I want to prepare 7.5 M Ammonium acetate solution. The recipe stated that I need to dissolved 57.81 g ammonium acetate in water to final volume of 100 ml, then sterilize by filtration (0.2 micro meter filter). The final pH will be 5.5. I don't understand and don't know how to sterilize by filtration? I hope someone can explain and show how to prepare this solution. Thank you
The math is correct, add 57.8 grams to 100 mL = 7.5 M
What is meant by sterilize by filtration is that most bacteria is larger than 0.2 microns, so after preparation if you filter your 7.5M solution through a 0.2uM filter that will "sterilize" your solution.
What are you using this for? Is this filtration step necessary? - It's important to know that ammonium acetate solutions are considered volatile buffers and have a short shelf life (less than a week - in some cases
Thank you for the answer. I'm doing dna extraction of moss. I'm trying to follow my supervisor method during her phd study. In her protocol, she add 80 microliter of 7.5 M ammonium acetate and 540 microliter of cold isopropanol to the lysate of aqueous phase in tube. So, does it means i could not use this solution (7.5 m ammonium acetate) when i also add isopropanol? Thank you so much for the reply.
sorry, actually i didn't understand the protocol better. in her protocol actually, she add 0.54 vol of cold isopropanol and 0.08 vol of 7.5 M ammonium acetate to the lysate in tube.
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