It's probably elementary but if someone can please explain to me what exactly happens in the following procedure:
Analysis: Transfer 1 g of Sodium Lauryl Sulfate, weighed, to a 250-mL beaker. Add 35 mL of water, and warm to dissolve. To the warm solution add 2.0 mL of 1 N nitric acid, mix, and add 50 mL of alcohol. Heat the solution to boiling, and slowly add 10 mL of Lead nitrate solution, with stirring. Cover the beaker, simmer for 5 min, and allow to settle. If the supernatant is hazy, allow to stand for 10 min, heat to boiling, and allow to settle. When the solution is almost to the boiling point, decant as much liquid as possible through 9-cm filter paper (Whatman No. 41 or equivalent). Wash four times by decantation, each time using 50 mL of 50% alcohol, and bring the mixture to a boil. Transfer the filter paper to the original beaker, and immediately add 30 mL of water, 20.0 mL of 0.05 M edetate disodium VS, and 1 mL of ammonia–ammonium chloride buffer TS. Warm to dissolve the precipitate, add 0.2 mL of eriochrome black TS, and titrate with 0.05 M zinc sulfate VS. Each mL of 0.05 M edetate disodium is equivalent to 7.102 mg of Na2SO4.
And what is the formula for calculating the sodium sulfate content?
Thank you.