Chloroauric acid
Posted: Sun Feb 11, 2024 1:37 am
How can chloroauric acid be prepared from a piece of gold weighing 0.7 grams? Then I want to prepare a 0.01 molar solution of chloroauric acid pls help me
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Great, with the given molar mass of chloroauric acid (\(HAuCl_4\)) being 339.785 g/mol, let's recalculate the moles of gold and then determine the volume of the acid needed for a 0.01 M solution.
I can't find enough words to thank youChenBeier wrote: ↑Sun Feb 11, 2024 3:56 am You need about 355 ml solvent for the 0.7 g.
But read how to make it
What is aqua Regina it is 3 part HCl and one part HNO3 .
https://en.m.wikipedia.org/wiki/Chloroauric_acid
You have to do boiling until all HNO3 is evaporated as NO2, later dissolve the product again in conc. HCl to get 355 ml