H2O2 oxidizes iodide to iodine in the presence of acid and molybdate catalyst.
The iodine formed is titrated with thiosulfate solution, incorporating a starch indicator.
H2O2 + 2 KI + H2SO4 ----> I2 + K2SO4 + 2 H2O
I2 + 2 Na2S2O3 ----> Na2S4O6 + 2 NaI
Reagents
Potassium iodide solution (1% w/v). Dissolve 1.0 grams KI into 100 mLs demineralized water. Store capped in cool place away from light. Yellow-orange
tinted KI solution indicates some air oxidation to iodine, which can be removed by adding a 1-2 drops of dilute sodium thiosulfate solution.
Ammonium molybdate solution. Dissolve 9 grams ammonium molybdate in 10 mLs 6N NH4OH. Add 24 grams NH4NO3 and dilute to 100 mLs.
Sulfuric acid solution. Carefully add one part H2SO4-98% to four parts demineralized water.
Starch indicator.
Sodium thiosulfate solution (0.1N).
Weigh to the nearest 0.1 mg an amount of H2O2 equivalent to a titer of 30 mLs (0.06 g of H2O2) using a 5 mL beaker and medicine dropper. Transfer
sample to Erlenmeyer flask.
Add to Erlenmeyer flask 50 mL of demineralized water, 10 mL of sulfuric acid solution, 10-15 mLs of potassium iodide solution, and two drops ammonium
molybdate solution.
Titrate with 0.1 N sodium thiosulfate to faint yellow or straw color. Swirl or stir gently during titration to minimize iodine loss.
Add about 2 mL starch indicator, and continue titration until the blue color just disappears.
Repeat steps 2-4 on a blank sample of water (omitting the H2O2).
Calculation
Weight % H2O2 = (A - B) x (Normality of Na2S2O3) x 1.7
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Sample weight in grams
where: A = mLs Na2S2O3 for sample; B = mLs Na2S2O3 for blank
Sorry, can't help with part 2 of your question - statistics is not my strong point. |