Redox Titration of Bleach Chemistry Tutorial

Key Concepts

• Commercially available solutions of bleach contain sodium hypochlorite, NaClO(aq).
• Sodium hypochlorite, NaClO, is a very pale yellow liquid.
• Hypochlorite, ClO^-(aq), is the bleaching agent in bleach.

  It acts by oxidizing the "stain molecules" to less-coloured forms of the molecules.
  In the process, chlorine is reduced from an oxidation state of +1 in ClO^- to -1 in Cl^-

• It is possible to determine the concentration of hypochlorite in bleach in the laboratory using a 2 step process:

  1. Hypochlorite is converted to chloride in the presence of excess iodide, producing brown-coloured iodine molecules.
  2. Redox titration of iodine molecules using aqueous sodium thiosulfate solution.

• Note that sodium thiosulfate, Na₂S₂O₃(aq), is not generally used as a primary standard, so it should be standardised using a suitable primary standard such as potassium iodate, KIO₃(aq).

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Theory

Note that the reaction goes to completion so the moles of brown I₂ produced in this reaction is equal to the moles of hypochlorite, ClO^-, present in the bleach solution.

We can determine the amount in moles of I₂ in the solution by titration with standardised sodium thiosulfate, Na₂S₂O₃(aq).

With each addition of thiosulfate, the concentration of I₂ decreases, and the brown colour will fade to a pale yellow colour then to colourless, making detection of the end point difficult.
So, a small amount of starch solution is added near the end point of the titration, when the solution is pale yellow².
Iodine forms a deep blue coloured complex with starch that is easy to see.
At the end point of the titration, the solution will change colour from deep blue to colourless.

Note that at the end point of the titration, moles I₂ = ½ × moles S₂O₃^2-

Calculations

1. Calculate the average titre of S₂O₃^2- in litres, V_(S2O3^2-)
2. Calculate moles of S₂O₃^2-, n_(S2O3^2-) = c_(S2O3^2-) x V_(S2O3^2-)
3. Calculate moles of I₂: n_(I2) = ½ x n_(S2O3^2-)
4. Calculate moles of ClO^- in bleach sample: n_(ClO^-) = n_(I2)
5. Calculate concentration of ClO^- in bleach sample: c_(ClO^-) = n_(ClO^-) ÷ V_(ClO^-)

Preparing the Standard Solution of Potassium Iodate (KIO₃)

Procedure
Place about 2 g of potassium iodate, KIO3(s), on a watchglass in an oven at 120oC for 2 hours. Transfer the dry potassium iodate to a dessicator to cool. Accurately weigh out about 1.75 g of the cool, dry potassium iodate. Transfer this potassium iodate to a clean 250.00 mL volumetric flask and make up to the mark with deionised water.

Calculate the concentration of the potassium iodate standard solution in mol L^-1.

1. Calculate the moles of potassium iodate in solution:

   moles of solute = mass of solute in grams ÷ molar mass of solute in g mol^-1

   n = m ÷ M

   where:
   n = moles of solute in mol
   m = mass of solute in g
   M = molar mass of solute in g mol^-1

   In our experiment:
   m = mass of potassium iodate (KIO₃) = 1.750 g
   M = molar mass of potassium iodate (KIO₃) = 39.10 + 126.9 + (3 x 16.00) = 214.0 g mol^-1
   n = moles of potassium iodate (KIO₃) = m ÷ M = 1.750 ÷ 214.0 = 8.178 × 10^-3 mol

2. Calculate the concentration of the aqueous potassium iodate solution in mol L^-1:

   concentration of solution = moles of solute ÷ volume of solution in litres
   c = n ÷ V

   where:
   c = concentration of solution in mol L^-1
   n = moles of solute in mol
   V = volume of solution in L

   In our experiment:
   n = moles of KIO₃ = 8.178 x 10^-3 mol
   V = 250.00 mL = 250.00 ÷ 1000 mL/L = 0.2500 L
   c(KIO_3(aq)) = n ÷ V = 8.178 x 10^-3 ÷ 0.2500 = 0.03271 mol L^-1

Standardising the Sodium Thiosulfate Solution (Na₂S₂O₃)

Standardise the sodium thiosulfate solution prepared above by titration with the potassium iodate primary standard.

Procedure
Fill a 50.00 mL burette (buret) with the sodium thiosulfate solution prepared above. Pipette a 25.00 mL aliquot of the standard potassium iodate solution into a 250 mL conical (erlenmeyer) flask. Add 10 mL of 10% sulfuric acid solution and 2 g of potassium iodide (KI) to the flask. The solution will appear yellowy-orange. Add sodium thiosulfate solution from the burette to the potassium iodate solution in the flask until the colour starts to fade and becomes straw-coloured. Add a little starch to the conical flask. The solution will turn deep blue. Continue adding sodium thiosulfate solution drop-wise to the conical flask while swirling the flask until the solution turns colourless. Repeat the titration until concordant titres are achieved.
	Safety: Iodide and thiosulfate are mildly toxic, avoid contact. Acids are corrosive, avoid contact.

Use the results of your titration to calculate the concentration of the sodium thiosulfate solution.

1. Calculate the average titre of sodium thiosulfate solution:

   Sample Results
   Trial No.	titre / mL	comments
   1	25.69	rough: disregard
   2	24.96
   3	24.98
   4	24.97

   Calculate the average titre using the three concordant titres:

   average titre = (24.96 + 24.98 + 24.97) ÷ 3 = 24.97 mL

2. Calculate moles of iodate ion, IO₃^-(aq), using known concentration of potassium iodate solution, KIO₃(aq), and the aliquot volume:

   moles IO₃^-(aq) = moles of KIO₃(aq) = concentration KIO₃(aq) in mol L^-1 x volume in L
   n(IO₃^-) = c(KIO₃) x V(KIO₃)
   where
   V(KIO₃) = aliquot volume in L = 25.00 mL = 25.00 ÷ 1000 = 0.02500 L
   c(KIO₃) = 0.03271 mol L^-1
   n(IO₃^-) = 0.03271 x 0.02500 = 8.178 x 10^-4 mol

3. Calculate moles of thiosulfate ions consumed in reaction:
   The balanced chemical equation for this redox reaction is:

   IO₃^-(aq) + 6H⁺(aq) + 6S₂O₃^2-(aq) → I^-(aq) + 3S₄O₆^2-(aq) + 3H₂O(l)

   1 mole IO₃^-(aq) reacts with 6 moles S₂O₃^2-(aq)
   therefore 8.178 x 10^-4 mol IO₃^-(aq) reacts with 6 x 8.178 x 10^-4 = 4.907 x 10^-3 mol S₂O₃^2-(aq)

4. Calculate concentration of Na₂S₂O₃(aq) using the average titre:

   c(Na₂S₂O_3(aq)) = n(Na₂S₂O_3(aq)) ÷ V(Na₂S₂O_3(aq))

   where:
   n(Na₂S₂O_3(aq)) = 4.907 x 10^-3 mol
   V = average titre = 24.97 mL = 24.97 ÷ 1000 = 0.02497 L
   c(Na₂S₂O_3(aq)) = 4.907 x 10^-3 ÷ 0.02497 = 0.1965 mol L^-1

Titrating the Bleach

Step 1: Dilute the Bleach Sample

Step 2: React Hypochlorite with Iodide to Produce Iodine

Procedure
Pipette a 25.00 mL aliquot of diluted bleach sample into a 250 mL conical (erlenmeyer) flask. Use a measuring cylinder to add 50 mL of acidified potassium iodide solution3 to the conical flask. The solution will turn brown.

Note that 1 mole of iodine molecules, I₂, is produced for every 1 mole of hypochlorite, ClO^-, present in the dilute bleach solution.

Step 3: Titrate the Iodine in Diluted Bleach Sample

Procedure

Fill a 50.00 mL burette (buret) with the standardised sodium thiosulfate solution, Na2S2O3(aq), prepared above. Add sodium thiosulfate solution from the burette to the dilute bleach solution in the conical flask prepared above until it changes colour from brown to a straw-yellow. Add a couple of drops of 1% starch solution to the conical flask. The solution will turn deep blue as the starch forms a complex with the remaining iodine. Add sodium thiosulfate from the burette drop-wise to the conical flask until the solution changes colour from deep blue to colourless. Repeat the titrations until concordant titres are achieved.

The iodine, I₂, produced in Step 2 has now been titrated with thiosulfate according to the balanced redox equation shown below:

Step 4: Calculating the Concentration of Hypochlorite Ions in Bleach

Sample results for titration of iodine using 0.1965 mol L^-1 thiosulfate solution.

1. Calculate the moles of thiosulfate used in the titration with iodine:
   n(S₂O₃^2-) = c(S₂O₃^2-) x V(S₂O₃^2-)

   c(S₂O₃^2-) = 0.1965 mol L^-1 (standardised thiosulfate solution)
   V(S₂O₃^2-) = average titre = (20.12 + 20.16 + 20.14) ÷ 3 = 20.14 mL = 20.14 ÷ 1000 = 0.02014 L

   n(S₂O₃^2-) = 0.1965 x 0.02014 = 3.958 x 10^-3 mol

2. Calculate moles of iodine, I₂, that reacted with the thiosulfate:

   2S2O32-(aq)

   +

   I2(aq)

   →

   2I-(aq)

   +

   S4O62-(aq)

   1 mole thiosulfate, S₂O₃^2-, reacts with ½ mole iodine, I₂
   Therefore 3.958 x 10^-3 mol of S₂O₃^2- reacted with ½ x 3.958 x 10^-3 = 1.979 x 10^-3 mol of I₂

3. Calculate the moles of hypochlorite, ClO^-(aq), present in the 25.00 mL aliquot of diluted bleach sample:

   ClO-(aq)

   +

   2I-(aq)

   +

   2H+(aq)

   →

   I2(aq)

   +

   Cl-(aq)

   +

   H2O(l)

   1 mole hypochlorite, ClO^-, produced 1 mole iodine, I₂.

   Since the solution contained 1.979 x 10^-3 mol of I₂, there must have been 1.979 x 10^-3 mol of ClO^- in the dilute bleach sample.

4. Calculate the concentration of hypochlorite in the dilute bleach sample:

   c(ClO^-_(diluted)) = n(ClO^-_(diluted)) ÷ V(ClO^-_(diluted))

   where:
   n(ClO^-_(diluted)) = 1.979 x 10^-3 mol
   V(ClO^-_(diluted)) = 25.00 mL = 25.00 ÷ 1000 = 0.02500 L
   c(ClO^-_(diluted)) = 1.979 x 10^-3 ÷ 0.02500 = 0.07916 mol L^-1

5. Calculate the concentration of hypochlorite in the original, undiluted bleach

   c_(dilute)V_(dilute) = c_(undilute)V_(undilute)
   c_(undilute) = c_(dilute)V_(dilute) ÷ V_(undilute)
   where:
   c_(dilute) = 0.07916 mol L^-1
   V_(dilute) = 250.00 mL = 250.00/1000 = 0.2500 L
   V_(undilute) = 25.00 mL = 25.00/1000 = 0.02500 L
   c_(undilute) = 0.07916 x 0.2500 ÷ 0.02500 = 0.7916 mol L^-1

The concentration of hyprochlorite in the bleach is 0.7916 mol L^-1

Exam Question Example (StoPGoPS method for problem solving)

1. What are you being asked to do?

   Determine the concentration of hypochlorite ions in mol L^-1 in bleach
   c(ClO-) = ? mol L^-1

2. What information (data) have you been given in the question?

   V(bleach) = volume of bleach = 20.00 mL = 20.00/1000 = 0.02000 L
   n(I^-) = moles iodide to react with hypochlorite in bleach = excess mol
   c(S₂O₃^2-) = concentration of thiosulfate solution = 1.950 mol L^-1
   V(S₂O₃^2-) volume of thiosulfate solution used in titration = 12.58 mL = 12.58/1000 = 0.01258 L

3. What is the relationship between what you know and what you need to find out?

   Reaction 1: hypochlorite reacts with excess iodide to produce iodine:

   hypochlorite		iodide				iodine		chloride
   ClO-(aq)	+	2I-(aq)	+	2H+(aq)	→	I2(aq)	+	Cl-(aq)	+	H2O(l)
   faintly yellow		colourless				brown		colourless		colourless

   n(ClO^-) = n(I₂)

   Reaction 2: titration of iodine produced in reaction 1 with standardised aqueous sodium thiosulfate solution:

   thiosulfate	+	iodine	→	iodide	+	dithionate
   2S2O32-(aq)	+	I2(aq)	→	2I-(aq)	+	S4O62-(aq)

   n moles thiosulfate, S₂O₃^2-, reacts with ½ x n moles iodine, I₂

   n(I₂) = ½ x n(S₂O₃^2-)

   and n(S₂O₃^2-) = c(S₂O₃^2-) x V(S₂O₃^2-)

   therefore n(I₂) = ½ x c(S₂O₃^2-) x V(S₂O₃^2-)

   Since n(ClO^-) = n(I₂) = ½ x c(S₂O₃^2-) x V(S₂O₃^2-)

   and c(ClO^-) = n(ClO^-) ÷ V(ClO^-)

   it follows that
   

   c(ClO-) =
   	½ x c(S2O32-) x V(S2O32-)
   	V(ClO-)

4. Calculate the concentration of ClO^- in the bleach:

   c(ClO-)	=	½ x c(S2O32-) x V(S2O32-)
   		V(ClO-)
   	=	½ x 1.950 x 0.01258
   		0.02000
   c(ClO-)	=	0.6133 mol L-1

5. Is your answer plausible?

   Check your calculations: use your calculated concentration of ClO^- to calculate how much S₂O₃^2- would be required in the titration.
   n(ClO^-) = c(ClO^-) x V(ClO^-) = 0.6133 x 20.00/1000 = 0.012266 mol
   and n(I₂) = n(ClO^-) = 0.012266 mol
   n(S₂O₃^2-) used in titration = 2 x n(I₂) = 2 x 0.012266 = 0.024532 mol
   volume of S₂O₃^2- titre = n(S₂O₃^2-) ÷ V(S₂O₃^2-) = 0.024532 ÷ 1.95 = 0.01258 L = 12.58 mL
   Since this volume is the same as that given in the question, we feel confident that the concentration of ClO^- in the bleach is 0.6133 mol L^-1.

   Have we answered the question that was asked?
   Yes, we have determined the concentration of hypochlorite ions in bleach in mol L^-1.

6. State the solution to the problem:

   The concentration of hypochlorite ions in the bleach solution is 0.6133 mol L^-1