Dear Reasearchers,
I am working on a project in which I need to prepare 38% silver diamine fluoride for in-vitro testing related to dentistry. Since my background is dentistry, and not chemistry, could anyone please guide me regarding the best method for preparing 38% SDF?
Thank you for your suggestions in advance.
This is only a suggestion, it might work.
Add dilute NaOH to a solution of AgNO3 to get Ag2O.
Filter the solid and add HF to it to completely dissolved it( in non -glass beaker
Then add NH3 solution for each mol of Ag + you need two mold of NH3. Hopefully you get Ag(NH3)2F.
PS: HF reacts with glass!!
Mohammad Kooti Thank you very much Prof. for the detailed answer. I shall work on it and let you know.
Do you think if I add AgF into ammonia solution, it will form SDF?
Dear Mehmood
AgF is the only silver halide that is soluble in water. It will produce
Ag + and F- in solution. Ag+ has high tendecy to to form complex with NH3. So, if you add NH3 to AgF you can have the required SDF complex but not very pure unless the addition of NH3 is controlled.
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