I am preparing reagents for the sequential extraction procedure of heavy metal fractionation developed by Tessier et al. 1979. How I can prepare 0.04M NH2OH.HCl in 25% HOAc?
Bhushan Sonawane
2.7796 gm of Hydroxylammonium chloride in 25 % acetic acid( 25 ml of acetic in 100ml flask and make up to water i.e 25% acetic acid v/v
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Muhammad Amjad
Dissolve 2.739 g of Hydroxylammonium chloride in 25 % acetic acid and make the volume to one litre with same 25% acetic acid.
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Bhairo Prasad Ahirvar
Bhushan Sonawane Thank you for your answer. How have you calculated for 2.7796 g of salt? Please describe.
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Bhushan Sonawane
Bhairo Prasad Ahirvar The molecular wt of Hydroxylammonium chloride is 69.49
required 0.04 M then 0.04 X 69.49 = 2.7796
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