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Preparing buffer solns.

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Preparing buffer solns.

I have an assignment to prepare certain buffer solns from certain chemicals and I'm struggling with it, so some help would be greatly appreciated. I've looked in my old chem books to try to find guidance, but unsuccessfully. Anyway, here's an example of what I'm trying to do.
Given: 10x TAE- 0.4M tris, 0.2N acetic acied, 10mM EDTA
Prepare: 500mL of tris (MW=121.1g), glacial acetic acid (17.46N), 0.5M EDTA
Also, prepare 1L of 1xTAE.
So far, I have just converted the molarity of Tris to moles and then to grams by using the MW.
0.4mol/L x 0.5L= 0.2mol Tris x 121.1g/mol= 24.22g Tris
As for EDTA, I'm guessing to use the M1V1=M2V2 equation; therefore, 0.5MV1=.01M(0.5L) and V1= .01L
I do not understand the acetic acid and I'm not even sure what the N stands for; all that comes to mind is Newtons, but that's my physics brain talking and my chemistry brain is slow coming back online.
As for the 1x TAE, would it be correct just to take a tenth of each of the measurements?
Furthermore, when do you know to titrate it? Is that only when you're trying to reach a certain pH of a soln?
sunnidaze
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Re: Preparing buffer solns.

Hola in the case of acetic acid ,normality is the same that molarity so you can use 500x0.2=Vx17.46. Once you have the 10X TAE prepared , yes take a tent part and dilute with 9 parts of the final volume. This is a recipe of luckow and als manuals and I don´t remenber if you have to check the pH at any point, but recheks pHs the day after when tris is in the formula and use solid. Buena suerte
protold
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BTW it's c1V1 = c2V2 ! M is molecular weight, so it has nothing to do with that.

Also, for making 1× TAE you do not take one tenth of your solution, but you take some specified volume and add 9-times the same volume of water (better would be to fill up to some volume. E.g. to take 100 ml and fill up to 1l).
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JackBean
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Re: Preparing buffer solns.

Muchisimas gracias!
What about if you were trying to make a buffer and were given the percent weight/volume of one of the components?
sunnidaze
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what's the problem? If you have e.g. 20% w/v and need 2.5 l, thenjust simply 20%*2.5 = 50 g
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JackBean
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Re:

JackBean wrote:what's the problem? If you have e.g. 20% w/v and need 2.5 l, thenjust simply 20%*2.5 = 50 g

And you would be wrong 20% is 20g/100ml not by 1L. So in this case you would need 500g.
Patrick

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any proof. (Ashley Montague)

canalon
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