1. http://www.lamotte.com/images/pdf s/instructions/7414.pdf

2. Step 1: Manganous (Manganese II) Sulfate (Manganous Sulfate Solution) reacts with Potassium Hydroxide (Alkaline Potassium Iodide Azide Solution [KOH & KI]) to form Manganous (Manganese II) Hydroxide and Potassium Sulfate. MnSO 4 + 2KOH → Mn(OH) 2 + K 2 SO 4 A precipitate is formed… Mn(OH) 2 (s)

3. Step 2: Manganous (Manganese II) Hydroxide reacts with Oxygen & Water to produce Manganic (Manganese III) Hydroxide, (a redox reaction). The ratio of oxygen to Manganese II Hydroxide is 1:4. 4Mn(OH) 2 (s) + O 2 (aq) + 2H 2 O (l) → 4Mn(OH) 3 (s) Product is brown precipitate.

4. Step 3: Manganic (Manganese III) Hydroxide reacts with Sulfuric Acid to produce Manganic (Manganese III) Sulfate and Water. 2Mn(OH) 3 (aq) + 3H 2 SO 4 (aq) → Mn 2 (SO 4 ) 3 (s) + 6H 2 O (l) Oxygen is considered “fixed” at this point.

5. The previous reaction is immediately followed by Manganic (Manganese III) Sulfate reacting with Potassium Iodide (from the step 1 addition of Alkaline Potassium Iodide Azide Solution) to form Manganous (Manganese II) Sulfate and Potassium Sulfate and Iodine. (another redox reaction) Mn 2 (SO 4 ) 3 (s) + 2KI (aq) → 2MnSO 4 (aq) + K 2 SO 4 (aq) + I 2 (aq) The iodine causes the solution to appear yellow-brown. The amount of iodine is proportional to the amount of oxygen in Step 2.

6. Step 4: Sodium Thiosulfate (S 2 O 3 -2 ) reacts with the Iodine to form Sodium Tetrathionate (S 4 O 6 -2 ) and Sodium Iodide. 2Na 2 S 2 O 3 (aq) + I 2 (aq) → Na 2 S 4 O 6 (aq) + 2NaI (aq) When all of the iodine has finished reacting, the solution changes color from yellow-brown to colorless. Starch indicator may be added to enhance this endpoint. It will form a blue-black complex with iodine (not iodide ion) when present.