Hello everyone;
I want to prepare 0.02 M ammonium acetate buffer (pH:9); this needs 7.719 g of sodium acetate (anhydrous) to 800 mL distilled water, and 0.353 g of acetic acid. Now the problem is, i have glacial acetic acid (99% pure) in liquid form in 500 mL bottle. How much i should take from this glacial acetic acid to meet the desired requirement. Please guide
If your acetic acid is 99%, you have to remember that this is in bulk (grams). That is, in 100 g of solution, 99 g will be acetic acid and 1 g will be water. You will have to know the density of acetic acid and density of water. So you can do the conversion of grams to volume, in order to know how many grams of acetic acid in each mL of your solution. So you can know how many mL of acetic acid will give you the amount of grams you need. However, in this amount of mL, there will be some water that will have to be calculated, to discount the total volume of water to be added (800mL). But since your acetic acid is 99%, I think it will be somewhat despicable to be considered. But if it was a 50% solution, the amount of water would have to be considered.
But now I saw that Omar gave you the answer. But, it is good for you to know how to calculate, and to do your calculations alone, so you will always solve this in other similar situations with acids that are mixture in water.
First things first.
1. You wont prepare ammonium acetate buffer from sodium acetate. (you can have it after titration with NH4OH though)
2. To prepare 800 mL of 0.02M ammonium acetate you have to dissolve 1.312 g of ammonium acetate in water, not 7.719 g - that would give you 0.1M solution.
3. Why do you add acid if you are going to titrate it to the pH 9 Anyway?
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