W_sample = sample mass (g); 0.03545 g/meq = mass of one meq of Cl (or 0.0355 g/meq with Cl = 35.5 g/mol).
Worked Examples
Question: A 2.50 g sample of bleaching powder is suspended in water and made up to 250.0 mL. A 25.0 mL aliquot is acidified with dilute acetic acid and treated with excess potassium iodide. The liberated iodine requires 24.00 mL of 0.100 M sodium thiosulphate solution to reach the starch end point. Taking atomic mass of Cl as 35.5 g/mol, calculate the percentage of available chlorine in the sample.
Step-by-step solution:
1. Moles of Na2S2O3 = 0.100 mol/L × (24.00 / 1000) L = 0.002400 mol
2. Moles of Cl2 in 25.0 mL aliquot = 0.002400 mol / 2 = 0.001200 mol
3. Mass of Cl2 in aliquot = 0.001200 mol × 71.0 g/mol = 0.08520 g
4. Total mass of Cl2 in 250.0 mL = 0.08520 g × (250.0 / 25.0) = 0.8520 g
5. Percentage of available chlorine = (0.8520 g / 2.50 g) × 100 = 34.08%
Second route (Milliequivalents):
meq of Na2S2O3 = 0.100 N × 24.00 mL = 2.400 meq; Total meq in 250.0 mL = 2.400 × 10 = 24.00 meq.
Mass of Cl = 24.00 meq × 0.0355 g/meq = 0.8520 g; % Available Cl = (0.8520 / 2.50) × 100 = 34.08%.
Answer: 34.08%
Question: A 5.00 g sample of bleaching powder reacts with excess dilute sulphuric acid. The evolved chlorine gas is dried and collected at STP. The volume of gas collected is 784 mL. Calculate the percentage of available chlorine using traditional textbook STP (molar volume = 22.4 L/mol at 0 °C, 1 atm) and compare with modern IUPAC STP (molar volume = 22.7 L/mol at 0 °C, 1 bar). Take atomic mass of Cl = 35.5 g/mol.
Step-by-step solution:
1. Traditional STP (22.4 L/mol = 22400 mL/mol):
Moles of Cl2 = 784 mL / 22400 mL/mol = 0.0350 mol
Mass of Cl2 = 0.0350 mol × 71.0 g/mol = 2.485 g
% Available Chlorine = (2.485 g / 5.00 g) × 100 = 49.70%
Check by proportion: Mass = (71.0 g × 784 mL) / 22400 mL = 2.485 g; % = (2.485 / 5.00) × 100 = 49.70%.
2. Modern IUPAC STP (22.7 L/mol = 22700 mL/mol):
Moles of Cl2 = 784 mL / 22700 mL/mol = 0.03454 mol
Mass of Cl2 = 0.03454 mol × 71.0 g/mol = 2.452 g
% Available Chlorine = (2.452 g / 5.00 g) × 100 = 49.04%
Check by proportion: Mass = (71.0 g × 784 mL) / 22700 mL = 2.452 g; % = (2.452 / 5.00) × 100 = 49.04%.
Answer: 49.70% (traditional STP with 22.4 L/mol) and 49.04% (IUPAC STP with 22.7 L/mol)
Common Mistakes to Avoid
1. Confusing equivalent weight with molecular weight: Equivalent weight of chlorine is 35.5 g/eq, while 1 mole of Cl2 (71.0 g) corresponds to 2 moles of thiosulphate. Mixing these causes a two-fold error.
2. Forgetting aliquot dilution factor: Titrating 25 mL from a 250 mL suspension represents one-tenth of the sample. Omitting the 10-fold multiplier gives an incorrect percentage.
3. Using hydrochloric acid in titration: Concentrated HCl causes aerial oxidation of iodide to iodine, giving an artificially high titre and an inflated percentage.
4. Adding starch indicator too early: Starch must be added only when the solution fades to pale yellow. Adding it early causes iodine to bind irreversibly, obscuring the end point.
Exam Relevance and Question Patterns
Available chlorine connects redox titrations, the mole concept, and halogen chemistry. The table below outlines how this topic appears in examinations.
| Exam | Question Format | Tested Skills |
|---|---|---|
| CBSE Class 11 and 12 | Descriptive problems and practical viva | Redox equations, equivalent weights, and indicator theory |
| NEET | Objective multiple-choice questions | Direct formula use, dilution ratios, and redox ratios |
| JEE Main | Numerical value questions | Aliquot scaling, milliequivalent balances, and STP gas conversions |
| CUET | Objective chemistry problems | Relating commercial compositions to oxidising capacity |